7 concepts21 worked examples32 exercises5 exam-level7 figures
What are you here for?
09Conservation of Energy
Hold a ball at shoulder height, 1.80 m above the floor, and let go: it lands doing 5.94 m/s. Now let the same ball roll from rest down a smooth chute whose top is at exactly 1.80 m and whose bottom is level with the floor, so that it leaves the chute moving sideways instead of straight down. It leaves at 5.94 m/s. Nothing you have been taught so far can integrate a force along a chute that bends, and yet the second answer agrees with the first to three digits.
By the end of this section you can produce that 5.94 m/s in one line for a track of any shape, name the single word in the question that makes the one-line answer legal, and still get an answer when the word is missing and the track is rough.
In 60 seconds
Gravity and an ideal spring hand back every joule they take, so instead of computing their work along the path you can keep a stored amount that depends only on where the body is; add that store to $\tfrac12 mv^{2}$ and the sum stays put unless something like friction is in the room, in which case the sum falls by exactly the friction force times the distance travelled.
Stored energy from the work a conservative force does
$$\Delta U = -W_{\rm cons}$$
the definition every stored energy in this section comes from: what the force gives up, the store gains
Gravitational store near the ground
$$U = mgy$$
heights small compared with the size of the Earth; $y$ is measured upwards from a level you choose and state
Elastic store in an ideal spring
$$U = \tfrac12 k x^{2}$$
$x$ is the stretch or the squash measured from the natural length; the sign of $x$ never matters
a rough surface; $d$ is the distance travelled along the surface, not the straight line between the ends
Force read off the potential curve
$$F_x = -\frac{dU}{dx}$$
you have $U(x)$ as a graph or a formula and want the direction and size of the force without a free body diagram
Gravitational store far from a planet
$$U = -\frac{GMm}{r}$$
distances comparable with the planet's radius, so $mgy$ is no longer usable; the zero sits at infinity, which is why $U$ is negative
$$v_{\rm esc} = \sqrt{\frac{2GM}{R}}$$
the launch speed that leaves nothing over at infinity; independent of the mass launched and of the direction, if there is no air
Power, average and instantaneous
$$\bar P = \frac{W}{t}, \qquad P = Fv$$
the question asks how fast the work is delivered rather than how much; the second form needs the force along the velocity
Three most common mistakes
Putting the length of the slope into $mgy$ instead of the vertical drop. A 4.00 m slide down a $30^{\circ}$ ramp is a drop of 2.00 m, and using 4.00 m doubles the released energy and multiplies the final speed by 1.41.
Treating friction as if it had a stored energy. It has none: it depends on the length of the path, so there is no number attached to the position of the body. Friction enters as $f_k d$ subtracted from the total, once, at the end.
Measuring the two heights from different levels. The zero of $U$ is free to choose but it must be chosen once; a start measured from the floor and an end measured from the table top is the most common wrong line in this section.
The assessment table gives 20% to each midterm, 25% to the final, 10% to quizzes in total, 5% to homework and 20% to the laboratory. It says nothing about which topic sits on which paper, so nothing is claimed here about that; what can be said is that this section reuses the forces, the ramps and the friction of the previous ones, so a mistake made there is made again here.
How much time do you have?
10 minutes
You leave with the one equation that does most of the work, $\tfrac12 mv_1^{2}+U_1 = \tfrac12 mv_2^{2}+U_2$, with the two stores that go into it, and with the sentence that decides whether you are allowed to use it at all.
In 60 seconds · Formula card · Potential energy: the work a conservative force has banked · Conservation of mechanical energy: the sum that refuses to move · Mistake ledger
45 minutes
You add the part that separates a pass from a good mark: the ledger with friction in it, where the energy that leaves the total is accounted for rather than waved away, and the fading ladder that makes you write the reasoning yourself.
In 60 seconds · Conventions used here · Conservative forces: when the work forgets the path · Potential energy: the work a conservative force has banked · Conservation of mechanical energy: the sum that refuses to move · When friction is in the room: the total still balances · Method boxes · Scaffolding comes off · Practice B (computation) · Check yourself
full read
Everything above plus the energy landscape, which answers where a body can and cannot go without solving anything, the store that runs out at infinity with the escape speed that follows from it, power, and the interleaved set that makes you choose the tool before you use it.
The opening pages · Recall first · Try it yourself first · Notation · All seven concept blocks · Method boxes · Contrast pairs · Scaffolding comes off · Full exam-style question · Practice A to D · Mistake ledger · Formula card · Glossary · Check yourself
By the end of this section
Decide whether a given force is conservative by testing whether its work depends on the path, and show that the work of a conservative force around any closed trip is zero while friction's is not.
Compute the gravitational and the of a body, state the zero level you chose, and show that a change in the store does not depend on that choice.
Apply conservation of mechanical energy to find a speed, a height or a spring compression on a smooth track of any shape, and say which word in the question licensed it.
Read a : locate the for a given total energy, mark the region the body cannot enter, and get the direction of the force from the slope.
Balance the energy ledger when friction or another acts, using the distance along the path, and find the produced.
Use the store that goes as $-GMm/r$ when the height is no longer small, and derive an escape speed from it.
Calculate an average power from work and time and an instantaneous power from force and speed, and convert between joules, and .
Syllabus coverage
Conservation of Energy
Conservative and , potential energy for gravity and for a spring, the conservation of mechanical energy and the problem solving that comes with it, energy landscapes and turning points, the ledger when friction dissipates energy, and the gravitational store far from a planet with the escape speed that follows
The week line carries no chapter numbers, so no chapter number is quoted anywhere in this section. The scope taken is the standard content of that title in the set textbook, split across six of the seven blocks.
covered
Power
The rate at which work is delivered, in watts, and the two forms $W/t$ and $Fv$
The week line names conservation of energy only, so power is flagged here rather than folded in silently. It is not an outsider: the course catalogue lists work, energy and power in one breath, the previous section sent it forward to this one, and it is the closing section of the same textbook chapter. One block, at the end, with its own practice.
off_syllabus
Energy of a swinging or bouncing system followed over time
How the store and the motion trade places again and again once a body is left to oscillate
Deferred to the oscillations section later in the course. Every spring problem here is a single trip between two stated instants, which needs nothing beyond this page; anything that asks how long the trip took, or how the motion repeats, belongs there.
deferred
Energy in collisions between two bodies
What happens to the mechanical total when two bodies hit each other
Deferred to the linear momentum section. Everything on this page follows one body between two instants; the moment a second body enters and the two exchange energy, the missing tool is momentum, not energy.
deferred
Where the energy that friction removes actually goes
Thermal energy as the microscopic destination of the work friction does
Named in three sentences, because a total that only ever falls is not a conservation law and the student is right to be suspicious. It is not developed: no temperature, no specific heat, no calculation asks for it beyond the joules that left the mechanical account.
off_syllabus
Recall first
Work done by a constant force
$W = Fd\cos\theta$, where $\theta$ is the angle between the force and the displacement. A force at $90^{\circ}$ to the motion does no work at all, which is why a normal force never appears in an energy equation.
Every stored energy on this page is defined as the work a particular force does, so the definition of work is the raw material.
Work done by a varying force
$W = \int_{x_1}^{x_2} F(x)\,dx$, the area under the graph of force against position. For a spring obeying $F = -kx$ this gives $\tfrac12 kx^{2}$ for a stretch $x$ from the natural length.
The elastic store and the $-GMm/r$ store are both areas under a curved force graph; without this integral neither can be built.
Kinetic energy and the work-energy principle
$KE = \tfrac12 mv^{2}$ and $W_{\rm net} = \Delta KE$. The work counted here is the net work: every force, added with its sign.
Conservation of mechanical energy is this principle with the work of the conservative forces moved to the other side and renamed.
Kinetic friction
$f_k = \mu_k N$, directed against the sliding, with $N$ found from the direction perpendicular to the motion. On a slope of angle $\theta$ with no other forces across it, $N = mg\cos\theta$.
The only nonconservative force in this section is friction, and its size still has to come from a force equation across the surface.
Newton's law of universal gravitation
$F = GmM/r^{2}$ between two point masses a distance $r$ apart, with $r$ measured from centre to centre.
The store that replaces $mgy$ far from the Earth is the area under this force curve, so the force law has to be on the page first.
Speed from a constant acceleration
$v_2^{2} = v_1^{2} + 2a(y_2-y_1)$ for motion with constant acceleration along a line.
Used only to check the energy answers by a second, independent route in the simplest cases, where the acceleration really is constant.
Try it yourself first (3 questions)
1§08 recall — work done by a force at an angle●○○○○
Three questions from the sections behind this one, to find out what needs rereading before you start. Getting one wrong costs nothing and only tells you which recall above to read again. First: a crate is pulled 4.00 m along a level floor by a rope at $60.0^{\circ}$ to the floor, with a tension of 50.0 N.
Given
rope tension $50.0\ \mathrm{N}$ at $60.0^{\circ}$ to the horizontal
displacement $4.00\ \mathrm{m}$ horizontally
$\cos 60.0^{\circ} = 0.500$
Find
(a) Find the work done by the rope on the crate.
Hint 1/4
The rope pulls partly forwards and partly upwards, and the crate moves only forwards. Decide which part of the pull is being paid for.
Hint 2/4
$W = Fd\cos\theta$, where $\theta$ is the angle between the force and the displacement.
Hint 3/4
Here $F = 50.0$ N, $d = 4.00$ m and $\theta = 60.0^{\circ}$ with a cosine of 0.500.
Hint 4/4
The work is $(50.0)(4.00)(0.500) = 100$ J, half of what the same rope would deliver pulled along the floor.
Show solution
Resolving first makes the vertical component visible, which is the part the mistake ignores.
the other component, $50.0\sin 60.0^{\circ} = 43.3$ N, is vertical and the crate does not move vertically
$$W = F_\parallel d = (25.0)(4.00) = 100\ \mathrm{J}$$
identical to $Fd\cos\theta$, which is the same calculation written in a different order
Answer $$\boxed{\;W = 100\ \mathrm{J}\;}$$
Check
Limit check: at $\theta = 0$ the same rope would do 200 J and at $\theta = 90^{\circ}$ it would do nothing, and 100 J sits between them, as a $60^{\circ}$ pull should.
2§08 recall — the work needed to speed a car up●●○○○
Second question. A 1200 kg car speeds up from 10.0 m/s to 20.0 m/s on a level road.
Given
$m = 1200\ \mathrm{kg}$
speed from $10.0$ to $20.0\ \mathrm{m/s}$
level road
Find
(a) Find the net work done on the car.
Hint 1/4
The quantity that changes is not the speed but something built from it. Write that quantity down at both ends before subtracting anything.
Hint 2/4
The work-energy principle: $W_{\rm net} = \Delta KE = \tfrac12 mv_2^{2} - \tfrac12 mv_1^{2}$.
Hint 3/4
The numbers are $m = 1200$ kg, $v_1 = 10.0$ m/s and $v_2 = 20.0$ m/s.
Hint 4/4
$W = \tfrac12(1200)(400-100) = 1.80\times10^{5}$ J, three times what the first 10.0 m/s cost.
Show solution
Computing both ends separately is slower but immune to the classic error of squaring the change in speed.
Check by ratio: $KE \propto v^{2}$, so going from 10 to 20 m/s multiplies the kinetic energy by 4 and adds three times the original, and three times $6.00\times10^{4}$ is $1.80\times10^{5}$ J.
3§06 recall — the normal force on a slope●●○○○
Third question, and this one catches people. A 12.0 kg block rests on a slope inclined at $30.0^{\circ}$ to the horizontal, with no forces on it other than gravity, the normal force and friction.
Given
$m = 12.0\ \mathrm{kg}$
slope at $30.0^{\circ}$
$\cos 30.0^{\circ} = 0.866$
$g = 9.80\ \mathrm{m/s^{2}}$
Find
(a) Find the normal force the slope exerts on the block.
Hint 1/4
Draw the two directions that matter: along the slope and across it. The normal force lives in one of them, and so does only part of the weight.
Hint 2/4
Across the slope there is no acceleration, so $N = mg\cos\theta$; the remaining part of the weight, $mg\sin\theta$, lies along the slope.
Hint 3/4
The numbers: $m = 12.0$ kg, $g = 9.80$ m/s$^{2}$ and $\cos 30.0^{\circ} = 0.866$.
Hint 4/4
$N = (12.0)(9.80)(0.866) = 102$ N, smaller than the 118 N weight because the slope only has to hold part of it.
Show solution
The across-slope equation is the only one that contains $N$, so it is the only one worth writing for this question.
Resolve across the slope
$$N - mg\cos\theta = 0$$
no acceleration across the surface, since the block stays on it
$$N = (12.0)(9.80)(0.866) = 102\ \mathrm{N}$$
13% less than the weight, and the missing part is what pulls the block along the slope
Answer $$\boxed{\;N = 102\ \mathrm{N}\;}$$
Check
Limit check: at $\theta = 0$ the formula gives $N = 118$ N, the full weight on a level floor, and at $\theta = 90^{\circ}$ it gives zero, which is right for a vertical wall the block simply falls past.
Notation
symbol
reads as
means
watch out
$U$
you
potential energy, the amount stored by the position of the body or the state of the spring, in joules
$U$ belongs to a configuration, not to a force in general: there is no $U$ for friction.
$\Delta U$
delta you
the change in the store, final value minus initial value
This is the quantity with physical meaning. $U$ on its own depends on the zero level you chose.
$E$
ee
the , $KE + U$, of the body and whatever it is attached to
In this section $E$ is mechanical only; the thermal energy that friction produces is written separately.
$y$
why
the height above the chosen zero level, positive upwards
$y$ is a vertical height, never the distance travelled along a slope.
$h$
aitch
a vertical drop or rise between two stated points, always a positive number
Used only in the shorthand $v=\sqrt{2gh}$; if you are unsure of a sign, go back to $U = mgy$ with the axis.
$x$
ex
for a spring, the stretch or the squash measured from the natural length
Measured from the natural length, not from the floor and not from the hand holding it.
$k$
kay
the spring constant, in newtons per metre
A property of the spring alone; it does not change when you push harder.
$f_k$
eff kay
the kinetic friction force, $\mu_k N$
The energy it removes is $f_k d$ with $d$ the distance along the path, which is why a longer route costs more.
$W_{\rm NC}$
double you en see
the work done by all the nonconservative forces together
Negative for friction. A pushing hand is also nonconservative, and its work is positive.
$\bar P$
pee bar
average power, the work divided by the time it took
Distinguish it from the instantaneous $P = Fv$, which is the value at one moment.
Conventions used here
Where the zero of stored energy sits in these problems
Every problem here states its own zero level, and it is stated in the first line of the solution, before any number: usually the lowest point the body reaches, sometimes the natural length of a spring. Once chosen it holds for both sides of the equation. A stored energy on its own is not a physical fact; a change in one is.
Two heights measured from two different levels is the most common wrong line in an energy solution, and it is invisible in the algebra.
Up is positive, and what that does to the signs
The $y$ axis points up, so a body above the chosen zero has $U = mgy$ positive and one below it has $U$ negative, and the sign takes care of itself if you keep the axis. Speeds enter these equations squared, so a body moving down and a body moving up at the same speed carry the same kinetic energy, and an energy equation can never tell you which way something is going.
This is the price of the method: it is blind to direction, which is exactly why it is so short.
The constants used throughout this section
$g = 9.80\ \mathrm{m/s^{2}}$ near the ground, $G = 6.674\times10^{-11}\ \mathrm{N\cdot m^{2}/kg^{2}}$, and for the Earth $GM_E = 3.98\times10^{14}\ \mathrm{N\cdot m^{2}/kg}$ with a radius of $6.38\times10^{6}\ \mathrm{m}$. The same values were used in the earlier sections and no block here quietly rounds $g$ to 10.
A number that changes between blocks makes two correct answers disagree in the third digit and destroys any check you try to run.
What the word smooth is doing in a question
Smooth, frictionless, ideal spring, light string, negligible air resistance: each of these is a licence to use the conserved total. If a question omits them and mentions a rough surface, a coefficient, or a drag force, the total is not conserved and the ledger with $f_k d$ in it is the tool. Read the adjectives before writing the equation.
The method is chosen by one word in the question, and that word is usually an adjective nobody reads twice.
Digits kept in an energy answer
Answers are given to three significant figures and intermediate values are carried at more, so a printed 39.2 J may be 39.204 J inside the next line. Where a final speed comes out as 4.71 m/s, the underlying number is 4.7144; nothing here is rounded to two digits and then squared.
Squaring a rounded number is how a check that should agree ends up disagreeing in the second digit.
What is idealised away in an energy problem here
Bodies are treated as points, so no energy goes into spinning them; strings and springs have no mass of their own; air resistance is absent unless the question puts it there. Where one of these would change the answer noticeably it is said in the example rather than left for you to find.
Spinning and air are real, and a student who is not told they were dropped cannot tell an idealisation from an error.
9.1Conservative forces: when the work forgets the path
Gravity and an ideal spring give back every joule they take, and that is what makes this section's shortcut legal.
Last section we computed each force's work separately. Two of those forces never needed the path at all.
Solvable with what we have
A 180 N push over 3.50 m straight up a ramp: one angle, one distance.
The energy in a spring squashed 12 cm: the area under a straight graph.
The speed at the foot of a straight ramp: constant net force.
Not solvable yet
The work gravity does on a ball rolling down a bending chute.
The speed of that ball where the chute flattens out at the floor.
How far up a valley wall a cart on a wiggly track gets.
The chute is 2.80 m long measured along the metal, so put the weight and that length into the constant-force formula. For a 0.500 kg ball, $W = (0.500)(9.80)(2.80) = 13.7$ J, and if all of it goes into motion, $v = \sqrt{2(13.7)/0.500} = 7.41\ \mathrm{m/s}$.
Why it fails
The ball actually arrives at 5.94 m/s, so the estimate is 25% high. In $Fd\cos\theta$ the angle is between the force and the displacement, and on a bend it changes at every point. Chop the chute into short straight pieces: only the vertical part of each earns anything, because the weight points straight down. Those parts add up to the 1.80 m of drop, not the 2.80 m of metal.
DefinitionDefinition 9.1: a conservative force
Conditions
The test is applied to one force at a time, not to the whole set acting on the body
Every route compared runs between the same two points, for the same body
The two forms of the test are equivalent, and the proof below is two lines
Gravity and the force of an ideal spring pass the test; kinetic friction and air drag fail it
$$\boxed{\;W_{\rm path\,1}(A\to B) = W_{\rm path\,2}(A\to B)\ \ \text{for every pair of routes}\iff W(\text{closed trip}) = 0\;}$$
A force is conservative when its work between two points is the same on every route, which is the same statement as doing no work on a trip that ends where it started. Gravity passes: only the drop matters. Friction fails: what matters to it is how far you dragged the thing.
Why the two forms of the test say the same thing
Take two routes from A to B and call the work along them $W_1$ and $W_2$. Build a closed trip: out along route 1, back along route 2 reversed. Reversing a route reverses every little displacement while the force at each point is unchanged, so every contribution flips sign and the return leg contributes $-W_2$. The whole loop therefore does $W_1 - W_2$. That is zero for every pair of routes exactly when $W_1 = W_2$ for every pair of routes, which is the claim.
Three routes for the same 2.0 kg book from the shelf at 1.60 m down to the floor: a straight line, a staircase and a sagging curve. The routes have three different lengths, and $\textcolor{#8250df}{\text{gravity}}$ does the same $+31.4\ \mathrm{J}$ along each of them, because the only thing it is paid for is the 1.60 m of drop.
Looks like this, but is not
A force that always points the same way is conservative; one that changes direction is not.
A spring reverses direction inside one problem and is conservative; friction on a floor points along one fixed line and is not. The test is whether the work cares about the route, and friction cares because it is paid by the metre: 36.8 J along the 5.00 m diagonal, 51.5 J along the 7.00 m of wall, same two corners.
route
length travelled (m)
work by gravity (J)
work by a 4.0 N friction (J)
straight
3.53
+31.4
-14.1
staircase
4.75
+31.4
-19.0
sagging curve
3.70
+31.4
-14.8
Read it by columns, not by rows. The gravity column is constant while the lengths differ by a third: that is what looks like in numbers. The friction column tracks the length exactly, because that force is charged per metre. Anything attachable to the position of the book must behave like the third column, and no bookkeeping can make the fourth do so.
Gravity on three routes from the shelf to the floor
A 2.0 kg book is taken from a shelf 1.60 m above the floor down to the floor by three routes: straight down the diagonal, down a staircase of four equal steps, and along a sagging curve. Find the work done by gravity along each.
Given
$m = 2.0\ \mathrm{kg}$
drop from $y_1 = 1.60$ m to $y_2 = 0$
route lengths 3.53 m, 4.75 m and 3.70 m
$g = 9.80\ \mathrm{m/s^{2}}$
Find
the work done by gravity along each of the three routes
Solution
Cutting the path into pieces first, before any arithmetic, is what makes one calculation serve three routes; substituting numbers first would have meant three separate sums and would have hidden the point.
Cut any route into short pieces and see what gravity is paid for
the weight points down and has no horizontal part, so a horizontal step earns nothing at all no matter how long it is
$$\sum \Delta y = y_2 - y_1 = -1.60\ \mathrm{m}$$
the vertical steps add up to the total drop whatever order they came in, which is where the shape of the route disappears
Put the numbers in, once, for all three routes
$$W = -mg(y_2-y_1) = -(2.0)(9.80)(-1.60)$$
same expression for every route, because nothing about the route survived the previous step
$$W = \boxed{+31.4\ \mathrm{J}}\ \text{on all three}$$
positive: the weight points down and the book goes down, so gravity does positive work on it; what turns negative here is the change in the store, $\Delta U = -W = -31.4$ J, and that is the number the minus sign belongs to
Answer $$\boxed{\;W_{\rm gravity} = +31.4\ \mathrm{J}\ \text{on every route}\;}$$
Check
Independent route: drop the book from rest through the same 1.60 m and use kinematics. It arrives at $v=\sqrt{2(9.80)(1.60)} = 5.60$ m/s, so its kinetic energy is $\tfrac12(2.0)(5.60)^{2} = 31.4$ J, gained from nothing but gravity. The two numbers agree, and the second one never mentioned a route.
One line of algebra covered three routes, and it would have covered three hundred.
Whenever the weight is the only force being asked about, the shape of the path is decoration. Look for the vertical drop and ignore everything else in the picture.
Friction there and back: the round trip that does not cancel
A 3.0 kg box is dragged 5.00 m across a level floor with $\mu_k = 0.25$, then dragged straight back to where it started. Find the work done by friction on the way out, on the way back and on the whole trip, and the work done by gravity on the same trip.
Given
$m = 3.0\ \mathrm{kg}$
$\mu_k = 0.25$ on a level floor
5.00 m out and 5.00 m back
$g = 9.80\ \mathrm{m/s^{2}}$
Find
the friction work on each leg and on the closed trip, and the gravity work
Solution
Getting the friction force first and only then walking the legs keeps the two sign decisions separate; doing both at once is where the second minus sign usually goes missing.
Size of the friction force
$$N = mg = (3.0)(9.80) = 29.4\ \mathrm{N}$$
the floor is level and nothing else pushes across it, so the normal force carries the whole weight
$$f_k = \mu_k N = (0.25)(29.4) = 7.35\ \mathrm{N}$$
kinetic friction, because the box is sliding rather than about to slide
Independent check by a different question: where did the 73.5 J go? The floor and the box are warmer, and nothing about the box's position changed, so no stored amount could have absorbed it. If friction had a store, returning the box to its starting point would have had to return the store to its starting value, which would demand zero net work.
Any force whose work on a closed trip is not zero cannot be given a potential energy, and that single sentence is the whole reason the next block exists.
Checkpoint
§09.1 — testing a force for path independence●●○○○
A crate is pushed from the bottom corner of a room to the opposite top corner of a ramp system, once by a short steep route and once by a long shallow one, arriving at the same final point both times.
Given
the two routes start and finish at the same two points
the long route is twice as long as the short one
the surfaces are rough and the crate slides on them
Find
(a) Which quantity is guaranteed to be the same for both routes?
Hint 1/4
You are not being asked to compute anything. You are being asked which of the listed quantities can be settled by the definition of a conservative force alone.
Hint 2/4
The definition: the work of a conservative force between two fixed points is the same on every route, and gravity is conservative while kinetic friction is not.
Hint 3/4
The two routes here share their endpoints, so they share the same vertical rise; they do not share their lengths, and the long route is twice the short one.
Hint 4/4
Only the work done by gravity is fixed by the endpoints; the friction work and the push both depend on the route taken.
Show solution
Sorting the forces before looking at the options is faster than testing the options one by one, and it is the habit the section is built on.
Sort the three forces by the test
$$W_{\rm gravity} = -mg(y_2-y_1)$$
depends on the endpoints only, so it is the same on both routes
$$W_{\rm friction} = -f_k d$$
$d$ is the distance travelled along the surface, so doubling the route doubles the cost
whatever the other two do, the push makes up the difference, so it inherits the route dependence of friction
Answer $$\boxed{\;\text{only } W_{\rm gravity}\ \text{is the same on both routes}\;}$$
Check
Limiting case: make the surfaces smooth. Then the friction term vanishes and the push work becomes route independent too, which is consistent with the claim that it was friction that carried the route dependence.
⚠ Calling a round trip free because the body came back
Gravity really does return what it took on a round trip, and the habit spreads to friction, where the two legs add instead of cancelling.
wrong$$W_{\rm friction}(\text{out and back}) = -36.8 + 36.8 = 0$$
9.2Potential energy: the work a conservative force has banked
A number attached to where the body is, holding the work the conservative force will hand back when it is allowed to.
A force whose work depends only on the endpoints can have that work tabulated once against position, and then never computed again.
DefinitionDefinition 9.2: potential energy
Conditions
Defined only for a conservative force; friction cannot have one
It belongs to a configuration, a height or a stretch, not to an instant in time and not to a force in isolation
The zero level is a free choice which must be stated and then kept for the whole problem
$U = mgy$ needs a height small enough that $g$ has not changed; near the ground that means anything under a few kilometres
$$\boxed{\;\Delta U = U_2 - U_1 = -W_{\rm cons}\;;\qquad U_{\rm grav} = mgy\;;\qquad U_{\rm el} = \tfrac12 k x^{2}\;}$$
Whatever work the conservative force gives up, the store gains, and whatever the store loses, the force has done. Lifting a body slowly means gravity does negative work on it and the gravitational store rises by the same number of joules; letting it fall spends the store again. The minus sign is the whole content of the definition: the store is an account that runs opposite to the force's own account.
Where mgy and one half k x squared come from
Raise a body from $y_1$ to $y_2$. Gravity does $W = -mg(y_2-y_1)$, computed in the previous block for any route, so $\Delta U = +mg(y_2-y_1)$, and writing $U = mgy$ with any additive constant reproduces it. Stretch a spring from its natural length to $x$. The spring force is $-kx$ and the work it does is the area under that line, $W = -\tfrac12 kx^{2}$, so $\Delta U = +\tfrac12 kx^{2}$ from a natural length where the store is taken as zero. Both formulas are the same construction: integrate the force, flip the sign, and choose where zero sits.
One 2.0 kg book lifted 0.30 m off a table, with the zero of height put at the floor, at the table top and at the ceiling in turn. The $\textcolor{#8250df}{\text{three stored values}}$ disagree, including in sign, and all three change by the same 5.88 J.
Looks like this, but is not
Friction takes 36.8 J out of the box, so the box has $-36.8$ J of friction potential energy.
Write down what that number would be a function of. A potential energy is a value you can look up once you know where the body is, which is why $mgy$ needs only the height. To find this one you would have to know how far the box has been dragged, so the same box at the same point on the same floor would carry $-36.8$ J on its first trip and $-73.5$ J after being dragged back and forth, and $-147$ J after four crossings. That is not a property of the position; it is a history. Energy still leaves the mechanical account, and this section handles it in the friction block, but never as a store attached to the box.
The same book, three zero levels, one honest answer
A 2.0 kg book lies on a table whose top is 0.75 m above the floor, in a room with a ceiling 2.40 m above the floor. Somebody lifts it 0.30 m. Find the before and after with the zero taken at the floor, at the table top and at the ceiling, and find the change in each case.
Given
$m = 2.0\ \mathrm{kg}$
table top at 0.75 m above the floor
ceiling at 2.40 m above the floor
the book is raised by 0.30 m
$g = 9.80\ \mathrm{m/s^{2}}$
Find
the six stored values and the three changes
Solution
Doing all three frames before taking any difference is deliberate: it shows that the disagreement is real and that it cancels, which a single frame could never demonstrate.
Zero at the floor
$$U_1 = (2.0)(9.80)(0.75) = 14.7\ \mathrm{J}$$
$y$ is measured up from the chosen zero, and the book starts on the table
$$U_2 = (2.0)(9.80)(1.05) = 20.6\ \mathrm{J}$$
0.75 m plus the 0.30 m lift, in the same frame
Zero at the table top
$$U_1 = (2.0)(9.80)(0) = 0$$
the book starts at the level we have just called zero, which is legal and often the least work
$$U_2 = (2.0)(9.80)(0.30) = 5.88\ \mathrm{J}$$
the height above the new zero is now the lift itself
still negative, but less so, because the book moved towards the zero
Compare the changes, which is what physics actually claims
$$\Delta U = 20.6-14.7 = 5.88\ \mathrm{J}$$
floor frame
$$\Delta U = 5.88-0 = 5.88\ \mathrm{J}$$
table frame, identical
$$\Delta U = -26.5-(-32.3) = 5.88\ \mathrm{J}$$
ceiling frame, identical again, which is the point of the example
Answer $$\boxed{\;\Delta U = mg\,\Delta y = (2.0)(9.80)(0.30) = 5.88\ \mathrm{J}\ \text{in every frame}\;}$$
Check
Independent check by force and distance: lifting the book slowly needs an upward force of $mg = 19.6$ N through 0.30 m, which is 5.88 J of work done by the hand, and with no change of speed all of it has to have gone into the store. No zero level was mentioned anywhere in that sentence.
Six values computed to produce one number that was available in one line.
State the zero once, in words, before the first equation. Nobody loses marks for choosing an unusual zero; people lose marks for using two.
Getting k from a hanging mass, then the store at two squashes
A spring hangs from a hook. A 2.5 kg mass hung on it stretches it 7.0 cm and rests there. The spring is then taken off, laid flat and squashed by 12 cm, and later by 18 cm. Find the spring constant, the energy stored at each squash, and how much more it takes to go from 12 cm to 18 cm.
squashes of 0.12 m and 0.18 m from the natural length
$g = 9.80\ \mathrm{m/s^{2}}$
Find
the spring constant, the two stored energies and the extra energy between them
Solution
The hanging mass is a force measurement and the squash is an energy one; keeping them in separate subgoals stops the weight of the hanging mass from wandering into the energy line, where it does not belong.
Read the stiffness off the hanging mass
$$kx = mg \;\Rightarrow\; k = \frac{(2.5)(9.80)}{0.070}$$
at rest the spring's pull balances the weight, which is a force statement and not an energy one
$$k = 350\ \mathrm{N/m}$$
newtons per metre: 350 N would be needed to hold a full metre of stretch, which is why nobody stretches a spring a full metre
Ratio check that avoids the arithmetic entirely: the store goes as $x^{2}$, so going from 12 to 18 cm multiplies it by $(18/12)^{2} = 2.25$, and $2.52 \times 2.25 = 5.67$ J, matching the direct calculation.
A spring question almost always begins with a hidden force step that delivers $k$. Find $k$ first, then stop thinking about forces.
Checkpoint
§09.2 — what the zero level can and cannot change●●○○○
A student solving a ramp problem chooses the top of the ramp as the zero of gravitational potential energy, so every height below it comes out negative. A classmate says the answer will be wrong because energy cannot be negative.
Given
the same problem, two candidate zero levels
one choice makes $U$ negative over the whole ramp
Find
(a) Decide whether the classmate is right, and give the reason in one sentence.
Hint 1/4
Ask what the final answer of a ramp problem actually is: a speed, or a height, or a distance. Then ask which of the quantities in the equation that answer depends on.
Hint 2/4
The rule is $\Delta U = -W_{\rm cons}$: only differences in $U$ carry physical content, and adding the same constant to every $U$ in a problem leaves every difference alone.
Hint 3/4
Here the two choices differ by a constant $mgh_{\rm top}$ added to every value; the start and the end both shift by that same amount.
Hint 4/4
The classmate is wrong: negative values of $U$ are ordinary, and the speed the problem asks for comes out identical.
Show solution
Shifting the zero symbolically settles every case at once; testing one numerical example would only have settled that example.
the statement that will be used, whatever the zero
$$y_i \to y_i + c \;\Rightarrow\; mgc \text{ on both sides}$$
shifting the zero adds the same constant to both sides, so it cancels before any answer appears
Say what could not be shifted away
$$\tfrac12 mv^{2} \ge 0$$
kinetic energy has no free constant: its zero is the state of rest, and that is physics, not a choice
Answer $$\boxed{\;\text{a negative } U \text{ is legal; the final speed is unchanged}\;}$$
Check
Concrete check: on a 2.0 m ramp with a 1.0 kg body released from rest at the top, the floor-zero frame gives $19.6 = \tfrac12 v^{2}$ and the top-zero frame gives $0 = \tfrac12 v^{2} - 19.6$. Both give $v = 6.26$ m/s.
⚠ Using the distance along the slope in mgy
The slope length is the number printed in the question and the vertical drop usually is not, so the hand reaches for the one that is written down.
wrong$$\Delta U = mg L = (2.0)(9.80)(4.00) = 78.4\ \mathrm{J}$$
right$$\Delta U = mg L\sin\theta = (2.0)(9.80)(4.00)(0.500) = 39.2\ \mathrm{J}$$
⚠ Measuring the spring deformation from the wrong place
A diagram gives the position of the block on a table, and that number is used as $x$, although $x$ has to be counted from the spring's natural length.
wrong$$U = \tfrac12 k\,(\text{position of the block})^{2}$$
The same spring at three squashes. Doubling the squash from 6 cm to 12 cm multiplies the store by four, and the last 6 cm of the 18 cm squash costs 3.15 J on its own, more than the whole first 12 cm.
9.3Conservation of mechanical energy: the sum that refuses to move
Add the store to the motion and, on a smooth track, the same number comes out at every point of the journey.
We have a store that depends on position and a kinetic energy that depends on speed; the work-energy principle turns out to say that their sum does not change.
TheoremTheorem 9.3: conservation of mechanical energy
Conditions
Only conservative forces do work on the body: smooth surfaces, ideal springs, free flight, light strings
Forces perpendicular to the motion are allowed and change nothing, because they do no work: a normal force, the tension in the string of a pendulum
The same zero level for $U$ on both sides
It is one body between two chosen instants; states 1 and 2 must both be named before writing the line
$$\boxed{\;E = KE + U = \text{constant}\;;\qquad \tfrac12 mv_1^{2} + U_1 = \tfrac12 mv_2^{2} + U_2\;}$$
Take the kinetic energy the body has at the start and the store its position gives it at the start, and add them. Do the same at the end. The two totals are equal. Nothing in between matters, not the shape of the track, not the time taken, not the direction the body was moving; what came out of the store went into the motion, joule for joule, and if the body climbs again the motion pays it back.
Two lines from the work-energy principle
The work-energy principle says $W_{\rm net} = \Delta KE$. If every force that does work is conservative, then $W_{\rm net} = W_{\rm cons} = -\Delta U$ by the definition of the store. So $-\Delta U = \Delta KE$, that is $\Delta KE + \Delta U = 0$, and a quantity whose change is zero is a constant. The theorem adds no new physics: it is Newton's second law with the work of gravity and springs moved to the other side of the equation and given a name.
A 0.500 kg ball on a smooth slide starting from rest at 1.80 m, drawn at three points. The $\textcolor{#8250df}{\text{stored}}$ column empties as the $\textcolor{#1f6feb}{\text{kinetic}}$ column fills, and the $\textcolor{#116329}{\text{total}}$ stays at 8.82 J.
Looks like this, but is not
The ball is heavier, so it arrives faster.
Put the mass in and watch it leave: $\tfrac12 mv^{2} = mgh$ divides by $m$ on both sides and gives $v=\sqrt{2gh}$ with no mass in it. A 0.500 kg ball and a 5.00 kg ball released from 1.80 m on the same smooth slide both arrive at 5.94 m/s; the heavy one carries ten times the energy and needs ten times as much to reach that speed. The cancellation is not a general licence: it happens only because every term in that particular equation carried one factor of $m$. Put a spring in the problem, where the store is $\tfrac12 kx^{2}$ with no mass in it, and the mass stops cancelling immediately.
The ball on the bending chute, finally
A 0.500 kg ball is released from rest at the top of a smooth chute whose top is 1.80 m above the floor and whose bottom is level with the floor. The metal of the chute is 2.80 m long and it bends. Find the speed of the ball as it leaves the bottom, and say which word in the question made the calculation possible.
Given
$m = 0.500\ \mathrm{kg}$, released from rest
top of the chute 1.80 m above the bottom
length of the chute 2.80 m
the chute is smooth
$g = 9.80\ \mathrm{m/s^{2}}$
Find
the speed at the bottom, and the licence for the method
Solution
Energy is the only available route: the chute bends, so the acceleration is not constant and no kinematic formula applies to it. The word that licensed the shortcut is smooth.
Name the two instants and the zero level before anything else
$$\text{state 1: at rest at the top};\quad \text{state 2: at the bottom}$$
an energy statement links two instants, so both have to exist on paper before the equation does
$$U = 0 \text{ at the floor}$$
the lowest point of the motion, which keeps every number positive
Write the conserved sum and cross out what is zero
$$\tfrac12 m v_1^{2} + mgy_1 = \tfrac12 m v_2^{2} + mgy_2$$
legal because the chute is smooth: the normal force does no work and gravity is conservative
or equivalently $v=\sqrt{2gh}$, since every term carried the mass
$$v_2 = \boxed{5.94\ \mathrm{m/s}}$$
the 2.80 m of metal never entered the calculation, and neither did the shape of the bend
Answer $$\boxed{\;v_2 = 5.94\ \mathrm{m/s}\;}$$
Check
Independent check by kinematics on a different problem with the same drop: a ball dropped straight from 1.80 m has constant acceleration, so $v^{2} = 2(9.80)(1.80) = 35.3$ and $v = 5.94$ m/s. The chute and the free drop agree, which is the hook resolved. Order of magnitude: 5.94 m/s is about 21 km/h, a brisk cycling speed, and reasonable for something falling the height of a door.
Three lines, of which two were bookkeeping.
If a question gives you a height and asks for a speed, and says smooth, the answer is $\sqrt{2gh}$ before you have finished reading the sentence. The shape of the track is there to frighten you.
How fast the pendulum bob is going at the bottom
A pendulum bob of mass 0.300 kg hangs on a light string of length 2.00 m. It is pulled aside until the string makes $37.0^{\circ}$ with the vertical and released from rest. Find its speed at the lowest point, and state what would change if the bob were twice as heavy.
Given
$m = 0.300\ \mathrm{kg}$
$L = 2.00\ \mathrm{m}$
released from rest at $37.0^{\circ}$ from the vertical
$g = 9.80\ \mathrm{m/s^{2}}$
$\cos 37.0^{\circ} = 0.7986$
Find
the speed at the lowest point, and the effect of doubling the mass
Solution
Conservation beats forces here because the tension changes size throughout the swing and the acceleration is never constant; a Newton approach would need calculus and would give the same number.
Turn the angle into a height, which is the only geometry in the problem
$$h = L - L\cos\theta = L(1-\cos\theta)$$
the bob hangs $L\cos\theta$ below the pivot when the string is at $\theta$, and $L$ below it at the bottom
$$h = 2.00(1-0.7986) = 0.403\ \mathrm{m}$$
a swing that looks wide is a rise of only 40 cm, which is worth knowing before trusting an answer
Apply the conserved sum between release and the lowest point
$$mgh = \tfrac12 m v^{2}$$
the string tension is perpendicular to the motion at every instant and does no work, so only gravity is left, and it is conservative
$$v = \sqrt{2gh} = \sqrt{2(9.80)(0.4027)}$$
the mass cancels, so the 0.300 kg was never needed
$$v = \boxed{2.81\ \mathrm{m/s}}$$
about 10 km/h at the bottom of the swing
Double the mass
$$v = \sqrt{2gh}\ \text{is unchanged}$$
no mass appears; the heavier bob carries twice the energy and needs twice as much to reach the same speed
Answer $$\boxed{\;v = 2.81\ \mathrm{m/s},\ \text{unchanged if the mass doubles}\;}$$
Check
Limit check: let $\theta \to 90^{\circ}$, a string held horizontal. Then $h \to L = 2.00$ m and $v \to \sqrt{2(9.80)(2.00)} = 6.26$ m/s, the same as for a 2.00 m free fall, which is right because the bob has fallen the full length of the string. Let $\theta \to 0$ and $v \to 0$, also right.
In any pendulum question the whole physics is in the line $h = L(1-\cos\theta)$. Get that height right and the rest is $\sqrt{2gh}$.
A spring gun firing a ball straight up
A spring of stiffness 350 N/m is squashed 12.0 cm, a 50.0 g ball is placed on it and it is released, firing the ball vertically. Find how high the ball rises above the point of release, ignoring air resistance.
Given
$k = 350\ \mathrm{N/m}$
squash $x = 0.120\ \mathrm{m}$
$m = 50.0\ \mathrm{g} = 0.0500\ \mathrm{kg}$
fired vertically, no air resistance
$g = 9.80\ \mathrm{m/s^{2}}$
Find
the height gained above the release point
Solution
Choosing states where the body is at rest at both ends removes both kinetic terms and makes the problem a single division; a solution that computes the muzzle speed first works but does twice the algebra.
Choose the zero at the squashed position and list the two states
$$\text{state 1: spring squashed, ball at rest};\quad \text{state 2: at the top, at rest}$$
both states have zero kinetic energy, which is why no speed ever has to be computed
measured from the squashed position, because that is where the zero was put; above the natural length of the spring it is 12 cm less
Answer $$\boxed{\;h = 5.14\ \mathrm{m}\ \text{above the release point}\;}$$
Check
Independent check in two stages: at the moment the spring reaches its natural length the ball has $2.52 - (0.0500)(9.80)(0.120) = 2.46$ J of kinetic energy, so it leaves at 9.92 m/s, and a body launched at 9.92 m/s rises $v^{2}/2g = 5.02$ m further. Adding the 0.120 m of spring travel gives 5.14 m above the squashed position, the same answer by a route that stops in the middle.
One equation, two terms, no speeds.
When both ends of the journey are moments of rest, energy turns a two-stage motion problem into one line. Look for those instants before choosing your states.
Checkpoint
§09.3 — spotting when the shortcut is legal●●○○○
Four short problems arrive on the same page, and only one of them can be solved by setting the mechanical energy at the start equal to the mechanical energy at the end, with no extra term.
Given
each problem gives a start and an end state
in each case the body is a single object
Find
(a) Pick the problem in which mechanical energy is conserved.
Hint 1/4
Do not solve anything. Go through the four situations and ask, for each force that acts, whether it is conservative and whether it does any work at all.
Hint 2/4
Mechanical energy is conserved when every force that does work is conservative; a force perpendicular to the motion does no work and never spoils it.
Hint 3/4
So a normal force on a smooth surface is harmless, a string tension perpendicular to the motion is harmless, a friction force is fatal, and a hand pushing along the motion is also fatal.
Hint 4/4
The swinging bob on the light string is the one: the tension does no work because it is always perpendicular to the motion, and gravity is conservative.
Show solution
Testing forces is quicker and more reliable than testing situations, because a situation can look complicated while its force list is short.
Apply the same two questions to every force
$$\text{Does it do work?}\quad \vec F\cdot d\vec l \ne 0?$$
a perpendicular force is out of the discussion immediately, whatever its size
$$\text{If it does, is it conservative?}$$
gravity and ideal springs are; friction, drag and any push or pull applied from outside are not
nothing left that spoils it, so the total is constant
$$\text{the other three: } f_k, \ \text{drag or a push does work}$$
each of those moves energy across the boundary of the mechanical account
Answer $$\boxed{\;\text{the pendulum bob on the light string}\;}$$
Check
Sanity test on the rejected cases: in each of them the body would end with a different speed if the offending force were switched off, so its presence cannot be irrelevant to an energy statement.
⚠ Measuring the two heights from different levels
The start is read off one part of the diagram and the end off another, and each looks natural on its own.
wrong$$mg(1.80) + 0 = \tfrac12 mv^{2} + mg(0.60)\ \text{with } y \text{ from two zeros}$$
right$$mg(1.80) + 0 = \tfrac12 mv^{2} + mg(0.60)\ \text{with both } y \text{ from the same floor}$$
⚠ Cancelling the mass when a spring is in the equation
The cancellation is real in $mgh=\tfrac12 mv^{2}$ and gets remembered as a rule rather than as an accident of that equation.
wrong$$\tfrac12 kx^{2} = \tfrac12 mv^{2} \;\Rightarrow\; v = \sqrt{k}\,x$$
right$$\tfrac12 kx^{2} = \tfrac12 mv^{2} \;\Rightarrow\; v = x\sqrt{\frac{k}{m}}$$
9.4The energy landscape: where a body is allowed to go
Draw the store against position, draw the total as a flat line, and the gap between them is the motion the body has left.
Once the total is fixed, the store at each position decides how much motion is left over there, and that turns a graph into an answer.
RuleRule 9.4: reading a potential energy curve
Conditions
The total $E$ is a horizontal line on the same axes as $U(x)$, because it does not change with position
Valid where mechanical energy is conserved; with friction the line sinks as the body goes and the reading is only local
$KE = E - U$ can never be negative, which is the whole content of the rule
The slope form $F_x = -dU/dx$ gives the component of the conservative force along the axis, not the net force on the body
$$\boxed{\;KE(x) = E - U(x) \ge 0\;;\qquad F_x = -\frac{dU}{dx}\;}$$
The curve is the store, the flat line is the total, and the gap between them is the kinetic energy at that position. Where the curve rises to meet the line the gap closes and the body turns round; above the line it cannot be at all, since that would need negative kinetic energy. The force points downhill on this graph, which is what the minus sign in the slope rule says.
Why the force is minus the slope
Move the body a short distance $\Delta x$ with only the conservative force acting. That force does $W = F_x \Delta x$, and by the definition of the store $\Delta U = -W$, so $\Delta U = -F_x\Delta x$ and $F_x = -\Delta U/\Delta x$. Shrink the step and it is the derivative. Test it on the two stores we have: $U = mgy$ gives $F_y = -mg$, correct, weight downwards; $U = \tfrac12 kx^{2}$ gives $F_x = -kx$, correct, the spring pulls back towards the natural length.
A cart released from rest at the 12.0 m end of a smooth track. The $\textcolor{#116329}{\text{total energy line}}$ sits at the height it started from, and the $\textcolor{#cf222e}{\text{crossing on the last rise}}$ at 9.6 m along is where the cart stops and comes back. The 14.0 m summit beyond it is out of reach with the energy it has.
Looks like this, but is not
The curve is the shape of the track, so the body rolls along the curve.
Here they coincide, because $U = mgy$ is proportional to height and the graph is a scaled copy of the track profile. That is a special case. Draw the same graph for a mass on a horizontal spring and it is a parabola, while the mass slides along a flat table and never rises at all. The vertical axis always means joules, not metres, and reading it as a picture of the path stops working the moment gravity is not the store.
point on the track
height (m)
stored (J)
kinetic (J)
speed (m/s)
release point
12.0
47.0
0
0
first valley
2.0
7.84
39.2
14.0
middle summit
9.0
35.3
11.8
7.67
second valley
0
0
47.0
15.3
far summit
14.0
54.9
not available
never gets there
The stored column is just the height column multiplied by $mg = 3.92$ N, and the kinetic column is what is left of 47.0 J after that subtraction. The last row is the interesting one: the store it would need is larger than the total the cart has, so the kinetic energy there would have to be negative, and the cart turns round before it arrives. Notice also that the cart is faster in the second valley than in the first, by exactly the amount the extra 2.0 m of drop is worth.
How far along the track does the cart get
The track in the figure has its release point at 12.0 m, a first valley at 2.0 m, a summit at 9.0 m, a second valley at 0 m and then a final rise towards 14.0 m. A 0.400 kg cart is released from rest at the release point and the track is smooth. Find its speed in each valley and at the summit, and find the height at which it turns round on the final rise.
Given
$m = 0.400\ \mathrm{kg}$, released from rest at 12.0 m
heights along the track: 12.0, 2.0, 9.0, 0 and rising to 14.0 m
the track is smooth
$g = 9.80\ \mathrm{m/s^{2}}$
Find
the speeds at the three marked points and the turning height
Solution
Computing the total once and reusing it is what makes four questions into one calculation; treating each point as a fresh conservation problem would mean four equations and four chances to slip.
Fix the total once, at the only point where it is easy
Independent check on the 14.0 m speed: the first valley is 10.0 m below the release point and the cart started from rest, so $v=\sqrt{2(9.80)(10.0)}=14.0$ m/s, obtained without ever computing the total. The mass never entered either route, which is another consistency point.
One total, three subtractions and one division.
On a smooth track a body always returns to its starting height, wherever it goes in between. If your turning point comes out higher than the release point, you have gained energy from somewhere and the arithmetic is wrong.
Turning points for a mass on a spring, read two ways
A 0.600 kg block on a smooth horizontal table is attached to a spring of stiffness 250 N/m. It is set moving so that its total mechanical energy is 5.00 J, with the store measured from the natural length of the spring. Find how far from the natural length the block can get, its greatest speed, and the force on it at the extremes.
Given
$m = 0.600\ \mathrm{kg}$
$k = 250\ \mathrm{N/m}$
$E = 5.00\ \mathrm{J}$, store zero at the natural length
smooth horizontal table, so gravity does no work
Find
the extreme displacements, the greatest speed and the force there
Solution
Both extremes and the middle come from the same conserved line; hunting for the force first would have needed the acceleration, which changes at every point.
Turning points: all of the total is in the spring
$$E = \tfrac12 kx^{2} \;\Rightarrow\; x = \pm\sqrt{\frac{2E}{k}}$$
the block stops where the store has swallowed the whole total, and the $\pm$ is real: there is a turning point on each side
Independent check by an intermediate point: at $x = 0.100$ m the store is $\tfrac12(250)(0.100)^{2} = 1.25$ J, leaving 3.75 J of motion, so $v = \sqrt{2(3.75)/0.600} = 3.54$ m/s. That is less than 4.08 and more than zero, and it sits where the curve says it should.
Fast where the curve is low, slow where it is high, stopped where it meets the line, and pushed hardest where it is steepest. Those four sentences work for every store you will meet.
Checkpoint
§09.4 — reading a turning point off a curve●●○○○
A particle moves along the $x$ axis in a landscape whose store rises steadily from 2.0 J at $x=1$ m to 10.0 J at $x=5$ m. The particle is at $x=1$ m with a kinetic energy of 4.0 J, moving towards larger $x$, and no friction acts.
Given
$U(1\ \mathrm{m}) = 2.0\ \mathrm{J}$
$U(5\ \mathrm{m}) = 10.0\ \mathrm{J}$, rising steadily in between
$KE(1\ \mathrm{m}) = 4.0\ \mathrm{J}$
no friction
Find
(a) At what value of the store does the particle turn round?
Hint 1/4
You are not being asked where it turns round in metres. You are being asked for the value of the store there, which needs one number you can get in one line.
Hint 2/4
The total is fixed: $E = KE + U$, computed anywhere. The particle turns round where the kinetic energy has fallen to zero, so where $U = E$.
Hint 3/4
At $x=1$ m the two known parts are $U = 2.0$ J and $KE = 4.0$ J.
Hint 4/4
The total is 6.0 J, so the particle turns round where the store has climbed to 6.0 J.
Show solution
Working in joules and only afterwards in metres avoids reading a position off a curve that has not been drawn.
Fix the total from the one point where both parts are known
$$E = KE + U = 4.0 + 2.0 = 6.0\ \mathrm{J}$$
the total is the same everywhere, so one point is enough to pin it down
Impose the condition that defines a turning point
$$KE = 0 \;\Rightarrow\; U = E = 6.0\ \mathrm{J}$$
not 10.0 J, which is where the curve happens to end, and not 4.0 J, which was the kinetic energy rather than the total
Answer $$\boxed{\;U = 6.0\ \mathrm{J}\ \text{at the turning point}\;}$$
Check
Consistency check: 6.0 J lies between the 2.0 J at the start and the 10.0 J at the far end, so the turning point is inside the region described, roughly halfway along it. An answer outside that range would have meant the particle either never turned or turned before it started.
⚠ Reading the vertical axis of a store graph as a height
For a body on a track the two are proportional, so the habit works until the first spring question and then fails silently.
wrong$$U(x)\ \text{axis in metres}$$
right$$U(x)\ \text{axis in joules};\quad y = U/(mg)\ \text{only for gravity}$$
⚠ Dropping the minus sign in the slope rule
The formula is written down as a derivative and the sign looks like decoration, until the force comes out pointing uphill.
wrong$$F_x = \frac{dU}{dx} \;\Rightarrow\; F = +kx$$
right$$F_x = -\frac{dU}{dx} \;\Rightarrow\; F = -kx$$
9.5When friction is in the room: the total still balances
The mechanical total falls by exactly the friction force times the distance travelled, and that amount can be named rather than lost.
Every question so far had the word smooth in it; take the word away and the sum stops being constant in a way we can still measure exactly.
TheoremTheorem 9.5: the ledger with nonconservative forces
Conditions
$W_{\rm NC}$ is the work of every force that is not conservative: friction, drag, a hand pushing, a motor pulling
For kinetic friction of constant size, $W_{\rm NC} = -f_k d$ with $d$ the distance travelled along the surface, which is why the route matters again
$\Delta KE$ and $\Delta U$ are still final minus initial, with one zero level throughout
Nothing here is a new law: it is the work-energy principle with the conservative part renamed
Whatever the nonconservative forces do, the mechanical total changes by exactly that amount. Friction always does negative work, so it always makes the total fall, and the amount it takes is the friction force multiplied by the distance the body actually travelled, not the straight line between the ends. Read the second form as a sentence: what you started with equals what you ended with plus what the floor took.
Where the second form comes from
Split the net work into two piles, conservative and not: $W_{\rm cons} + W_{\rm NC} = \Delta KE$. The first pile is $-\Delta U$ by the definition of the store, so $W_{\rm NC} = \Delta KE + \Delta U$. For sliding friction of constant size, $W_{\rm NC} = -f_k d$, and moving that term to the other side gives the ledger form with a plus sign in front of $f_k d$, which is easier to use because every term in it is positive.
The forces on a 2.00 kg block sliding 4.00 m down a rough $30^{\circ}$ slope. The $\textcolor{#1f6feb}{\text{normal force}}$ is what sets the size of the $\textcolor{#8250df}{\text{friction}}$, and only the $\textcolor{#cf222e}{\text{weight}}$ has anything to do with the 2.00 m of drop.
Looks like this, but is not
Energy is not conserved when there is friction.
Mechanical energy is not conserved; energy is. The 17.0 J that leave the block in the example below do not stop existing, they end up in the block and the slope as thermal energy, and if you could measure carefully enough you would find both slightly warmer. This is why the wider statement is called the law of conservation of energy and why the phrase used here is always mechanical energy. In this course you are never asked for the temperature rise, only for the joules that left the mechanical account, but the difference between lost and moved is the difference between a bookkeeping error and physics.
A block down a rough slope, with the smooth answer for comparison
A 2.00 kg block slides 4.00 m from rest down a slope inclined at $30.0^{\circ}$ with $\mu_k = 0.250$. Find its speed at the bottom, the energy given to thermal energy, and the speed it would have had on a smooth slope.
Given
$m = 2.00\ \mathrm{kg}$, released from rest
slope $30.0^{\circ}$, distance along it $d = 4.00\ \mathrm{m}$
the speed at the bottom, the thermal energy produced, and the smooth-slope speed
Solution
The energy route needs no acceleration and would still work if the slope curved; the force route is used here only as a check, and it would fail on a curved slope.
Turn the slope length into a drop, and get the friction force across the slope
Independent check with Newton's second law, which is available because the slope is straight and the acceleration is constant: $a = g(\sin\theta - \mu_k\cos\theta) = 9.80(0.500-0.2165) = 2.78$ m/s$^{2}$, and $v = \sqrt{2(2.7783)(4.00)} = 4.71$ m/s. The two routes agree to three digits.
Two force lines, one energy line. The force lines were unavoidable: friction's size is not an energy question.
Every rough-slope problem has this shape: convert the slope length to a drop for the store, keep the slope length for the friction, and do not let the two swap places.
How far the block slides across the floor before stopping
The same 2.00 kg block leaves the bottom of the slope at 4.71 m/s and slides onto a level floor with the same coefficient $\mu_k = 0.250$. Find how far it travels before stopping, and how the answer changes if it arrives twice as fast.
Given
$m = 2.00\ \mathrm{kg}$, arriving at $v = 4.71\ \mathrm{m/s}$
level floor, $\mu_k = 0.250$
$g = 9.80\ \mathrm{m/s^{2}}$
Find
the sliding distance, and the effect of doubling the arrival speed
Solution
Writing $f_k$ as $\mu_k mg$ only at the end shows the mass cancelling, which is the fact worth remembering: on the level, the sliding distance does not depend on how heavy the block is.
Set the ledger up on a level floor, where the store never changes
$$\tfrac12 mv^{2} = f_k d$$
no height change, so the store cancels from both sides and every joule of motion has to be taken by the floor
the kinetic energy quadruples while the friction force is unchanged, which is the same square law that governs stopping distances for cars
Answer $$\boxed{\;d = 4.54\ \mathrm{m},\quad \text{and } 18.1\ \mathrm{m} \text{ at twice the speed}\;}$$
Check
Independent check without energy: the deceleration is $a = \mu_k g = 2.45$ m/s$^{2}$, and $v^{2}=2ad$ gives $d = 22.19/(2\times 2.45) = 4.53$ m, agreeing to the rounding of the arrival speed. Note also that the mass cancelled in the second route, and in the first it cancels too once $f_k = \mu_k mg$ is substituted.
A body sliding to rest on a rough level surface always travels $v^{2}/(2\mu_k g)$, and that is worth carrying as a formula in its own right.
Checkpoint
§09.5 — which distance goes into the friction term●●○○○
A box is dragged from one corner of a rough room to the opposite corner, once along the diagonal, a distance of 5.00 m, and once along two walls, a distance of 7.00 m. The floor is level and the coefficient is the same everywhere.
Given
level floor, same $\mu_k$ everywhere
diagonal route 5.00 m, route along the walls 7.00 m
same start and finish points
Find
(a) Compare the thermal energy produced on the two routes.
Hint 1/4
Nothing here needs a number. Ask what the friction term is multiplied by, and then ask which of the two routes makes that factor larger.
Hint 2/4
The friction term in the ledger is $f_k d$, and $d$ is the distance travelled along the surface, not the displacement between the endpoints.
Hint 3/4
Here $f_k$ is the same on both routes because the floor and the box are unchanged; the distances are 5.00 m and 7.00 m.
Hint 4/4
The longer route produces more thermal energy, in the ratio 7 to 5, so 40% more.
Show solution
Working with the ratio avoids inventing a mass and a coefficient the question never gave, and shows that the answer does not depend on them.
Write the term that differs
$$E_{\rm th} = f_k d$$
the only route-dependent quantity in the whole ledger
the friction force cancels because the surface and the box are the same
Say what did not change
$$\Delta U = 0\ \text{on both routes}$$
the floor is level and the endpoints are shared, so the store is untouched either way
Answer $$\boxed{\;40\%\ \text{more thermal energy along the walls}\;}$$
Check
Numerical check with a 20.0 kg box and $\mu_k = 0.300$: $f_k = 58.8$ N, giving 294 J along the diagonal and 412 J along the walls, and $412/294 = 1.40$ as claimed.
⚠ Using the vertical drop in the friction term
The height was just computed for the store and it is the freshest number on the page, so it gets reused one line later where the path length belongs.
wrong$$f_k d = (4.24)(2.00) = 8.49\ \mathrm{J}$$
right$$f_k d = (4.24)(4.00) = 17.0\ \mathrm{J}$$
⚠ Taking the normal force as mg on a slope
It is $mg$ on a level floor, which is where the formula was first met, and the slope quietly changes it to $mg\cos\theta$.
The same 2.00 kg block on the same 4.00 m slope with three roughnesses. The 39.2 J released by the 2.00 m drop is fixed; what changes is how it is divided between the block's motion and the warmed slope.
9.6Far from the ground: the store that runs out at infinity
When the height is no longer small the store stops being a straight line and becomes a curve that flattens out at infinity.
The formula $U=mgy$ was built on a constant $g$; the moment a rocket climbs a few thousand kilometres that assumption is gone and the store has to be rebuilt from the force law.
TheoremTheorem 9.6: the gravitational store far from a body
Conditions
$M$ is the mass of the planet and $m$ of the small body; $r$ is measured from the centre of the planet, never from its surface
The zero is fixed at infinity, which is what makes every finite $r$ give a negative value
Both bodies treated as points, or as spheres, which is enough for a planet
Reduces to $mgy$ for heights small compared with the radius, and that is shown below rather than asserted
The store of a body at a distance $r$ from the centre of a planet is minus $G$ times the two masses divided by that distance. Negative does not mean anything is wrong: it means the body has less than it would have infinitely far away, which is where the zero was put. Climbing away makes the store less negative, which is to say it rises, and reaching zero means having exactly enough to arrive at infinity with nothing left over.
Building the store from the force, and recovering mgy
The radial component of the pull on the small body is $F_r = -GMm/r^{2}$, negative because it points inwards. The work done in moving from $r_1$ to $r_2$ is the integral of that, $W = GMm\left(\frac{1}{r_2}-\frac{1}{r_1}\right)$, and the store changes by minus this, which is exactly what $U = -GMm/r$ produces. For the small-height limit take $r_1 = R$ and $r_2 = R+h$ with $h \ll R$: the change is $GMm\,h/[R(R+h)] \approx (GM/R^{2})mh = mgh$, since $GM/R^{2}$ is precisely the $g$ at the surface. The old formula is the new one seen from close up.
The gravitational store per kilogram against distance from the centre of the Earth. A launch fixes a horizontal $\textcolor{#cf222e}{\text{total line}}$; where that line cuts the $\textcolor{#8250df}{\text{curve}}$ the body turns round, and a line at $\textcolor{#116329}{\text{zero}}$ never cuts it at all.
Looks like this, but is not
The store is negative, so the body has negative energy and something has gone wrong.
The sign is a consequence of putting the zero at infinity, and that choice was made because it is the only place that is the same distance from every planet. Exactly the same freedom was used in the block on the zero level near the ground, where taking the ceiling as zero made a book on a table carry $-32.3$ J. What is physical is the total: a body whose total is negative is trapped and will come back, a body whose total is zero or more is not. That is a statement about which side of zero the sum sits on, and it survives any relabelling because both $KE$ and $U$ were computed in the same frame.
The launch speed that never comes back
Find the speed at which a projectile would have to leave the surface of the Earth, ignoring air resistance and the rotation of the planet, in order never to return. Use $GM_E = 3.98\times10^{14}\ \mathrm{N\cdot m^{2}/kg}$ and $R_E = 6.38\times10^{6}\ \mathrm{m}$, and say what the answer would be for a body of twice the mass.
launched from the surface, no air, planet not rotating
Find
the escape speed, and its dependence on the mass launched
Solution
Setting $E=0$ is shorter than following the projectile: any attempt to track the flight needs an acceleration that changes with height, and no kinematic formula in this course handles that.
Say in energy language what never returning means
$$E = \tfrac12 mv^{2} - \frac{GM_Em}{R_E}$$
the total at the moment of launch, with the store taken from the theorem and the zero at infinity
$$E \ge 0$$
a body with a negative total has a turning point at a finite distance and must come back; zero is the cheapest ticket out
Set the total to zero and solve
$$\tfrac12 mv^{2} = \frac{GM_Em}{R_E}$$
the borderline case, arriving at infinity with no speed left
Independent check by a different grouping: the store per kilogram at the surface is $-GM_E/R_E = -62.4$ MJ/kg, so escaping needs 62.4 MJ of kinetic energy per kilogram, and $\tfrac12 v^{2} = 62.4\times10^{6}$ gives $v = 11.2$ km/s. Order of magnitude: this is about thirty times the speed of sound, which is why escaping is hard and orbiting, at 7.9 km/s, is merely expensive.
Two lines, and the mass cancelled in the first.
Escape speed is a property of the planet, not of the thing leaving it. A pebble and a rocket need the same 11.2 km/s; what differs is the fuel bill for reaching it.
A projectile fired at 8.0 km/s: how far does it get
A projectile is fired vertically from the surface of the Earth at $8.00\times10^{3}$ m/s, which is less than the escape speed. Ignoring air resistance, find the greatest distance from the centre of the Earth that it reaches, and the altitude above the surface.
Given
$v_1 = 8.00\times10^{3}\ \mathrm{m/s}$ at $r_1 = R_E$
the maximum distance from the centre and the altitude above the surface
Solution
Working per kilogram is the trick that keeps the numbers small enough to check by eye; carrying an unknown $m$ through four lines adds nothing and costs concentration.
about 6720 km of altitude, comparable with the radius of the planet, which is exactly why $mgy$ could not have been used
Answer $$\boxed{\;r_{\max} = 1.31\times10^{7}\ \mathrm{m},\quad h = 6.72\times10^{3}\ \mathrm{km}\;}$$
Check
Check against the wrong tool, deliberately: $mgy$ would have predicted $h = v^{2}/2g = 3.27\times10^{6}$ m, about 3270 km, less than half the real answer. It underestimates because it charges the projectile the full surface weight all the way up, while the real pull weakens with height. The sign of the discrepancy is the check: the true height must be larger, and it is.
Four short lines, two of which were unit bookkeeping in megajoules.
Once the total per kilogram is negative, the turning distance is a single division. If it comes out smaller than the planet's radius, the projectile never left the ground and something is wrong.
Checkpoint
§09.6 — what escape speed does and does not depend on●●○○○
Two probes are launched from the surface of the same airless planet, one of mass 200 kg and one of mass 2000 kg. An engineer claims the heavier probe needs a larger launch speed to escape.
Given
same planet, same launch point, no atmosphere
probe masses 200 kg and 2000 kg
Find
(a) Decide whether the engineer is right, and give the reason.
Hint 1/4
This is a question about which symbols survive to the end of the derivation. Write the escape condition and look for the probe's mass.
Hint 2/4
The condition is $\tfrac12 mv^{2} = GMm/R$, and $M$ is the planet's mass while $m$ is the probe's.
Hint 3/4
Both terms carry exactly one factor of the probe's mass $m$, so it divides out before any number is substituted.
Hint 4/4
The engineer is wrong about the speed: both need 11.2 km/s on Earth. The heavier probe needs ten times the energy to reach it.
Show solution
Cancelling symbolically settles both probes at once and shows exactly which mass the answer is sensitive to.
Write the condition and cancel
$$\tfrac12 mv^{2} = \frac{GMm}{R}$$
the borderline total of zero, with the store from the theorem
$$v = \sqrt{\frac{2GM}{R}}$$
one factor of $m$ on each side, so the probe's mass leaves and only the planet's stays
Separate the two questions the engineer merged
$$E_{\rm needed} = \tfrac12 m v_{\rm esc}^{2} \propto m$$
the energy does scale with the probe's mass, ten to one here, even though the speed does not
Answer $$\boxed{\;\text{same escape speed; ten times the energy}\;}$$
Check
Independent check by the limit that does matter: increase the planet's mass instead and the speed does rise, as $\sqrt{M}$. So the formula is not blind to mass in general, only to the mass of the thing being launched.
⚠ Measuring r from the surface instead of the centre
Altitudes are what questions quote, and the height above the ground is the number sitting in the given list.
wrong$$U = -\frac{GMm}{h}$$
right$$U = -\frac{GMm}{R+h}$$
⚠ Using mgy for a climb comparable with the planet's radius
The formula is familiar and works everywhere in the rest of the course, and nothing in the arithmetic complains when it is misused.
right$$r_{\max} = \frac{GM}{\frac{GM}{R}-\frac{v^{2}}{2}}\;\Rightarrow\; h = 6.72\times10^{6}\ \mathrm{m}$$
9.7Power: the same work, delivered at different rates
How many joules per second are being delivered, which is what an engine or a person is actually rated by.
Every question so far asked how much energy moved; this last block asks how quickly, which is a different number and a different unit.
DefinitionDefinition 9.7: power
Conditions
$\bar P$ is an average over a stated interval; $P = Fv$ is the value at one instant
In $P = Fv$ the force is the component along the velocity, exactly as in the definition of work
The unit is the watt, one joule per second; a kilowatt hour is an energy, not a power
Nothing new is being conserved here; power is a rate, and rates are not conserved quantities
$$\boxed{\;\bar P = \frac{W}{t}\;;\qquad P = \vec F\cdot\vec v = Fv\cos\theta\;;\qquad 1\ \mathrm{W} = 1\ \mathrm{J/s}\;}$$
Power is work per unit time. Two people who carry the same load up the same stairs do the same work and the one who arrives first delivered more power. The second form says the same thing for a body already moving: multiply the force by the speed at that instant and you get the joules per second being delivered right then, which is why a car needs more power to hold a speed on a hill than on the flat although the speed is the same.
Why P = Fv follows from the definition
Over a short time $\Delta t$ a body moving at speed $v$ along a force $F$ covers $\Delta x = v\Delta t$, so the work done is $W = F v \Delta t$ and the rate is $W/\Delta t = Fv$. If the force is at an angle, only its component along the velocity does work, which puts the cosine back. Nothing else is needed: $P = Fv$ is the definition of work divided by the time it took, written for one instant.
One student, one staircase, three speeds. The work done against gravity is $\textcolor{#116329}{\text{2646 J in all three cases}}$ and the $\textcolor{#1f6feb}{\text{power}}$ is what the clock decides.
Looks like this, but is not
A kilowatt hour is a big kilowatt, so it is a unit of power.
It is a power multiplied by a time, so it is an energy: $1\ \mathrm{kWh} = (1000\ \mathrm{W})(3600\ \mathrm{s}) = 3.6\times10^{6}$ J. The electricity bill is charged in kilowatt hours because what you buy is energy, while the rating stamped on the heater, 2 kW, is the rate at which it will spend it. Run that heater for three hours and you have bought 6 kWh, which is 21.6 MJ, enough to lift a small car about 1.6 km if you could couple it to a rope. The same trap catches the horsepower, which is a power, and the horsepower hour, which is not.
A student on the stairs, and how the answer changes with the clock
A 60.0 kg student climbs a staircase of vertical height 4.50 m in 6.00 s at a steady pace. Find the work done against gravity and the average power delivered, express it in horsepower, and find the power if the same climb takes 12.0 s or 3.50 s.
Given
$m = 60.0\ \mathrm{kg}$
vertical rise $4.50\ \mathrm{m}$
times of 6.00 s, 12.0 s and 3.50 s
$1\ \mathrm{hp} = 746\ \mathrm{W}$
$g = 9.80\ \mathrm{m/s^{2}}$
Find
the work, the average power at three paces, and the power in horsepower
Solution
Computing the work once and dividing three times is the whole lesson: students who recompute $mgh$ for each pace usually end up believing the work changed.
The work is a height question and has nothing to do with the clock
$$W = mgh = (60.0)(9.80)(4.50)$$
the vertical rise, not the length of the staircase, because that is what the store responds to
$$W = 2646\ \mathrm{J}$$
the same for every pace, which is the point of the example
Divide by each time
$$\bar P = \frac{2646}{6.00} = 441\ \mathrm{W}$$
joules per second, and a sustainable output for a fit person over a short climb
$$\frac{441}{746} = 0.591\ \mathrm{hp}$$
roughly half a horsepower, which is a fair account of what a horse is
Independent check by the other formula: at a steady pace the student rises at $4.50/6.00 = 0.750$ m/s while pushing with a force equal to the weight, 588 N, so $P = Fv = (588)(0.750) = 441$ W. The two definitions agree, as they must when the pace is steady.
If a question gives you a time, the answer is probably a power. If it gives you a height, the answer is probably an energy. Reading which one is wanted is half the marks.
The power a car needs to hold its speed up a hill
A 1400 kg car climbs a hill inclined at $6.00^{\circ}$ to the horizontal at a steady 22.0 m/s. Air resistance and rolling resistance together amount to a constant 700 N opposing the motion. Find the power the engine must deliver at the wheels, and compare it with the power needed on level ground at the same speed.
Given
$m = 1400\ \mathrm{kg}$
slope $6.00^{\circ}$
steady $v = 22.0\ \mathrm{m/s}$
resistance $700\ \mathrm{N}$
$g = 9.80\ \mathrm{m/s^{2}}$
$\sin 6.00^{\circ} = 0.1045$
Find
the power on the hill and on the level
Solution
$P=Fv$ is the natural tool when the speed is steady and the force is constant; the work-per-second check is the same calculation rearranged, which is why it agrees exactly rather than approximately.
Steady speed means the forces balance, so find the driving force first
$$F = mg\sin\theta + F_{\rm res}$$
no acceleration, so the drive exactly matches the two resisting contributions; this is a force step, and energy methods will not produce it
Independent check by energy over one second: in 1.00 s the car covers 22.0 m and climbs $22.0\sin 6.00^{\circ} = 2.30$ m, gaining $(1400)(9.80)(2.30) = 31.6$ kJ of store, while the resistance takes $(700)(22.0) = 15.4$ kJ. The total, 47.0 kJ in one second, is 47.0 kW.
One force line and one multiplication.
At a steady speed the engine's power is spent entirely on climbing and on resistance. Nothing goes into kinetic energy, because the kinetic energy is not changing.
Checkpoint
§09.7 — telling a power from an energy●○○○○
Four quantities appear in a utility bill and a car brochure: a 2.0 kW heater rating, a 6.0 kWh monthly figure, a 63 hp engine output and a 2646 J staircase climb.
Given
heater rating 2.0 kW
monthly figure 6.0 kWh
engine output 63 hp
staircase climb 2646 J
Find
(a) Which pair of these are energies rather than powers?
Hint 1/4
Sort the four by their units before thinking about what they describe. Two of them contain a time in the numerator's favour and two do not.
Hint 2/4
A power is joules per second: the watt and the horsepower are powers. An energy is joules: the joule and the watt multiplied by a time are energies.
Hint 3/4
So 2.0 kW is a power, 63 hp is a power, 2646 J is an energy, and a kilowatt hour is a kilowatt multiplied by an hour.
Hint 4/4
The two energies are the 6.0 kWh figure and the 2646 J climb.
Show solution
Converting everything to joules and joules per second removes the need to reason about what the quantities mean at all.
Scale check on the result: 21.6 MJ is about 8000 staircase climbs, which sounds right for a month of heating and hopeless as a description of a person.
⚠ Dividing by the time twice
The speed already contains a time, so multiplying $Fv$ and then dividing by the duration looks like extra care and is a second division.
wrong$$P = \frac{Fv}{t}$$
right$$P = Fv \quad\text{or}\quad \bar P = \frac{Fd}{t}$$
⚠ Treating a kilowatt hour as a power
It is written with a power in front of it and it appears on a bill next to numbers that are rates.
Solving a problem with conservation of mechanical energy
The question links a height, a speed or a spring deformation to another one of the same three, mentions no time, and describes the surfaces as smooth or the motion as free.
Name two instants.
Write state 1 and state 2 in words before any symbol: at rest at the top, just leaving the spring, at the highest point. Most wrong solutions are wrong here, not in the algebra.
Choose the zero and say so.
Usually the lowest point the body reaches, or the natural length of the spring. Write the sentence down; it costs one line and it is the line that keeps the two heights in the same frame.
Check the licence.
List the forces that do work. If they are all conservative you may use the conserved total; if a rough surface or an external push appears, go to the next method box instead.
Write the full line, then delete the zeros.
$\tfrac12 mv_1^{2} + mgy_1 + \tfrac12 kx_1^{2} = \tfrac12 mv_2^{2} + mgy_2 + \tfrac12 kx_2^{2}$, and cross out the terms that are zero. Deleting is safer than remembering which terms to include.
Solve in symbols where you can.
If every term carries the mass, cancel it before substituting; you learn something from seeing that it goes and you avoid arithmetic.
Check by a limit or a second route.
Set an angle to zero, make a surface smooth, or reach the same number with kinematics if the acceleration happens to be constant.
Where it goes wrong
Two heights measured from two different zeros.
The length of a slope used where the vertical drop belongs.
Using the method on a rough surface because the word rough was in a clause at the end of the question.
Trying to answer a question about time. This method cannot see time at all.
When the surface is rough
A coefficient of friction, the word rough, a drag force, or any external push or pull that acts along the motion.
Get the normal force from a force equation.
Across the surface, where there is no acceleration: $N = mg$ on the level, $N = mg\cos\theta$ on a slope, and something else if another force has a component across the surface. Energy methods never produce $N$ for you.
Then $f_k = \mu_k N$, and it is constant if $N$ is.
Constant $N$ means the friction term is a simple product later; if the surface curves, $N$ changes and the problem is beyond this course.
Identify the distance travelled along the surface.
Not the drop, not the displacement between the endpoints, and not half of the round trip: the length of the actual path over the rough part.
Write the ledger with every term positive.
$\tfrac12 mv_1^{2} + U_1 = \tfrac12 mv_2^{2} + U_2 + f_k d$, read as what we started with equals what we ended with plus what the floor took.
Solve, then say where the missing energy went.
It is thermal energy in the surface and the body. Saying so is the difference between a total that vanished and a total that moved.
Where it goes wrong
$N = mg$ used on a slope.
The friction term given the vertical drop instead of the path length.
A round trip treated as though friction cancels between the legs.
A minus sign in front of $f_k d$ on the side where it is already subtracted, putting the energy back in.
Choosing between Newton's second law and an energy method
At the start of every mechanics question, before any equation is written.
Read what is asked for.
A speed or a height or a distance points to energy. A time, an acceleration or a force points to Newton.
Look at the path.
If it bends, kinematics is out and energy is the only survivor. If it is straight with a constant force, both work and Newton is often shorter.
Count the instants that matter.
Energy compares two instants and says nothing about what happened in between. If the question is about the middle, you need forces.
Use the loser as the check.
When both methods are available, solve with one and verify with the other. That is where several of the checks in this section came from.
Where it goes wrong
Reaching for $F=ma$ on a curved track, where the acceleration is not constant and the kinematic formulas do not apply.
Reaching for energy when the question asks how long something took.
Trying to get a normal force out of an energy equation.
The same slope, done with Newton's second law
A 2.00 kg block slides from rest down a smooth slope of length 4.00 m inclined at $30.0^{\circ}$. Find its speed at the bottom using forces.
Given
$m = 2.00\ \mathrm{kg}$
$L = 4.00\ \mathrm{m}$ at $30.0^{\circ}$
smooth, released from rest
Find
the speed at the bottom
Solution
Two steps, and it also delivers the acceleration, which the energy route never sees.
Resolve along the slope and find the acceleration
$$ma = mg\sin\theta \;\Rightarrow\; a = g\sin\theta$$
the normal force is perpendicular to the motion and drops out of the along-slope equation
$$a = (9.80)(0.500) = 4.90\ \mathrm{m/s^{2}}$$
constant, because the slope is straight and the angle never changes
Use a kinematic formula
$$v^{2} = v_0^{2} + 2aL = 0 + 2(4.90)(4.00)$$
legal only because the acceleration is constant
$$v = 6.26\ \mathrm{m/s}$$
three digits, matching the data
Answer $$\boxed{\;v = 6.26\ \mathrm{m/s}\;}$$
Check
Substituting $a = g\sin\theta$ and $L$ symbolically gives $v=\sqrt{2gL\sin\theta}$, and $L\sin\theta$ is the vertical drop, which is the energy answer in disguise.
The same slope, done with energy
The same 2.00 kg block, the same smooth 4.00 m slope at $30.0^{\circ}$, released from rest. Find the speed at the bottom using energy.
Given
$m = 2.00\ \mathrm{kg}$
$L = 4.00\ \mathrm{m}$ at $30.0^{\circ}$
smooth, released from rest
Find
the speed at the bottom
Solution
One line once the height is found, and it survives a change of shape that would destroy the force route.
the only geometry the method needs, and the step where marks are usually lost
Set the totals equal
$$mgh = \tfrac12 mv^{2} \;\Rightarrow\; v = \sqrt{2gh}$$
the mass cancels, so it never has to be substituted
$$v = \sqrt{2(9.80)(2.00)} = 6.26\ \mathrm{m/s}$$
identical to the force answer, as it must be
Answer $$\boxed{\;v = 6.26\ \mathrm{m/s}\;}$$
Check
Change the slope into a smooth curve of the same height and this solution is unchanged, while the force solution collapses. That is the real difference between them.
Both routes give 6.26 m/s on a straight slope; the force route additionally hands you the acceleration, and the energy route additionally survives if the slope stops being straight.
How to tell them apart
Ask whether the path is straight and whether the question wants an acceleration or a time. If either answer is yes, use forces. If the path bends or only speeds and heights are involved, use energy.
How fast at the bottom: a question energy can answer
A 2.00 kg block slides from rest down a smooth $30.0^{\circ}$ slope 4.00 m long. How fast is it going at the bottom?
Given
smooth slope, 4.00 m at $30.0^{\circ}$, from rest
Find
the speed at the bottom
Solution
Energy is enough because the question compares two instants and asks for a speed.
the drop is $4.00\sin 30.0^{\circ} = 2.00$ m, and no other feature of the slope matters
Answer $$\boxed{\;v = 6.26\ \mathrm{m/s}\;}$$
Check
Independent check: the force route gives $a=4.90$ m/s$^{2}$ and the same speed.
How long it takes: a question energy cannot answer
The same block on the same slope. How long does the slide take?
Given
smooth slope, 4.00 m at $30.0^{\circ}$, from rest
Find
the time taken
Solution
Nothing about energy was wrong; it simply does not contain the variable being asked for.
Try the energy route and watch it fail
$$\tfrac12 mv^{2} = mgh$$
this equation contains no time symbol at all, so no rearrangement of it can produce one
Go back to forces
$$a = g\sin\theta = 4.90\ \mathrm{m/s^{2}}$$
constant, because the slope is straight
$$L = \tfrac12 a t^{2} \;\Rightarrow\; t = \sqrt{\frac{2L}{a}} = \sqrt{\frac{8.00}{4.90}}$$
a kinematic formula, which is where time lives
$$t = 1.28\ \mathrm{s}$$
and if the slope curved, this would be beyond the course entirely
Answer $$\boxed{\;t = 1.28\ \mathrm{s}\;}$$
Check
Consistency: an average speed of $6.26/2 = 3.13$ m/s over 4.00 m takes $4.00/3.13 = 1.28$ s, matching, and the average of the speed is legal here only because the acceleration is constant.
The two questions differ by one word, fast against long, and that word decides which method can even be attempted.
How to tell them apart
Energy statements have no time in them. If the answer is measured in seconds, the method is forces and kinematics, whatever the rest of the question looks like.
Scaffolding comes off
The common skeleton
Name the two instants and write down what is known at each.
Choose the zero of the store and state it; use it on both sides.
List the forces that do work and decide whether any is nonconservative.
Write the full energy line and delete the terms that are zero.
Solve for the one unknown, keeping symbols while the mass may still cancel.
Check with a limit, an order of magnitude, or a second route.
1 · fully worked
Fully worked: a ball on a smooth curved ramp
A 0.300 kg ball is released from rest at the top of a smooth curved ramp whose top is 1.20 m above the bottom. Find its speed at the bottom.
Given
$m = 0.300\ \mathrm{kg}$, from rest
drop $h = 1.20\ \mathrm{m}$
the ramp is smooth and curved
$g = 9.80\ \mathrm{m/s^{2}}$
Find
the speed at the bottom
Solution
Energy rather than forces, because the ramp curves and no constant acceleration exists to feed a kinematic formula.
Two instants and a zero
$$\text{state 1: at rest, } y_1 = 1.20\ \mathrm{m}$$
the release, where the kinetic term is zero
$$\text{state 2: at the bottom, } y_2 = 0$$
with the zero of the store put at the bottom, so the store term dies there
The licence, then the line
$$\text{smooth} \Rightarrow \text{only gravity does work}$$
the normal force is perpendicular to the motion everywhere on the ramp, however the ramp curves
$$\tfrac12 m v_1^{2} + mgy_1 = \tfrac12 m v_2^{2} + mgy_2$$
the full line before deleting anything
$$0 + mg(1.20) = \tfrac12 mv_2^{2} + 0$$
two terms deleted, both for reasons written in the previous subgoal
the mass cancelled, so the 0.300 kg was never used
$$\text{check: } h = \frac{v^{2}}{2g} = \frac{23.52}{19.6} = 1.20\ \mathrm{m}$$
running the formula backwards returns the given height, which catches a misplaced factor of two
Answer $$\boxed{\;v = 4.85\ \mathrm{m/s}\;}$$
Check
Order of magnitude: 4.85 m/s is about 17 km/h, and the drop is roughly the height of a table, which is the right size for something falling that far.
2 · you write the reasoning
Now an easier problem with the reasoning removed. A 0.500 kg ball is dropped from rest through 0.800 m onto a floor. The steps are given; write the reason for each one before opening the model answer. The physics is easier than rung 1 on purpose: the work here is explaining, not solving.
State 1: at rest, $y_1 = 0.800$ m. State 2: at the floor, $y_2 = 0$.
reasoning
Both instants have to exist before an energy line can compare them, and the zero of the store is placed at the floor here so that the second store term will vanish.
Only gravity does work, so $\tfrac12 mv_1^{2}+mgy_1 = \tfrac12 mv_2^{2}+mgy_2$.
reasoning
The licence: with air resistance ignored there is no nonconservative force at all, so the mechanical total is constant and the full line may be written.
The line becomes $mg(0.800) = \tfrac12 mv_2^{2}$.
reasoning
Two terms are deleted, and each for its own reason: the ball starts at rest, and it ends at the level chosen as the zero.
So $v_2 = \sqrt{2(9.80)(0.800)} = 3.96$ m/s.
reasoning
The mass cancels because every surviving term carries one factor of it, so the 0.500 kg is never substituted and the answer would be the same for a brick.
3 · find the buried error
Harder than rung 2, and this solution has exactly two errors in it. A 0.500 kg block is held against a spring of stiffness 400 N/m squashed by 0.150 m at the foot of a ramp inclined at $25.0^{\circ}$, with $\mu_k = 0.200$ between block and ramp. The block is released and slides up the ramp. How far along the ramp does it travel before stopping, measured from the release point? Take $\sin 25.0^{\circ}=0.4226$ and $\cos 25.0^{\circ}=0.9063$. Find the two wrong steps.
Step 1. The spring stores $\tfrac12 kx^{2} = \tfrac12(400)(0.150)^{2} = 4.50$ J, and that is the whole budget for the trip.
Step 2. At the stopping point the budget has gone into the gravitational store and into friction: $4.50 = mgd + f_k d$.
Step 3. The friction force is $f_k = \mu_k mg = (0.200)(0.500)(9.80) = 0.980$ N.
Step 4. So $4.50 = d(4.90 + 0.980)$ and $d = 4.50/5.88 = 0.765$ m.
the two buried errors (2)
⚠ step 2
The gravitational store was written as $mgd$ with $d$ measured along the ramp. The store responds to the vertical rise, which is $d\sin 25.0^{\circ}$, so the term should be $mgd\sin 25.0^{\circ} = 2.07d$ rather than $4.90d$.
The length along the ramp is the unknown being solved for and the sine has to be remembered rather than read off the page, so it is the first thing to be dropped under exam pressure.
right
$4.50 = mgd\sin\theta + f_k d$, with $mg\sin\theta = (0.500)(9.80)(0.4226) = 2.07$ N.
⚠ step 3
The normal force on a slope is $mg\cos\theta$, not $mg$, so $f_k = (0.200)(0.500)(9.80)(0.9063) = 0.888$ N, not 0.980 N.
The formula $f_k=\mu_k mg$ is the one first met on a horizontal floor, where it is correct, and the cosine arrives only later.
right
$f_k = \mu_k mg\cos\theta = 0.888$ N, about 9% smaller than the value used.
4 · the bare problem
§09.5 — spring, ramp and friction with no scaffolding●●●●○
The scaffolding is gone. A 0.500 kg block is held against a spring of stiffness 400 N/m squashed by 0.150 m at the foot of a ramp inclined at $25.0^{\circ}$, with $\mu_k = 0.200$ between the block and the ramp. The block is released.
(a) How far along the ramp, measured from the release point, does the block travel before it stops?
(b) How much of the spring's energy ended up as thermal energy?
Hint 1/4
The trip has two named instants, and at both of them the block is at rest. Write down what is in the account at each of them before writing any equation.
Hint 2/4
The ledger with friction is $\tfrac12 kx^{2} = mgd\sin\theta + f_k d$, with $f_k = \mu_k mg\cos\theta$ and $d$ measured along the ramp.
Hint 3/4
The numbers: $k = 400$ N/m with a squash of 0.150 m, $m = 0.500$ kg, $\theta = 25.0^{\circ}$ with sine 0.4226 and cosine 0.9063, and $\mu_k = 0.200$.
Hint 4/4
The block travels 1.52 m along the ramp and 1.35 J of the 4.50 J ends up as thermal energy.
Show solution
Both unknown costs are proportional to the same distance, so factorising $d$ out turns the problem into one division; solving for the height first would need the same work and an extra conversion.
Independent check on the split: the store gained $mg\,d\sin\theta = (4.90)(1.521)(0.4226) = 3.15$ J and the floor took 1.35 J, and $3.15+1.35 = 4.50$ J, the whole spring budget, with nothing left over. Limit check: set $\mu_k = 0$ and the distance rises to $4.50/2.071 = 2.17$ m, which is longer, as removing friction must make it.
Whenever a body starts and ends at rest, every kinetic term disappears and the whole question becomes a division.
Full exam-style question
Ramp, rough patch and spring, in four partsexam format
A 0.800 kg block is released from rest at the top of a smooth curved ramp whose top is 1.50 m above the floor. At the bottom it slides across a rough horizontal patch 2.00 m long with $\mu_k = 0.300$, and then meets a spring of stiffness 600 N/m standing on smooth floor. Find (a) its speed at the bottom of the ramp, (b) its speed as it reaches the spring, (c) the greatest compression of the spring, and (d) the height it reaches back up the ramp after the spring pushes it away, given that it crosses the rough patch a second time.
Given
$m = 0.800\ \mathrm{kg}$, released from rest
ramp height $1.50\ \mathrm{m}$, ramp smooth and curved
the four quantities, in order, each using the answer before it
Solution
Each part hands its answer to the next, so it is worth carrying energies in joules rather than speeds: three of the four parts never need a speed at all, and squaring rounded speeds is where this kind of question usually goes wrong.
Check the whole account in one line: the block began with $(0.800)(9.80)(1.50) = 11.76$ J, the patch was crossed twice and took $2\times 4.704 = 9.408$ J, and $11.76 - 9.408 = 2.352$ J is exactly the energy sitting in the final height, $(0.800)(9.80)(0.300) = 2.352$ J. Nothing is missing and nothing was created.
Four parts, one energy account, and only two square roots in the whole thing.
A block that keeps crossing the same rough patch loses the same amount every crossing, so the number of crossings it can afford is the total divided by that amount: here 11.76/4.704 gives two full crossings and a half of a third.
Practice
A · concept 4 questions
1§09.3 — two balls thrown from the same roof●●○○○
Two identical balls are thrown from the edge of a roof 12.0 m above the ground with the same speed of 8.00 m/s. The first is thrown straight up, the second straight down. Air resistance is negligible.
Given
roof height 12.0 m above the ground
both launched at 8.00 m/s, one up and one down
air resistance negligible
$g = 9.80\ \mathrm{m/s^{2}}$
Find
(a) Compare the speeds with which the two balls hit the ground.
Hint 1/4
Do not follow either ball through its flight. Ask instead what the energy statement needs to know about a journey, and whether the two journeys differ in any of those things.
Hint 2/4
Mechanical energy is conserved for both: $\tfrac12 mv_1^{2} + mgy_1 = \tfrac12 mv_2^{2} + mgy_2$, and the kinetic term contains $v^{2}$, which is blind to direction.
Hint 3/4
Both start at $y = 12.0$ m with speed 8.00 m/s and both finish at $y=0$; the only difference between them is a direction, which never enters the equation.
Hint 4/4
Both land at 17.3 m/s. The ball thrown upward takes longer, but it returns to the roof moving at 8.00 m/s downward and from there the two are identical.
Show solution
Energy is chosen because the two flights differ in everything except the quantities energy actually uses.
Write the conserved statement for either ball
$$\tfrac12 mv_0^{2} + mgh = \tfrac12 mv_f^{2}$$
the same two instants for both balls, with the zero of the store at the ground
$$v_f = \sqrt{v_0^{2} + 2gh}$$
the mass cancels and only the size of the launch speed survives, since it entered squared
the same arithmetic serves both balls, which is the content of the answer
$$v_f = 17.3\ \mathrm{m/s}\ \text{for both}$$
about 62 km/h, a believable landing speed from the third floor
Answer $$\boxed{\;v_f = 17.3\ \mathrm{m/s}\ \text{for both balls}\;}$$
Check
Check with kinematics on the upward ball: it rises $(8.00)^{2}/(2\times 9.80) = 3.27$ m to a height of 15.27 m, then falls the whole way, arriving at $\sqrt{2(9.80)(15.27)} = 17.3$ m/s. Same number by a route that took three times as long.
An energy equation cannot tell up from down. That is a limitation when you want a direction and a gift when the direction is a distraction.
2§09.2 — is a stored value a physical fact●●○○○
A student writes on an exam paper: a 2.0 kg book resting on a table 0.75 m above the floor has 14.7 J of gravitational potential energy. The marker is deciding whether the sentence is right as it stands.
Given
$m = 2.0\ \mathrm{kg}$
table top 0.75 m above the floor
$g = 9.80\ \mathrm{m/s^{2}}$
Find
(a) Decide whether the sentence is correct as written, and repair it if not.
Hint 1/4
The arithmetic in the sentence is fine. The question is whether the sentence contains everything it needs in order to mean something.
Hint 2/4
$U$ is defined only up to an additive constant: $\Delta U = -W_{\rm cons}$ fixes differences, not values, so a value requires a stated zero.
Hint 3/4
With the zero at the floor, $U = (2.0)(9.80)(0.75) = 14.7$ J; with the zero at the table top it is 0 J; with the zero at a 2.40 m ceiling it is $-32.3$ J.
Hint 4/4
The sentence is incomplete rather than wrong: it becomes correct the moment the words measured from the floor are added.
Show solution
Showing three frames is more convincing than quoting the rule, because the disagreement is the argument.
Show that the value moves and the difference does not
three legal frames, three different values for one book on one table
$$\Delta U = mg\,\Delta y = 5.88\ \mathrm{J}\ \text{in all three}$$
the quantity that appears in every answer is untouched by the choice
Answer $$\boxed{\;\text{incomplete: the zero level has to be stated}\;}$$
Check
Independent test: no experiment can measure the 14.7 J directly. What can be measured is the 5.88 J of work needed to raise the book by 0.30 m, or the speed it arrives at when it falls, and both are differences.
3§09.5 — what conservation of energy still claims●●○○○
A crate slides across a rough floor and comes to rest. A student says this proves that energy is not conserved when friction acts.
Given
the crate loses all of its kinetic energy
the floor is level, so no height changes
the surface is rough
Find
(a) Choose the statement that describes the situation correctly.
Hint 1/4
The disagreement is about the word energy, not about the crate. Ask which total the student is watching and whether it is the only total there is.
Hint 2/4
The rule is $W_{\rm NC} = \Delta E$: the mechanical total falls by the work of friction, and the energy that leaves it appears as thermal energy in the crate and the floor.
Hint 3/4
Here the mechanical total falls to zero, and the missing amount equals $f_k d$ exactly, with $d$ the distance the crate slid.
Hint 4/4
Mechanical energy is not conserved; energy is. The joules that left the mechanical account are in the warmed floor.
Show solution
Separating the two totals before arguing removes the disagreement entirely: both statements in the dispute are true of different totals.
Name the two totals separately
$$E_{\rm mech} = KE + U \;\to\; 0$$
the store never changes on a level floor, so the fall is the whole kinetic energy
$$E_{\rm mech,1} = E_{\rm mech,2} + f_k d$$
the missing amount is not an unknown: it is the friction force times the distance slid
Say where it went
$$E_{\rm thermal} = f_k d$$
in the crate and the floor, which are both very slightly warmer; measuring that rise is a thermodynamics question and not one for this course
Answer $$\boxed{\;\text{mechanical energy falls by } f_k d;\ \text{energy is conserved}\;}$$
Check
Test of the claim: a 20 kg crate at 3.0 m/s carries 90 J, and with $\mu_k = 0.30$ it slides $90/58.8 = 1.53$ m. The predicted thermal energy, $(58.8)(1.53) = 90$ J, is exactly the kinetic energy that disappeared.
4§09.7 — no change of speed, no work●●●○○
A cyclist rides up a long hill at a perfectly steady speed. A classmate argues that since the kinetic energy never changes, no work is being done on the bicycle and its rider.
Given
steady speed, so the kinetic energy is constant
the road climbs steadily
the cyclist keeps pedalling
Find
(a) Decide whether the classmate is right, and say what the pedalling is paying for.
Hint 1/4
Take the claim apart into two separate claims: that the net work is zero, and that no work is being done at all. Only one of them survives.
Hint 2/4
The work-energy principle says $W_{\rm net} = \Delta KE$, so a constant speed means the net work is zero; it says nothing about the individual works, which can be large and opposite.
Hint 3/4
Here gravity does negative work as the rider climbs, and the pedalling does positive work of the same size, plus whatever the resistances take.
Hint 4/4
The classmate is right about the net work and wrong about the rest: the cyclist is filling the gravitational store, which is why the effort is real.
Show solution
The confusion is entirely in the word work being used for two different quantities, so the solution names both before computing either.
Separate the net work from the individual works
$$W_{\rm net} = \Delta KE = 0$$
true, and it is the only thing a constant speed tells you
at least that much, and more once the resistances are paid
Answer $$\boxed{\;\text{net work zero, rider's work } \ge 73.5\ \mathrm{kJ}\;}$$
Check
Scale check: 73.5 kJ over a ten minute climb is 123 W of useful output, which is a realistic sustained figure for an amateur cyclist and a good sign the arithmetic is sound.
B · computation 8 questions
1§09.1 — two trails to the same summit●●○○○
A hiker of mass 65.0 kg climbs from a car park to a summit 400 m higher. One trail is 2.50 km long and gentle, the other is 1.20 km long and steep.
Given
$m = 65.0\ \mathrm{kg}$
vertical rise 400 m on both trails
trail lengths 2.50 km and 1.20 km
$g = 9.80\ \mathrm{m/s^{2}}$
Find
(a) Find the work done by gravity on the hiker along each trail.
(b) Find the gain in gravitational potential energy along each trail.
(c) State what would differ between the trails if the question had asked about the work done against friction in the hiker's boots.
Hint 1/4
Before computing anything, decide which of the two given lengths can possibly enter the answer, and why the other one is in the question at all.
Hint 2/4
Gravity is conservative, so its work depends only on the change in height: $W_{\rm gravity} = -mg\,\Delta y$, and $\Delta U = +mg\,\Delta y$.
Hint 3/4
The numbers that matter are $m = 65.0$ kg, $\Delta y = 400$ m and $g = 9.80$ m/s$^{2}$; the 2.50 km and the 1.20 km are there to be rejected.
Hint 4/4
Both trails give $-2.55\times10^{5}$ J of work by gravity and the same $+2.55\times10^{5}$ J of store; only a path dependent force would tell the two apart.
Show solution
Rejecting the trail lengths in the first line is the whole exercise; any solution that uses them has answered a different question.
Compute once for both trails
$$W_{\rm gravity} = -mg\,\Delta y = -(65.0)(9.80)(400)$$
path independence means one calculation covers both trails
Scale check: 255 kJ is about 61 kilocalories, roughly a quarter of a chocolate bar, for a 400 m climb. That the number is embarrassingly small is a real feature of human efficiency, not an arithmetic slip: most of the food energy a climber burns leaves as heat.
An exam question that gives you two lengths and one height is usually testing whether you know which of the three the store responds to.
2§09.2 — energy stored between two stretches●●○○○
A spring of stiffness 250 N/m is already stretched by 0.100 m from its natural length. It is then stretched further, to 0.200 m from the natural length.
Given
$k = 250\ \mathrm{N/m}$
initial stretch 0.100 m
final stretch 0.200 m
Find
(a) Find the energy stored at each stretch.
(b) Find the extra energy needed to go from the first stretch to the second.
(c) Compare that with the energy needed for the first 0.100 m.
Hint 1/4
The store is quadratic, so the answer to (b) is not the store at 0.100 m again. Decide what you have to subtract from what.
Hint 2/4
The elastic store is $U = \tfrac12 kx^{2}$ with $x$ measured from the natural length, and the extra energy is the difference of two such values.
Hint 3/4
With $k = 250$ N/m, the two stretches are 0.100 m and 0.200 m.
Hint 4/4
The stores are 1.25 J and 5.00 J, the step between them costs 3.75 J, which is three times the 1.25 J the first stretch cost.
Show solution
Computing both stores and subtracting is safer than trying to integrate the second stretch on its own, where the lower limit is easy to forget.
Independent check by the average force: over the second stretch the force runs from $(250)(0.100)=25.0$ N to $(250)(0.200)=50.0$ N, averaging 37.5 N over 0.100 m, which is 3.75 J. The average is legal here because the force is linear in the stretch.
Odd numbers: successive equal stretches cost 1, 3, 5, 7 units. That pattern is the signature of a quadratic store and a quick way to check an answer.
3§09.3 — a roller coaster car with no engine●●○○○
A roller coaster car is released from rest at the top of the first hill, 32.0 m above the lowest point of the ride. The track is smooth enough to ignore friction.
Given
released from rest at 32.0 m
smooth track
$g = 9.80\ \mathrm{m/s^{2}}$
Find
(a) Find its speed at a point 12.0 m above the lowest point.
(b) Find the height at which its speed is 15.0 m/s.
(c) State whether the mass of the loaded car matters to either answer.
Hint 1/4
Both parts are the same equation solved for different unknowns, so set the equation up once in symbols with the height and the speed both left as letters.
Hint 2/4
Conservation on a smooth track gives $mgh_0 = \tfrac12 mv^{2} + mgh$, and the mass divides out of every term.
Hint 3/4
Here $h_0 = 32.0$ m, and the two cases are $h = 12.0$ m in (a) and $v = 15.0$ m/s in (b), with $g = 9.80$ m/s$^{2}$.
Hint 4/4
The speed at 12.0 m is 19.8 m/s and the height where the speed is 15.0 m/s is 20.5 m; the mass cancels in both.
Show solution
Doing the algebra before the arithmetic means one rearrangement serves both parts; substituting first would have meant solving a quadratic twice.
the same relation rearranged, so no new physics is involved
$$h = 20.5\ \mathrm{m}$$
higher than the 12.0 m point, as it must be for a lower speed
Answer $$\boxed{\;v = 19.8\ \mathrm{m/s},\qquad h = 20.5\ \mathrm{m}\;}$$
Check
Consistency between the two parts: the answer to (b), 20.5 m, lies between the release height and the 12.0 m point, and the corresponding speed, 15.0 m/s, lies between 0 and 19.8 m/s. Both orderings are what the conserved statement demands.
On a smooth track only differences of height ever appear, so you may put the zero anywhere, including at the point you are asking about.
4§09.4 — reading speeds off a landscape●●●○○
A 0.250 kg particle moves along a straight line in a region where the stored energy depends on position as given in the table. Its total mechanical energy is 8.00 J and no friction acts.
Given
$m = 0.250\ \mathrm{kg}$
$E = 8.00\ \mathrm{J}$, constant
stored energy: 2.00 J at $x=1$ m, 5.00 J at $x=2$ m, 8.00 J at $x=3$ m, 11.0 J at $x=4$ m
no friction
Find
(a) Find the kinetic energy and the speed at $x = 2$ m.
(b) Say where the particle turns round, and why it never reaches $x = 4$ m.
(c) State at which of the listed positions the particle moves fastest.
Hint 1/4
Every part of this question is the same subtraction repeated. Decide what is being subtracted from what before touching the table.
Hint 2/4
The rule is $KE = E - U$, and $KE$ can never be negative, so any position where the table shows $U > E$ is simply not available to the particle.
Hint 3/4
Here $E = 8.00$ J throughout and the table gives $U$ as 2.00, 5.00, 8.00 and 11.0 J at $x = 1, 2, 3$ and 4 m.
Hint 4/4
At $x=2$ m the kinetic energy is 3.00 J and the speed is 4.90 m/s; the particle turns round at $x=3$ m and is fastest at $x=1$ m.
Show solution
Computing the whole column of kinetic energies first answers all three parts at once and makes the impossible row visible immediately.
Do the subtraction at every listed position
$$KE(1) = 8.00-2.00 = 6.00\ \mathrm{J}$$
largest gap, so this is where the particle is fastest
$$KE(2) = 8.00-5.00 = 3.00\ \mathrm{J}$$
the position the question asks about
$$KE(3) = 8.00-8.00 = 0$$
the gap has closed: this is the turning point by definition
$$KE(4) = 8.00-11.0 = -3.00\ \mathrm{J}$$
impossible, so the particle never gets there and the table row is a trap
Turn the two kinetic energies that matter into speeds
twice the kinetic energy is only $\sqrt2$ times the speed, which is worth noticing
Answer $$\boxed{\;v(2) = 4.90\ \mathrm{m/s};\ \text{turns at } x=3\ \mathrm{m};\ \text{fastest at } x=1\ \mathrm{m}\;}$$
Check
Structural check: the speeds fall as the store rises, monotonically, which is what a rising curve under a flat line must produce. If any speed had come out larger where the store was larger, a subtraction would have been reversed.
A negative kinetic energy in a table is not an error in your arithmetic; it is the landscape telling you the particle stopped earlier.
5§09.5 — a car skidding to a stop●●●○○
A 1200 kg car travelling at 25.0 m/s brakes so hard that all four wheels lock and it skids to rest on a level road where the coefficient of kinetic friction is 0.700.
Given
$m = 1200\ \mathrm{kg}$
$v = 25.0\ \mathrm{m/s}$
$\mu_k = 0.700$ on a level road
$g = 9.80\ \mathrm{m/s^{2}}$
Find
(a) Find the skid distance.
(b) Find the thermal energy produced.
(c) State what both answers would become at 50.0 m/s.
Hint 1/4
The car ends at rest on level ground, so one whole side of the ledger is empty. Decide which terms survive before writing anything.
Hint 2/4
On a level road the store never changes, so $\tfrac12 mv^{2} = f_k d$ with $f_k = \mu_k mg$; the thermal energy produced is that same $f_k d$.
Hint 3/4
The numbers are $m = 1200$ kg, $v = 25.0$ m/s, $\mu_k = 0.700$ and $g = 9.80$ m/s$^{2}$.
Hint 4/4
The skid is 45.6 m and the thermal energy is $3.75\times10^{5}$ J; at twice the speed both become four times as large, 182 m and $1.50\times10^{6}$ J.
Show solution
Keeping $f_k = \mu_k mg$ symbolic until the last step is what makes the cancellation of the mass visible, and that cancellation is the part worth remembering.
Write the ledger with the empty terms deleted
$$\tfrac12 mv^{2} = f_k d$$
level road, ends at rest: no store change and no kinetic energy left at the end
Independent check with kinematics: the deceleration is $\mu_k g = 6.86$ m/s$^{2}$ and $v^{2}=2ad$ gives $d = 625/13.72 = 45.6$ m. Order of magnitude on the energy: 375 kJ would heat about a kilogram of steel by several hundred degrees, which is why brake discs glow.
Every skid, every stopping distance and every runaway lane obeys $d = v^{2}/(2\mu_k g)$. Learn it as a formula and you will recognise it in three different disguises.
6§09.5 — across a rough patch and into a spring●●●○○
A 2.00 kg block is sliding at 3.00 m/s along a horizontal surface. It crosses a rough patch 1.50 m long where $\mu_k = 0.150$, and then, on smooth floor, it meets a spring of stiffness 800 N/m.
Given
$m = 2.00\ \mathrm{kg}$, arriving at $3.00\ \mathrm{m/s}$
rough patch 1.50 m long with $\mu_k = 0.150$
$k = 800\ \mathrm{N/m}$ on smooth floor
$g = 9.80\ \mathrm{m/s^{2}}$
Find
(a) Find the speed with which the block reaches the spring.
(b) Find the maximum compression of the spring.
(c) Say whether the block will make it back across the rough patch.
Hint 1/4
The journey has three stages and only two of them cost anything. Identify the stage where energy leaves the account and the stage where it merely changes form.
Hint 2/4
The ledger is $\tfrac12 mv_1^{2} = \tfrac12 mv_2^{2} + f_k d$ across the patch, with $f_k = \mu_k mg$, and then $\tfrac12 mv_2^{2} = \tfrac12 kx^{2}$ at the spring.
Hint 3/4
The numbers: 2.00 kg at 3.00 m/s, patch 1.50 m with $\mu_k = 0.150$, spring 800 N/m, all on level ground so no height enters.
Hint 4/4
The block reaches the spring at 2.14 m/s, compresses it 0.107 m, and comes back with 4.59 J, which is more than the 4.41 J the patch charges, so it just gets across.
Show solution
Working in joules from start to finish avoids three separate square roots; only the two speeds actually asked for are ever extracted.
it emerges, barely; a patch 1.56 m long would have stopped it inside
Answer $$\boxed{\;v_2 = 2.14\ \mathrm{m/s},\quad x = 0.107\ \mathrm{m},\quad \text{returns at } 0.42\ \mathrm{m/s}\;}$$
Check
Whole-account check: the block began with 9.00 J and the patch was crossed twice at 4.41 J each, which is 8.82 J, leaving 0.18 J. That matches the final kinetic energy exactly, and it also shows the block could not have crossed a third time.
Count the crossings before you start. Total energy divided by the cost per crossing tells you how many the body can afford.
7§09.6 — reaching an altitude of one Earth radius●●●●○
A projectile is to be fired vertically from the surface of the Earth so that it just reaches an altitude equal to one Earth radius, arriving there with zero speed. Air resistance is ignored.
Structural check: the answer must lie below the escape speed, since reaching a finite height is easier than never coming back, and it does. The factor $1/\sqrt2$ also follows from the algebra directly: half the energy is $1/\sqrt2$ of the speed.
Doubling the distance from the centre costs half the escape energy. The last half buys the whole of the rest of the universe.
8§09.7 — a pump lifting water●●○○○
A pump raises 250 kg of water through a vertical height of 15.0 m in 60.0 s, delivering it at a negligible speed.
Given
mass lifted $250\ \mathrm{kg}$
vertical height $15.0\ \mathrm{m}$
time $60.0\ \mathrm{s}$
water arrives with negligible speed
$g = 9.80\ \mathrm{m/s^{2}}$
Find
(a) Find the useful work done.
(b) Find the average useful power.
(c) Express the energy in kilowatt hours.
Hint 1/4
The water arrives with negligible speed, which means none of the work went into kinetic energy. Decide where all of it went before writing an equation.
Hint 2/4
The useful work is the gain in the gravitational store, $W = mgh$, the average power is $\bar P = W/t$, and $1\ \mathrm{kWh} = 3.6\times10^{6}$ J.
Hint 3/4
The numbers are 250 kg, 15.0 m, 60.0 s and $g = 9.80$ m/s$^{2}$.
Hint 4/4
The work is $3.68\times10^{4}$ J, the average power is 613 W, and the energy is 0.0102 kWh.
Show solution
Working in joules and converting only at the end keeps the unit change away from the physics, where it causes most of the errors.
the same energy in a unit chosen for electricity bills
Answer $$\boxed{\;W = 3.68\times10^{4}\ \mathrm{J},\quad \bar P = 613\ \mathrm{W},\quad E = 1.02\times10^{-2}\ \mathrm{kWh}\;}$$
Check
Independent check on the power by the other formula: the water rises at $15.0/60.0 = 0.250$ m/s and the pump pushes with a force equal to the weight of the column it is lifting, on average $2450$ N, giving $P = (2450)(0.250) = 613$ W.
A kilowatt hour is 3.6 MJ. Memorise that one number and every energy bill in your life becomes a physics problem.
C · exam level 5 questions
1§09.3 — a ball inside a vertical loop●●●●○
A small ball is released from rest on a smooth track that leads into a vertical circular loop of radius 0.800 m. Exam format: four parts, and the last two depend on the first two.
Given
loop radius $R = 0.800\ \mathrm{m}$
the whole track is smooth
the ball is released from rest at a height $h$ above the bottom of the loop
$g = 9.80\ \mathrm{m/s^{2}}$
Find
(a) Find the least speed the ball can have at the top of the loop while staying on the track.
(b) Find the least release height $h$ for which the ball completes the loop.
(c) Find the speed at the bottom of the loop for that release height.
(d) Decide whether a release from 1.80 m completes the loop.
Hint 1/4
Two different pieces of physics are needed and they are needed in a fixed order: one of them decides what must be true at the top, and only then can energy carry that condition down to the release point.
Hint 2/4
At the top, gravity alone can supply the centripetal requirement in the limiting case: $mg = mv_{\rm top}^{2}/R$. Then conservation on a smooth track: $mgh = mg(2R) + \tfrac12 mv_{\rm top}^{2}$.
Hint 3/4
The numbers are $R = 0.800$ m and $g = 9.80$ m/s$^{2}$, with the top of the loop at a height $2R = 1.60$ m above the bottom.
Hint 4/4
The least speed at the top is 2.80 m/s, the least release height is 2.00 m, which is $2.5R$, the bottom speed is then 6.26 m/s, and a release from 1.80 m does not complete the loop.
Show solution
Energy alone cannot answer (a), because it never mentions the track's push; forces alone cannot answer (b), because the speed at the top has to be brought down to the release point. The order is forced.
(a) The condition at the top is a force condition, not an energy one
$$mg + N = \frac{mv_{\rm top}^{2}}{R}$$
at the top both the weight and the track push towards the centre, which is downwards there
Independent check on (b) by a ratio that contains no numbers: the requirement is always $h = 2.5R$, so doubling the radius doubles the release height. With $R = 0.800$ m that gives 2.00 m, and the arithmetic above agrees. Check on (d) by size: 1.80 m is below 2.00 m, so the answer had to be no.
Any loop question splits into exactly these two halves. Do the top first, always, because it is the half that produces the condition everything else has to satisfy.
2§09.4 — the same track, a lower release point●●●○○
A 0.400 kg cart is released from rest on the smooth track of this section, but this time from a point 10.0 m above the lowest part rather than 12.0 m. The track runs down to a valley at 2.0 m, over a summit at 9.0 m, down to a second valley at 0 m and then up a final rise towards 14.0 m.
Given
$m = 0.400\ \mathrm{kg}$, released from rest at 10.0 m
track heights: 2.0 m, 9.0 m, 0 m, then rising to 14.0 m
the track is smooth
$g = 9.80\ \mathrm{m/s^{2}}$
Find
(a) Find the total mechanical energy, taking the store as zero at the lowest point.
(b) Find the speed at the 9.0 m summit.
(c) Find the speed in the second valley.
(d) State where the cart finally turns round, and the least release height that would carry it over the far summit.
Hint 1/4
Fix the total once and then treat every part as the same subtraction. Only one part needs anything more than that.
Hint 2/4
The rules are $E = mgy_0$ from rest, $KE = E - mgy$ at any point, and a turning point wherever $mgy = E$.
Hint 3/4
The numbers: $m = 0.400$ kg, release at 10.0 m, points of interest at 9.0 m, 0 m and a far summit at 14.0 m, with $g = 9.80$ m/s$^{2}$.
Hint 4/4
The total is 39.2 J, the speeds are 4.43 m/s and 14.0 m/s, the cart turns round at 10.0 m on the final rise, and it would have to be released above 14.0 m to clear the far summit.
Show solution
Fixing the total first is what allows four parts to share one calculation; each part re-derived from scratch would take four times as long and would risk four different totals.
Comparison check against the worked example in the landscape block, where the release was from 12.0 m: every speed there was larger, and the summit speed was 7.67 m/s against 4.43 m/s here. Lowering the release by 2.0 m must slow the cart everywhere, and it does.
A summit the body barely clears is the dangerous case: at 4.43 m/s the cart is still moving, but a release from 9.0 m exactly would have left it stuck there.
3§09.5 — a block fired up a rough incline●●●●○
A 1.50 kg block is given a speed of 6.00 m/s straight up an incline of $20.0^{\circ}$ where the coefficient of kinetic friction is 0.250. Exam format: four parts.
(a) Find how far up the incline the block travels.
(b) Find the thermal energy produced on the way up.
(c) Show that the block slides back down, and find its speed on returning to the starting point.
(d) Find the fraction of the original kinetic energy that survives the round trip.
Hint 1/4
Going up and coming down are two different problems with the same geometry, because friction changes direction while gravity does not. Set them up separately before substituting anything.
Hint 2/4
Up: $\tfrac12 mv_0^{2} = mgd\sin\theta + \mu_k mg\cos\theta\, d$. Down: $mgd\sin\theta - \mu_k mg\cos\theta\, d = \tfrac12 mv_f^{2}$. The block slides back only if $\tan\theta > \mu_k$.
Hint 3/4
The numbers: 1.50 kg at 6.00 m/s, $\theta = 20.0^{\circ}$ with sine 0.342 and cosine 0.940, and $\mu_k = 0.250$.
Hint 4/4
The block travels 3.18 m up, produces 11.0 J of thermal energy on the way, returns at 2.59 m/s, and keeps 18.6% of its original kinetic energy.
Show solution
Writing the two directions as separate equations is what keeps the sign of the friction term honest; a single equation with a sign to be decided later is where this question is usually lost.
(a) Going up: both costs act in the same direction
Whole-trip check: the round trip produces $2\times 11.0 = 22.0$ J of thermal energy and the block began with 27.0 J, leaving 5.0 J, which matches $\tfrac12(1.50)(2.585)^{2} = 5.01$ J. The store is back where it started, as it must be after a round trip on a conservative force.
The surviving fraction on a round trip is $(\tan\theta-\mu_k)/(\tan\theta+\mu_k)$ for any incline, which gives 0.186 here and explains why a shallow rough slope destroys almost everything.
4§09.6 — a probe fired at 9.0 km/s●●●●○
A probe is fired vertically from the surface of the Earth at $9.00\times10^{3}$ m/s. Air resistance and the rotation of the Earth are ignored. Exam format: four parts.
(a) Find the total mechanical energy per kilogram at launch.
(b) Find the greatest distance from the centre of the Earth that the probe reaches.
(c) Find the altitude above the surface at that point.
(d) Find the speed at which it passes a distance of two Earth radii from the centre.
Hint 1/4
Everything in this question comes from one number, computed once. Decide what that number is and get it before doing anything else.
Hint 2/4
Per kilogram, $E = \tfrac12 v^{2} - GM/r$, and it is the same at every point of the flight; a turning point is where $\tfrac12 v^{2} = 0$.
Hint 3/4
The numbers: $v_0 = 9.00\times10^{3}$ m/s, $GM = 3.98\times10^{14}$, $R_E = 6.38\times10^{6}$ m, and the point in part (d) is at $r = 1.276\times10^{7}$ m.
Hint 4/4
The total is $-21.9$ MJ/kg, the greatest distance is $1.82\times10^{7}$ m, the altitude is $1.18\times10^{7}$ m, and the speed at two radii is $4.31\times10^{3}$ m/s.
Show solution
Working per kilogram removes an unknown mass that would have cancelled anyway, and working in megajoules keeps every number between 9 and 63, where a slip is visible.
Consistency check between (b) and (d): the point in part (d) is at $1.276\times10^{7}$ m, which is closer than the turning distance of $1.82\times10^{7}$ m, so the probe must still be moving there, and it is. Check against the escape speed: 9.00 km/s is below 11.2 km/s, so a finite turning point had to exist.
With the far-field store, every question is the same: compute the total once, then set the kinetic part to whatever the question wants and solve for the remaining unknown.
5§09.7 — a lift with and without a counterweight●●●○○
A lift cabin of total mass 1200 kg rises 24.0 m in 20.0 s at a steady speed. Exam format: four parts, and the last one is where the engineering is.
Given
total mass $1200\ \mathrm{kg}$
rise $24.0\ \mathrm{m}$ in $20.0\ \mathrm{s}$
steady speed throughout
counterweight of 900 kg in part (d)
$g = 9.80\ \mathrm{m/s^{2}}$
Find
(a) Find the work done against gravity.
(b) Find the average power delivered by the motor, ignoring friction.
(c) Express the energy used in kilowatt hours.
(d) Find the average power needed if a 900 kg counterweight descends by the same 24.0 m as the cabin rises.
Hint 1/4
Three of the four parts are one calculation and a unit change. The fourth asks what the motor actually has to supply, which is not the same as what the cabin gains.
Hint 2/4
The rules are $W = mgh$, $\bar P = W/t$, $1\ \mathrm{kWh} = 3.6\times10^{6}\ \mathrm{J}$, and with a counterweight the net store change is $(m_{\rm cabin}-m_{\rm cw})gh$.
Hint 3/4
The numbers: 1200 kg through 24.0 m in 20.0 s, steady speed $24.0/20.0 = 1.20$ m/s, and a 900 kg counterweight in part (d).
Hint 4/4
The work is $2.82\times10^{5}$ J, the average power is 14.1 kW, the energy is 0.0784 kWh, and with the counterweight the power falls to 3.53 kW.
Show solution
Parts (a) to (c) share one number, so it is computed once at full precision and reused; part (d) is deliberately not a new energy problem but the same one with a different net mass.
Independent check on (d) by forces: with the counterweight the net force the motor must overcome is $(300)(9.80) = 2940$ N, and at 1.20 m/s that is $P = (2940)(1.20) = 3528$ W, matching. Ratio check: 300/1200 is a quarter, and 3.53/14.1 is a quarter.
A counterweight does not make the journey cheaper by magic: it stores energy on the way down and returns it on the way up, which is exactly what a conservative arrangement does.
D · interleaved 4 questions
1§09.5 — a lorry on a runaway ramp●●●○○
A lorry of mass $1.20\times10^{4}$ kg loses its brakes and enters a gravel escape ramp at 30.0 m/s. The ramp climbs at $10.0^{\circ}$ and the gravel gives an effective coefficient of friction of 0.500.
(a) Find the length of ramp needed to stop the lorry.
(b) Find the share of the kinetic energy that goes into the gravel rather than into height.
Hint 1/4
Decide first which tools this situation allows. There is a slope, a rough surface, a known speed and a wanted distance, and no time is mentioned anywhere.
Hint 2/4
The ledger is $\tfrac12 mv^{2} = mgd\sin\theta + \mu_k mg\cos\theta\, d$, with $d$ measured along the ramp and the normal force $mg\cos\theta$.
Hint 3/4
The numbers: 30.0 m/s, $\theta = 10.0^{\circ}$ with sine 0.1736 and cosine 0.9848, $\mu_k = 0.500$, and the mass which will cancel.
Hint 4/4
The ramp must be 68.9 m long, and the gravel takes 74% of the energy while the climb takes 26%.
Show solution
Energy rather than forces, because the question asks for a distance and gives a speed; the force route would need the acceleration first and then a kinematic formula, which is two steps rather than one.
the two costs are both proportional to $d$, so their shares are the ratio of their coefficients and no arithmetic on the total is needed
$$\text{gravel share} = 0.739$$
of the $\tfrac12 mv^{2} = 5.40\times10^{6}$ J, so about $4.0\times10^{6}$ J goes into the gravel
Answer $$\boxed{\;d = 68.9\ \mathrm{m},\quad 74\%\ \text{into the gravel}\;}$$
Check
Check by removing the gravel: a smooth $10.0^{\circ}$ ramp would need $900/[19.6(0.1736)] = 265$ m, nearly four times as long, which is why the gravel is there. Check the size: 5.40 MJ is about the energy of 1.5 kWh, and it appears as heat in a few seconds.
When two costs are both proportional to the same distance, their shares of the budget are fixed before you know the budget.
2§09.3 — a ball on a string, released from horizontal●●●●○
A 0.400 kg ball on a light string of length 0.900 m is held with the string horizontal and released from rest. It swings down to the lowest point of its arc.
Given
$m = 0.400\ \mathrm{kg}$
$L = 0.900\ \mathrm{m}$
released from rest with the string horizontal
$g = 9.80\ \mathrm{m/s^{2}}$
Find
(a) Find the speed of the ball at the lowest point.
(b) Find the tension in the string there.
(c) State the ratio of that tension to the weight of the ball.
Hint 1/4
Two different pieces of physics are needed, in a fixed order: one gets the speed at the bottom, the other converts that speed into a force. Neither can do both.
Hint 2/4
Energy first: $mgL = \tfrac12 mv^{2}$ for a drop equal to the length of the string. Then a force equation at the lowest point, where the motion is circular: $T - mg = mv^{2}/L$.
Hint 3/4
The numbers: $m = 0.400$ kg, $L = 0.900$ m, $g = 9.80$ m/s$^{2}$, and the drop from horizontal to lowest point is the full 0.900 m.
Hint 4/4
The speed is 4.20 m/s, the tension is 11.8 N, and the ratio to the weight is exactly 3.
Show solution
Energy cannot produce a tension because the tension does no work and never appears in the energy statement; forces cannot produce the speed without integrating a varying tangential force. Each tool does the half it can.
(a) Energy for the speed
$$mgL = \tfrac12 mv^{2} \;\Rightarrow\; v = \sqrt{2gL}$$
the string tension does no work at any point of the swing because it is always perpendicular to the motion
Independent check on the ratio by a limiting case: release the ball from just below the pivot, an almost zero drop, and the formula gives $T \to mg$, which is right for a ball hanging at rest. The ratio rises with the drop and reaches exactly 3 for a quarter turn.
Whenever a question asks for a tension or a normal force at a point on a curved path, expect this two-step shape: energy for the speed, then a circular motion equation at that point.
3§09.6 — escape from a smaller planet●●●○○
A planet has a surface gravitational field strength of 3.70 m/s$^{2}$ and a radius of $3.39\times10^{6}$ m. A probe is to be launched from its surface.
Given
surface field strength $g_p = 3.70\ \mathrm{m/s^{2}}$
radius $R_p = 3.39\times10^{6}\ \mathrm{m}$
no atmosphere worth mentioning
Find
(a) Find the escape speed from the surface.
(b) Compare it with the 11.2 km/s needed to escape from the Earth.
(c) State what would happen to the answer if the planet had the same radius but twice the mass.
Hint 1/4
The escape speed formula is written with $GM$ in it and the question gives you $g_p$ instead. Find the relation between those two before doing anything else.
Hint 2/4
At the surface $g_p = GM/R_p^{2}$, so $GM = g_p R_p^{2}$, and substituting into $v_{\rm esc} = \sqrt{2GM/R_p}$ gives $v_{\rm esc} = \sqrt{2g_p R_p}$.
Hint 3/4
The numbers: $g_p = 3.70$ m/s$^{2}$ and $R_p = 3.39\times10^{6}$ m, against the Earth's 11.2 km/s.
Hint 4/4
The escape speed is 5.01 km/s, which is 45% of the Earth's, and doubling the mass at fixed radius would multiply it by $\sqrt2$ to 7.08 km/s.
Show solution
Converting $g_p$ into $GM$ first is shorter than trying to find the planet's mass, which the question never asks for and which would need $G$ as an extra number.
Convert the given data into the grouping the formula needs
Independent check by the Earth: putting $g = 9.80$ m/s$^{2}$ and $R = 6.38\times10^{6}$ m into the same formula gives $\sqrt{2(9.80)(6.38\times10^{6})} = 11.2$ km/s, the known value, so the form $\sqrt{2g R}$ is sound.
$v_{\rm esc} = \sqrt{2gR}$ is the version to carry into an exam: surface gravity and radius are what questions usually give you.
4§09.3 — a ball thrown at an angle●●●○○
A ball is thrown from ground level at 20.0 m/s at $35.0^{\circ}$ above the horizontal. Air resistance is negligible.
Given
launch speed $20.0\ \mathrm{m/s}$ at $35.0^{\circ}$
(a) Find the speed of the ball at the highest point of its flight.
(b) Find the greatest height it reaches, using energy.
(c) Explain why the answer to (a) is not zero, although the ball is at the top.
Hint 1/4
A body at the top of its flight is not at rest, and the reason is a fact about one of its two velocity components. Settle that before writing any energy line.
Hint 2/4
At the top the vertical component is zero and the horizontal component is unchanged, since nothing acts horizontally: $v_{\rm top} = v_0\cos\theta$. Then $\tfrac12 mv_0^{2} = \tfrac12 mv_{\rm top}^{2} + mgh$.
Hint 3/4
The numbers: $v_0 = 20.0$ m/s, $\theta = 35.0^{\circ}$ with cosine 0.8192, and $g = 9.80$ m/s$^{2}$.
Hint 4/4
The speed at the top is 16.4 m/s and the greatest height is 6.71 m; the speed is not zero because only the vertical part of the velocity ever changes.
Show solution
Energy handles the whole speed at once, which is why the horizontal part has to be removed by hand; the component method separates them from the start and is shorter here.
the mass cancels, and the subtraction is where the leftover horizontal motion is accounted for
$$h = 6.71\ \mathrm{m}$$
about the height of a two storey building
Answer $$\boxed{\;v_{\rm top} = 16.4\ \mathrm{m/s},\quad h = 6.71\ \mathrm{m}\;}$$
Check
Independent check with the vertical component alone: $v_{0y} = (20.0)(0.5736) = 11.47$ m/s and $h = v_{0y}^{2}/(2g) = 131.6/19.6 = 6.71$ m. The two routes agree, and the second one shows why the energy version needed the subtraction: the horizontal part of the kinetic energy never took part.
At the top of any projectile flight the body still carries $\tfrac12 mv_x^{2}$ of kinetic energy. Forgetting it is the classic way to overestimate the height.
Mistake ledger (16 entries)
⚠ Calling a round trip free because the body came back
Gravity really does return what it took on a round trip, and the habit spreads to friction, where the two legs add instead of cancelling.
wrong$$W_{\rm friction}(\text{out and back}) = -36.8 + 36.8 = 0$$
⚠ Buried in the ladder, step 2: the slope length used as a height
The length along the ramp is the unknown being solved for and the sine has to be remembered rather than read off the page, so it is the first thing to be dropped under exam pressure.
wrong$$4.50 = mgd + f_kd$$
right$$4.50 = mgd\sin\theta + f_kd$$
⚠ Buried in the ladder, step 3: the normal force taken as the full weight
The formula $f_k=\mu_k mg$ is the one first met on a horizontal floor, where it is correct, and the cosine arrives only later.
no atmosphere, no rotation; independent of the mass launched
Average and instantaneous power
$$\bar P = \frac{W}{t},\qquad P = Fv$$
the force along the velocity in the second form; watts in both
The kilowatt hour
$$1\ \mathrm{kWh} = 3.6\times10^{6}\ \mathrm{J}$$
an energy, not a power
Check yourself
Close the page and write, from memory: the test that decides whether a force can have a stored energy, the two stores this section uses with their formulas, the equation that holds when every working force is conservative, the extra term that appears when a surface is rough and what multiplies what inside it, and the two forms of power. Then open the formula card and mark what you missed rather than what you got.
Given a force, decide whether it is conservative, and say what the work it does around a closed trip has to be?
c-path-independence
Compute a gravitational and an elastic store, state the zero you used, and show that a change in the store does not care which zero you chose?
c-potential-energy
Find a speed on a smooth track of any shape in one line, and name the word in the question that allowed it?
c-mechanical-energy
Take a store curve and a total, mark the turning points, shade the region the body cannot enter, and say which way the force points?
c-energy-landscape
Write the ledger for a rough slope, get the friction force from the right normal force, and use the right distance in the friction term?
c-friction-ledger
Use the $-GMm/r$ store to find a turning distance, and derive an escape speed from surface gravity and radius?
c-far-from-earth
Get an average power from work and time, an instantaneous one from force and speed, and convert a kilowatt hour into joules?
c-power
Glossary (19 terms)
conservative forcekorunumlu kuvvet
A force whose work between two points is the same along every route, equivalently one that does no work on a closed trip. Gravity and the force of an ideal spring are the two met in this course.
nonconservative forcekorunumsuz kuvvet
A force whose work depends on the route taken, so no stored energy can be attached to it. Kinetic friction and air drag are the examples here, and an external push is another.
path independenceyoldan bağımsızlık
The property that the work done between two points does not depend on which route was taken between them. It is what allows a force's work to be tabulated against position once and reused.
potential energypotansiyel enerji
Energy attached to the position of a body or the state of a spring, defined by the change rule that the store rises by whatever work the conservative force gives up.
The store held by a deformed spring, equal to one half the spring constant times the square of the deformation measured from the natural length.
referans düzeyi
The height chosen to be the zero of the gravitational store. It is free to choose, must be stated, and must be used on both sides of any energy equation.
mechanical energymekanik enerji
The sum of the kinetic energy of a body and the potential energy of its configuration. It is the total that stays constant when only conservative forces do work.
conservation of mechanical energymekanik enerjinin korunumu
The statement that the sum of kinetic and potential energy is the same at two instants, valid when every force doing work is conservative.
kapalı yol
A route that finishes where it started. The work done by a conservative force around one is zero, and this is the sharpest test for whether a force is conservative.
turning pointdönüm noktası
A position where the kinetic energy has fallen to zero because the store has risen to meet the total, so the body stops and reverses.
potential energy curvepotansiyel enerji eğrisi
A graph of the store against position, with the total drawn as a horizontal line. The vertical gap between them is the kinetic energy available at that position.
thermal energyısıl enerji
The energy left in two surfaces after they have rubbed against each other, equal to the friction force times the distance slid. It is where the mechanical total goes when it falls.
dissipative force
A nonconservative force that always takes energy out of the mechanical account, never returning it, such as kinetic friction or air drag.
escape speedkaçış hızı
The launch speed for which the total mechanical energy is exactly zero, so the body arrives infinitely far away with nothing left over. It depends on the planet and not on what is launched.
boundbağlı
Said of a body whose total mechanical energy is negative in the frame where the store vanishes at infinity, so a turning point exists at a finite distance and it must come back.
powergüç
The rate at which work is done or energy is delivered, measured in watts, and equal to the force along the motion times the speed at that instant.
wattvat
The unit of power, one joule per second. A person climbing stairs briskly delivers a few hundred of them.
kilowatt hourkilovatsaat
A unit of energy, not of power, equal to a kilowatt sustained for an hour, which is $3.6\times10^{6}$ joules.
What comes next
§10 · Review and consolidation II: applying Newton's laws, gravitation, work and energy as one decision
Three sections have now given you three different ways to attack the same picture of a block on a ramp: a force equation, a work calculation and an energy ledger. The next one puts them side by side on purpose, with questions written so that choosing the wrong one of the three costs you twenty minutes rather than a wrong answer, which is exactly what happens in an exam.
Sources
D. Giancoli, Physics for Scientists and Engineers with Modern Physics, 5th edition The set textbook. Its chapter on the conservation of energy covers the same ground as this section, in the same order, and its end-of-chapter problems are the right next step once this practice set feels comfortable.
Course syllabus, week 9 line and assessment table The scope comes from the week line, which reads Conservation of Energy and quotes no chapter numbers, so no chapter number is quoted here either. The weightings on the summary card come from the assessment table and nothing beyond them is claimed.
SI units: the joule, the watt and the kilowatt hour Every store, every kinetic energy and every thermal amount in this section is in joules, which is what makes them addable in one line. Power is in watts, one joule per second, and the kilowatt hour is an energy equal to 3.6 megajoules.