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10Review and consolidation II: applying Newton's laws, gravitation, work and energy as one decision
A 28.0 kg child comes down a playground slide 2.50 m high and leaves the bottom at 4.20 m/s. The paper asks how much energy the slide swallowed. You know how to resolve forces on a slope and you know how to write down kinetic energy, and you still lose ten minutes, because the slide is curved, its angle changes the whole way down, and every equation you were taught for a slope assumed one fixed angle.
By the end of this section you can look at any problem built out of the last five weeks, decide in one written line whether it is a force problem or an energy problem, carry it to a number with units, and defend that number with a check that does not repeat the calculation that produced it.
In 60 seconds
Weeks six to nine gave you two complete ways to get a speed: follow the forces instant by instant, or count the energy at two chosen instants. Almost every mark lost in this half of the course is lost by picking the expensive route, not by algebra.
a force, a tension, a normal force or an acceleration is wanted at one instant, or a body moves on a circle so that the acceleration points at the centre
The between two instants
$$K_i + U_i + W_{\text{other}} = K_f + U_f$$
two instants are described by position and speed, the force may vary or the path may curve, and no time is asked for
near the surface with a declared zero level, far from a planet with the zero at infinity, and for a spring measured from its natural length
Sliding friction in an energy line
$$W_{\text{fric}} = -f_k d = -\mu_k F_N d$$
a body slides over a distance d along a real surface; d is the path length travelled, never the height dropped
Power, average and instantaneous
$$\bar{P} = \frac{W}{\Delta t},\qquad P = F v\cos\theta$$
the question says per second, or names a motor, a pump, a cyclist or an engine
Three most common mistakes
Multiplying the friction force by the height dropped instead of by the distance slid. Friction is paid along the path; a 3.00 m slide that drops 1.50 m loses friction energy over 3.00 m, and using the height halves the loss without looking wrong.
Adding a centripetal force to a free body diagram. Nothing pushes a car around a bend except the forces already drawn; the circular motion tells you what their sum has to equal, not that another arrow exists.
Using mgy for at satellite heights. That form assumes the field strength does not change over the climb, which fails once the height is an appreciable fraction of an Earth radius, and satellite questions are exactly that case.
The published weights are Midterm 1 20 per cent, Midterm 2 20 per cent, quizzes 10 per cent, homework 5 per cent, the final 25 per cent and the laboratory 20 per cent. This week's line is a catch up and review week, so no new material is being introduced; what is being tested is whether the five weeks behind you can be used together, and whether you can choose between them.
How much time do you have?
10 minutes
The single decision that decides how long a problem takes, and the one line of energy bookkeeping that most of this half of the course reduces to. If you read nothing else, these two stop you from starting a curved track problem with a constant acceleration formula.
The 60-second card · Two routes to the same number, and the line that picks one · The ledger · Formula card
45 minutes
Everything that turns into a number on a script: choosing the route, writing the without inventing a force, running the energy ledger with and without friction, and the two forms of gravitation with the rule for which one is legal at which height.
The 60-second card · Two routes to the same number, and the line that picks one · When the acceleration points sideways · The ledger · Gravitation read twice · method boxes · Scaffolding comes off · practice B
full read
Adds what a formula sheet cannot carry: why the zero of potential energy can be put anywhere without changing an answer, why a satellite pushed to a higher orbit ends up slower while absorbing energy, and the interleaved set that hides the problem type so that the decision itself is what you practise.
The opening pages · recall first · Pretest · Two routes to the same number, and the line that picks one · When the acceleration points sideways · Gravitation read twice · The ledger · Where the zero of potential energy sits, and what it cannot change · Power, and the four checks that catch a wrong answer · method boxes · Scaffolding comes off · Full exam-style question · practice A to D · Mistake ledger · Check yourself
By the end of this section
Decide, in one written line and before any algebra, whether a given mechanics problem is faster along the force route or the energy route, and name the feature of the wording that settles it.
Write the radial component of Newton's second law for a body on a circular path, with the centripetal requirement on the right hand side and only real forces on the left.
Choose between the near surface and the general form of gravitational potential energy, and compute orbital speeds, periods and from the general form.
Balance an energy ledger between two named instants, including a sliding friction term written as the friction force times the distance slid.
Place the zero of potential energy anywhere convenient, justify that the answer cannot depend on the choice, and combine gravitational and in one line.
Compute an average or instantaneous power, and check any mechanics answer with a unit test, a and an order of magnitude comparison before handing it in.
Syllabus coverage
Catch up and Review
The five weeks already taught, used together: Newton's laws applied with friction, circular motion and drag; universal gravitation and orbits; work done by constant and varying forces; kinetic energy and the work energy principle; and conservation of energy with potential energy and nonconservative losses
The week line names no chapter and no section number, so no chapter number is quoted anywhere in this section. The scope is fixed as the material already introduced in the weeks before this one, and nothing beyond it is used.
covered
Recall first
Newton's second law on one axis
For one chosen body, the sum of the force components along an axis equals the mass times the acceleration component along that same axis: $\sum F_x = ma_x$. The sum runs over every force acting on that body and over no others.
Every force route problem in this section is this line written twice, once per axis, and the whole difficulty is deciding where the axes point.
Kinetic friction on a sliding surface
While a body slides, the surface pushes back along the surface with $f_k = \mu_k F_N$, opposing the sliding, where $F_N$ is the normal force at that surface. The normal force equals $mg$ only on level ground with nothing else acting vertically.
Both routes need it: the force route puts it in the axis equation, and the energy route multiplies it by the distance slid to get the energy removed.
The centripetal requirement
A body moving at speed $v$ on a circle of radius $r$ has an acceleration of size $v^{2}/r$ pointing at the centre, whatever is producing it. So the components of the real forces along the radius must add to $mv^{2}/r$ toward the centre.
Every bend, loop, orbit and swinging string in this section is settled by writing that one equation, and the commonest error is to write it as a force in its own right.
Universal gravitation
Two point masses attract each other along the line joining them with a force of size $F = GmM/r^{2}$, where $r$ is the distance between their centres. A uniform sphere pulls on outside bodies exactly as if all its mass sat at its centre.
It supplies the force in every orbit problem, and its energy partner supplies the potential energy used for escape speed and for moving between orbits.
Work done by a constant force, and by a spring
A constant force does work $W = Fd\cos\theta$ over a straight displacement $d$, with $\theta$ the angle between force and displacement. A spring stretched or compressed by $x$ from its natural length has had work $\tfrac{1}{2}kx^{2}$ done on it, which is the area under its force against extension line.
The first form appears in every ledger with a pull or a push in it; the second is the one varying force this course asks you to handle by area rather than by calculus.
The work energy principle
The total work done by all the forces on a body equals the change in its kinetic energy: $W_{\text{net}} = \tfrac{1}{2}mv_f^{2} - \tfrac{1}{2}mv_i^{2}$. It holds whether or not the forces are constant and whether or not the path is straight.
It is the bridge that lets a curved track be handled at all, because it never asks what the acceleration was at any particular instant.
Conservation of
If the only forces doing work are gravity and ideal springs, then $K + U$ has the same value at every instant of the motion, so $K_i + U_i = K_f + U_f$. Any other force that does work breaks the equality and appears as an extra term.
It is the version of the ledger used whenever a surface is called smooth, and recognising when it is not allowed is half of what this section drills.
Try it yourself first (3 questions)
1§10.0 — what a normal force is worth on a slope●●○○○
Before starting, three short questions to find out which of the last five weeks needs rereading. Getting them wrong is information, not a problem; each one names the block that fixes it.
Given
Mass 5.00 kg, ramp angle 30.0 degrees
The box is at rest on the ramp
$g = 9.80\ \mathrm{m/s^{2}}$
Find
(a) What is the size of the normal force on the box?
Hint 1/4
The normal force is whatever the surface has to push with so that the box does not sink into it or lift off it. Ask what the acceleration perpendicular to the surface is.
Hint 2/4
Take an axis perpendicular to the ramp surface. The weight has a component $mg\cos\theta$ into the surface, and the acceleration along that axis is zero.
Hint 3/4
With $m = 5.00\ \mathrm{kg}$ and $\theta = 30.0^{\circ}$, the perpendicular equation reads $F_N - mg\cos 30.0^{\circ} = 0$, and $mg = 49.0\ \mathrm{N}$.
Hint 4/4
So $F_N = 49.0 \times 0.866 = 42.4\ \mathrm{N}$.
Show solution
Axes are turned along and perpendicular to the surface, because that is the choice in which the normal force appears in only one of the two equations.
Perpendicular axis only
$$\sum F_{\perp} = F_N - mg\cos\theta = 0$$
the box neither sinks into the ramp nor lifts off it, so the perpendicular acceleration is zero
the cosine, not the sine, because the perpendicular direction is the one adjacent to the angle of inclination
Answer $$\boxed{F_N = 42.4\ \mathrm{N}}$$
Check
Limiting check on the formula: flatten the ramp to zero degrees and $F_N \to mg = 49.0\ \mathrm{N}$, the level ground answer; tip it to ninety degrees, a vertical wall, and $F_N \to 0$, since a box against a vertical wall is not held up by it at all. Both ends behave.
If this one was uncomfortable, reread the recall list at the top of this section before going further; every friction term in this whole section runs through the normal force.
2§10.0 — work done by a force perpendicular to the motion●●●○○
A trap by design: most students meet the true version of this claim first and generalise it one step too far. A satellite moves in a circular orbit around the Earth at constant speed.
Given
The satellite moves on a circle at constant speed
The only force on it is the Earth's gravitational pull, directed at the centre of the circle
Find
(a) True or false: gravity does no work on the satellite over one complete orbit, and therefore gravity does no work on it at any instant.
Hint 1/4
Two separate claims are being made here. Deal with them one at a time and ask whether the second really follows from the first.
Hint 2/4
Work done by a force over a displacement is $Fd\cos\theta$, with $\theta$ the angle between the force and the direction of motion at that moment. Ask what that angle is for a circular orbit.
Hint 3/4
For a circular orbit, gravity points at the centre and the velocity is tangent to the circle, so the angle is 90 degrees at every instant and the cosine is zero. Over one orbit the satellite also returns to the same radius.
Hint 4/4
So both halves happen to be true here, but the second does not follow from the first: an elliptical orbit also returns to its starting point with zero net work while gravity is doing work almost everywhere along it.
Show solution
A counterexample is used rather than a general argument, because the claim under test is an implication and one case where the premise holds and the conclusion fails settles it.
the satellite speeds up near the planet and slows down far from it, so gravity is doing work almost everywhere, yet one full circuit returns everything
Answer $$\boxed{\text{both halves true here; the implication is false in general}}$$
Check
Independent check by looking at a quantity the argument did not use: in a circular orbit the speed is constant, which is only possible if the net work is zero at every instant. In an elliptical orbit the speed demonstrably varies, which is only possible if work is being done. The kinetic energy is the witness in both cases.
This is the same distinction as gravity being a : the round trip costs nothing, and that says nothing about any single leg of it.
3§10.0 — energy stored in a compressed spring●●○○○
One computation, to check that the spring formula is available. A spring of stiffness 320 N/m is compressed 0.150 m from its natural length and held there.
Given
Spring stiffness $k = 320\ \mathrm{N/m}$
Compression from the natural length $x = 0.150\ \mathrm{m}$
Find
(a) How much energy is stored in the spring?
(b) What force does the spring push back with at that compression?
Hint 1/4
There are two different spring quantities here and they are not the same expression. One is a force and one is an energy, so their units differ and so must their formulas.
Hint 2/4
The force is $F = kx$ and the stored energy is $U = \tfrac{1}{2}kx^{2}$, the second being the area under the first plotted against the compression.
Hint 3/4
With $k = 320\ \mathrm{N/m}$ and $x = 0.150\ \mathrm{m}$, substitute into each: the energy needs the square of the compression, the force does not.
Hint 4/4
So $U = \tfrac{1}{2}(320)(0.0225) = 3.60\ \mathrm{J}$ and $F = (320)(0.150) = 48.0\ \mathrm{N}$.
Show solution
Both are computed, side by side, because the pair is the point: keeping them apart is the whole content of this question.
Force first, then energy as its area
$$F = kx = (320)(0.150) = 48.0\ \mathrm{N}$$
newtons per metre times metres gives newtons, which is the unit check on this line
the energy is the triangular area under the straight line from zero force to 48.0 N over 0.150 m, which is $\tfrac{1}{2}(48.0)(0.150)$
Answer $$\boxed{U = 3.60\ \mathrm{J},\qquad F = 48.0\ \mathrm{N}}$$
Check
Independent check by the area rather than the formula: the force grows linearly from 0 to 48.0 N over 0.150 m, so the average force is 24.0 N and the work done is $24.0 \times 0.150 = 3.60\ \mathrm{J}$. The two agree, and the second route never used the one half explicitly.
If the one half went missing here, it will go missing in every spring ledger later in this section, so it is worth fixing now: the force is linear, the energy is the area, and areas of triangles carry halves.
Notation
symbol
reads as
means
watch out
$\vec{F}$, $F$
vector F, magnitude F
a force and its size; the arrow is dropped only when a component along a named axis is being written and its sign carries the direction
Writing $F$ where $\vec{F}$ is meant hides a direction, and a size is never negative while a component often is.
$F_N$
F sub N
the normal force, the push of a surface perpendicular to itself
It equals $mg$ only on level ground with no vertical acceleration and nothing else acting vertically; on a slope, in a lift or under an angled pull it does not.
$f_k$, $\mu_k$
f sub k, mu sub k
the kinetic friction force and the coefficient that produces it from the normal force
The coefficient is a pure number with no unit; a coefficient quoted in newtons is a copied error.
$a_r$, $a_t$
a sub r, a sub t
the radial and tangential components of the acceleration of a body on a curved path
The radial one is $v^{2}/r$ and always points at the centre; the tangential one is what changes the speed and is zero in uniform circular motion.
$K$
K
kinetic energy, $\tfrac{1}{2}mv^{2}$, a positive number in joules
It depends on the square of the speed, so doubling a speed quadruples it; and it has no direction, so it is never resolved into components.
$U$
U
potential energy, stored by position: $mgy$ near the ground, $-GmM/r$ in general, $\tfrac{1}{2}kx^{2}$ for a spring
Only differences in $U$ mean anything; the general gravitational form is negative everywhere and rises toward zero as the separation grows.
$W_{\text{other}}$
W sub other
the work done between the two instants by every force that has no potential energy, which in this section means friction, an applied pull, or a motor
Gravity and springs must not appear here as well as in $U$; counting a force twice is the same mistake as leaving it out, with the opposite sign.
$r$
r
in circular motion, the radius of the path; in gravitation, the distance between the two centres
In an orbit problem $r$ is measured from the centre of the planet, so an altitude has to have the planet radius added to it before it is used.
$P$
P
power, the rate at which work is done, measured in
One watt is one joule per second; a power multiplied by a time is an energy, and a power quoted in joules is a unit error worth checking for.
$\Delta$
delta
the change in a quantity, final value minus initial value
The order is fixed. A height that falls gives a negative $\Delta y$, and reversing the subtraction is how a released body ends up gaining potential energy.
Conventions used here
Positive directions, axes and angles across this review
Every problem here begins by drawing axes and saying in words which way is positive. On a slope the axes are turned so that one runs along the slope; on a circular path one axis is radial and points at the centre, and that axis is called positive toward the centre. Angles of inclination are measured from the horizontal, and angles inside a work calculation are measured between the force and the displacement. The choice is fixed once inside a problem and never changed halfway through.
A sign that drifts between two lines produces an answer wrong by a factor of minus one which looks entirely reasonable on the page.
The numerical constants fixed for every calculation in this review
The free fall acceleration is the positive number 9.80 m/s squared, and its direction is written into the equation rather than into the symbol. The gravitational constant is 6.674 times ten to the minus eleven, in newton metre squared per kilogram squared. The mass of the Earth is taken as 5.97 times ten to the twenty four kilograms and its radius as 6.38 times ten to the six metres. No calculation on this page rounds the free fall acceleration to ten.
These four numbers appear in almost every question here, and a section that quietly changes one of them between blocks makes its own worked examples disagree.
Where the zero of potential energy is placed on this page
For a body near the ground the zero of gravitational potential energy is placed at a level named in words at the start of the problem, usually the lowest point of the motion, and the height is measured upward from it. For a body far from a planet the zero is at infinite separation, which is what forces the general form to be negative. For a spring the zero is at its natural length. A problem states its choice once and every energy in that problem is measured from it.
Only differences in potential energy appear in any answer, so the level is free; but it is free only if it is the same in the initial and the final line of the same ledger.
What the idealising words are allowed to mean in this review
Smooth means the friction force is exactly zero. A light string does not stretch and has no mass, so the tension is the same at both of its ends. A light pulley turns freely and does not change the size of the tension. Air resistance is absent unless a resisting force is given as a number in newtons or as a formula. An ideal spring obeys its force law for the whole range used and stores all the work done on it.
Each of these words removes a term from an equation, and a reader who does not know which term has been removed cannot tell whether a missing term is an idealisation or a mistake.
How friction losses are written in an energy line
Sliding friction enters the ledger as a negative work equal to the friction force times the distance actually slid along the surface, and never as a potential energy. The distance is the path length, so a body that goes up a slope and comes back down pays for both legs. The energy removed does not vanish: it appears as in the block and the surface, and this section writes it on the right hand side as a positive thermal term whenever the wording asks how much energy was lost.
Friction is the one force in this course whose work depends on the route taken rather than only on the endpoints, so it cannot be given a potential energy and it cannot be collapsed into a height.
Digits, units and rounding in every boxed answer here
Data are quoted to three significant figures and boxed answers are given to three. Guard digits are carried through intermediate lines and dropped only in the box. Every boxed number carries an SI unit, and an intermediate line that has lost its units is treated as an unfinished line. Angles are given in degrees, and any trigonometric function is evaluated with the calculator set to degrees unless the quantity is explicitly written in radians.
A number without a unit cannot be checked and cannot be marked. Rounding early is also a real source of disagreement between a worked example and its own verification line, and this section carries extra digits precisely so that the two agree.
10.1Two routes to the same number, and the line that picks one
Forces followed instant by instant, or energy counted at two instants: both are right, only one is short.
Two complete methods are on the table, and nobody says which one a question wants.
Solvable with what we have
Given a slope angle and a friction coefficient, get the acceleration and run the kinematic formulas.
Given a constant force and a straight displacement, compute its work for the ledger.
Given the speeds at two instants, find the total work done in between.
Given a smooth track, equate the mechanical energy at two named instants.
Not solvable yet
Find the speed at the bottom of a curved chute, whose angle differs at every point.
Find how much energy a playground slide removed, with no friction force given.
Decide quickly which of two legal methods takes two lines and which takes twelve.
Take the curved chute, see the 1.80 m drop, reach for $a = g\sin\theta$. No single $\theta$ exists, so call it thirty degrees, get $a = 4.90\ \mathrm{m/s^{2}}$, guess a 3.6 m path and finish with $v = \sqrt{2(4.90)(3.6)} = 5.94\ \mathrm{m/s}$.
Why it fails
The number is close, which is the dangerous part; the method is not. $a = g\sin\theta$ assumes one fixed angle, and a curve has a different one at every point, leaving no constant acceleration for a kinematic formula. Reshape the chute to the same drop and the guess moves; the honest answer does not.
MethodMethod 10.1: pick the route before writing anything
Conditions
the force route needs the acceleration to be constant over the interval if a kinematic formula is going to be used afterwards, which means the net force must not change during that interval
the energy route needs two clearly named instants, and every force that does work between them has to be accounted for exactly once
the energy route cannot answer a question about a time, and it cannot give a force at an instant; those two words send you back to the force route
both routes need a single chosen body, named out loud before either equation is written
Top line: sum the forces, get an acceleration, let it generate the motion instant by instant. Bottom line: what the body had at the first instant, plus what went in or out, equals what it has at the second.
Why the two routes cannot disagree
Start from the second law along the direction of motion, $F_{\text{net}} = ma$, and multiply both sides by a small displacement $ds$ along the path.
On the right, $a\,ds = \frac{dv}{dt}ds = v\,dv$, because $ds/dt$ is the speed. So $F_{\text{net}}\,ds = mv\,dv$.
Add up both sides over the whole path from the first instant to the second. The left side is the total work done by all forces; the right side integrates to $\tfrac{1}{2}mv_f^{2}-\tfrac{1}{2}mv_i^{2}$.
So the energy route is not a second law of nature. It is the second law with the time integrated out, which is why it can survive a path that curves and a force that varies, and why it can no longer tell you anything about when.
The decision the section is about, drawn as the two questions that settle it. Read the wording, answer the top question, and the route is chosen before any algebra has been written. The right hand branch is the one students skip past, and it is the one that survives a curved track and a force that changes.
Looks like this, but is not
The energy route working: a bead slides from rest down a smooth curved wire dropping 1.80 m. The wire pushes perpendicular to the motion, so it does no work: gravity is the only ledger entry, and the bottom speed is $\sqrt{2gh} = 5.94\ \mathrm{m/s}$ whatever the shape.
This looks like the same problem and is not: ask how long the bead takes. The ledger has no clock, so two wires with the same drop give the same final speed and different times. A time on a curved track needs the force route, and calculus this course skips.
Speed at the bottom of a smooth curved chute that drops 1.80 m
A small block is released from rest at the top of a smooth chute whose surface curves the whole way down. The bottom of the chute is 1.80 m below the release point. Find the speed of the block as it leaves the bottom, and say what would change if the chute were reshaped to the same drop.
Given
Released from rest, so $v_i = 0$
Vertical drop $h = 1.80\ \mathrm{m}$
The chute is smooth, so friction does no work
The surface pushes perpendicular to the motion at every point
$g = 9.80\ \mathrm{m/s^{2}}$
Find
The speed of the block at the bottom.
Solution
The energy route is taken because the question names two instants and no time, and because the force route is not merely longer here but unavailable: the angle of the surface changes continuously, so there is no constant acceleration to feed into a kinematic formula.
Choose the two instants and the zero level
$$i:\ \text{at rest at the top},\qquad f:\ \text{at the bottom}$$
the ledger needs exactly two instants, and these are the only two the question describes
putting the zero at the lowest point of the motion keeps both potential energies positive or zero, which removes one place a sign can go wrong
Write the ledger and drop the terms that are zero
$$K_i + U_i + W_{\text{other}} = K_f + U_f$$
the general line, written before anything is deleted, so that deletions are decisions rather than omissions
$$0 + mgh + 0 = \tfrac{1}{2}mv_f^{2} + 0$$
released from rest kills the first term; the normal force is perpendicular to the motion everywhere so it does no work, and the chute is smooth so there is no friction term
the mass cancels on both sides, which is why the answer will not mention it
$$v_f = 5.94\ \mathrm{m/s}$$
three significant figures, matching the three carried by the drop
Answer $$\boxed{v_f = 5.94\ \mathrm{m/s}}$$
Check
Independent check by a limiting case that has a known answer: if the chute were replaced by a vertical drop of the same 1.80 m, the block would be in free fall and $v^{2} = 2gh$ gives the same 5.94 m/s. A smooth chute cannot do better than free fall and cannot do worse, because neither one has anything to remove energy, so agreement is exactly what the physics demands. The size is also sensible, about the speed of a fast run.
Four lines, one of which was bookkeeping. The force route would have needed the surface angle as a function of position, which the problem does not even supply.
The shape of the chute never entered the calculation, so it cannot enter the answer. Any smooth track with a 1.80 m drop delivers 5.94 m/s at the bottom, and that single sentence answers a whole family of exam questions about ramps, loops, valleys and hills.
Speed at the bottom of a rough straight ramp, worked both ways
The chute is replaced by a straight ramp at 30.0 degrees with the same 1.80 m drop, so the sliding distance along the surface is 3.60 m. The block, of mass $m$, starts from rest and the coefficient of kinetic friction is 0.250. Find the speed at the bottom along the force route and then along the energy route, and compare the cost of the two.
Given
Ramp angle $\theta = 30.0^{\circ}$, straight
Vertical drop $h = 1.80\ \mathrm{m}$, so the path length is $L = h/\sin\theta = 3.60\ \mathrm{m}$
Coefficient of kinetic friction $\mu_k = 0.250$
Released from rest, $v_i = 0$
$g = 9.80\ \mathrm{m/s^{2}}$
Find
The speed at the bottom, by each route.
Solution
Both routes are legal here because the ramp is straight and the forces are constant, so this is the one case where they can be compared honestly.
Force route: get the acceleration, then use kinematics
$$\sum F_x = mg\sin\theta - \mu_k F_N = ma_x$$
axes turned along the slope, positive down the slope, so friction opposes the sliding and carries a minus sign
three lines, and the normal force never had to be named separately
Answer $$\boxed{v = 4.47\ \mathrm{m/s}\ \text{by either route}}$$
Check
The two routes were genuinely independent: one went through an acceleration of 2.78 m/s squared and a kinematic formula, the other never computed an acceleration at all. They agree to three figures. A second, cheaper check is the limit $\mu_k \to 0$, which turns the energy line into $v = \sqrt{2gh} = 5.94\ \mathrm{m/s}$, the smooth answer from the previous example; the rough ramp is slower, as it has to be.
Force route six lines, energy route three, for the same number. On a curved ramp the force route would have been not merely longer but impossible with this course's tools.
When both routes are available the energy route is usually shorter by about half, and the saving grows with the number of forces.
Checkpoint
§10.1 — choosing the route from the wording●●○○○
Thirty seconds, no arithmetic. A question reads: a 0.500 kg ball on the end of a 1.20 m string is swung in a vertical circle; find the tension in the string when the ball is at the lowest point, given that its speed there is 7.67 m/s.
Given
The body is a ball on a string of length 1.20 m
Its speed at the lowest point is given as 7.67 m/s
The quantity asked for is the tension at that one instant
Find
(a) Which route does this question force, and which single word in it settles the matter?
Hint 1/4
Look at what is being asked for rather than at what is given. Is the unknown a speed, a height, or something else entirely?
Hint 2/4
An energy ledger contains kinetic energies, potential energies and works. A tension at an instant is none of those, so it cannot be read off a ledger.
Hint 3/4
Here the data are a mass of 0.500 kg, a radius of 1.20 m and a speed of 7.67 m/s at the lowest point, and the unknown is a force at that instant. Write the radial component of the second law with the centre upward as positive.
Hint 4/4
The word tension forces the force route, specifically the radial equation $T - mg = mv^{2}/r$.
Show solution
The decision is made from the unknown, not from the data, because the data in an exam question are usually enough for either route while the unknown is not.
Name the unknown and test it against the ledger
$$K_i + U_i + W_{\text{other}} = K_f + U_f$$
the energy line contains only energies and works, and a tension is a force in newtons, so it has no slot here
$$\sum F_r = m\frac{v^{2}}{r} \;\Rightarrow\; T - mg = m\frac{v^{2}}{r}$$
the radial equation is the only line in this course that produces a force at a named instant
carried through for completeness; the answer to the question asked was the route, but the route is only convincing if it finishes
Answer $$\boxed{\text{force route, radial equation};\quad T = 29.4\ \mathrm{N}}$$
Check
Limiting check on the formula rather than the arithmetic: if the ball were hanging still, $v = 0$ and the expression collapses to $T = mg = 4.90\ \mathrm{N}$, which is what a stationary hanging ball must read. The swinging value is six times larger, which is the right order for a ball moving at nearly 8 m/s on a short string.
Energy is often still needed in these problems, but only as a supplier: it delivers the speed at the instant in question, and then the radial equation converts that speed into a force.
⚠ Using a slope formula on a curved surface
the result $a = g\sin\theta$ is memorised as a fact about ramps rather than as a consequence of one fixed angle, so a curve looks like a ramp with a slightly awkward angle
wrong$$v = \sqrt{2\,(g\sin\theta)\,L}\quad\text{on a curved chute}$$
right$$v = \sqrt{2gh}\quad\text{on any smooth track of drop } h$$
⚠ Trying to get a time out of an energy ledger
the ledger answers so many questions cheaply that it feels universal, and the missing clock is invisible because no symbol for it is absent from the page
right$$\text{a time needs } \sum F = ma \text{ and then } v = v_0 + at$$
10.2When the acceleration points sideways: bends, loops and swings
On a circle the acceleration points at the centre, so the real forces have to add up to that and nothing else.
The route chart says a force at an instant means Newton. Circular motion is the case where that instant is every instant, and where the axis has to be chosen with more care than usual.
RuleRule 10.2: the radial equation
Conditions
the radius must be the radius of the circular path the body actually follows, not the length of some other line in the picture
the left hand side contains only forces that something exerts on the body: gravity, a normal force, a tension, friction; nothing is added because the path is curved
the speed used must be the speed at the instant the equation is written, so on a vertical circle it changes from point to point
if the speed is changing there is a tangential component too, and it is handled on a separate axis at right angles to the radial one
$$\boxed{\;\sum F_r = m\,\frac{v^{2}}{r}\quad\text{(taking toward the centre as positive)},\qquad \sum F_t = m\,a_t\;}$$
Point an axis at the centre of the circle and call that direction positive. Add up the components of the real forces along that axis. Whatever they come to, they must equal the mass times the speed squared over the radius, because that is the acceleration a body on that circle at that speed is having. The equation is a demand, not a new force: it tells you what the existing forces are obliged to add up to, and if they cannot manage it the body leaves the circle.
Where the speed squared over the radius comes from
A body going round a circle at constant speed has a velocity of fixed length whose direction turns steadily. Over a short time the velocity vector turns through the same angle that the position vector turns through, because the velocity stays tangent to the circle.
Draw the two velocity vectors, before and after, tail to tail. They have the same length $v$ and the angle between them is $\Delta\theta$, so the change in velocity has length $v\,\Delta\theta$ for a small angle, and it points toward the centre.
In that same time the body has travelled an arc $v\,\Delta t = r\,\Delta\theta$, so $\Delta\theta = v\,\Delta t / r$.
The acceleration is the change in velocity over the time, which is $v\,\Delta\theta/\Delta t = v^{2}/r$, pointing at the centre. Nothing in this argument mentions what is producing the motion, which is exactly why the result is a requirement placed on the forces rather than one of them.
A car on a flat bend, seen from above and from behind. Nothing new has been added to the diagram on the right: the only horizontal force is the friction of the road, and the circle merely fixes what that friction has to equal. The radial axis is drawn pointing at the centre, which is why the right hand side of the radial equation comes out positive.
Looks like this, but is not
This is the rule working: a car goes round a flat bend of radius 106 m at 25.0 m/s. The only horizontal force is the friction of the road on the tyres, so $f_s = mv^{2}/r$, and the largest friction the road can supply, $\mu_s mg$, sets the fastest safe speed.
This looks like the same rule and is not: the same car on the same bend, but the driver is also braking hard. Now the speed is changing, so there is a tangential acceleration as well, and the friction has to do two jobs at once with a total that still cannot exceed $\mu_s mg$. The radial equation on its own is no longer the whole story, and a car that would have held the bend at steady speed can slide out of it while braking. The rule did not fail; it was only ever the radial half of a two axis problem.
Tightest bend a car can take at 25.0 m/s on a flat road
A car travels at a steady 25.0 m/s on a flat, unbanked road. The coefficient of static friction between the tyres and the road is 0.600. Find the smallest radius of bend the car can follow without sliding, and state what happens to that radius if the car is loaded with passengers.
Given
Steady speed $v = 25.0\ \mathrm{m/s}$
Flat road, so the normal force is vertical and the friction is horizontal
Coefficient of static friction $\mu_s = 0.600$
$g = 9.80\ \mathrm{m/s^{2}}$
Find
The smallest radius the car can hold, and whether the answer depends on the load.
Solution
Static friction is used rather than kinetic because the tyres are rolling, not skidding: the patch of rubber touching the road is instantaneously at rest with respect to it.
Answer $$\boxed{r_{\min} = 106\ \mathrm{m},\ \text{independent of the mass}}$$
Check
Order of magnitude check against something familiar: 25.0 m/s is 90 km/h, and a motorway bend taken at 90 km/h is a long sweeping curve a hundred metres or so in radius, not a corner. A second check is the limit $\mu_s \to 0$, an icy road, where the required radius runs off to infinity, meaning the only path the car can hold is a straight line. Both agree with the formula.
Two axis equations and one inequality. The whole difficulty was in noticing that the ceiling on static friction, not friction itself, is what sets the limit.
The mass cancelling is worth remembering because it is counter to intuition: a loaded lorry and an empty car slide off the same bend at the same speed, since the extra weight buys exactly as much extra friction as it costs in required force.
Slowest a 0.150 kg ball can be moving at the top of a vertical circle
A 0.150 kg ball is whirled in a vertical circle on a light string of length 1.20 m. Find the slowest speed it can have at the very top while the string stays taut, then find its speed at the bottom of that same swing and the tension there.
Given
Mass $m = 0.150\ \mathrm{kg}$
String length, and hence radius, $r = 1.20\ \mathrm{m}$
The string is light and does not stretch, and the circle is vertical
The condition at the top is that the string is on the point of going slack
$g = 9.80\ \mathrm{m/s^{2}}$
Find
The minimum speed at the top, the speed at the bottom, and the tension at the bottom.
Solution
The top is handled with the radial equation and the bottom is reached with an energy ledger, because between those two points the speed changes and no force at an intermediate instant is wanted.
The condition at the top
$$\sum F_r = T + mg = m\frac{v_t^{2}}{r}$$
at the top the centre of the circle is straight down, so both the weight and the tension point the same way, toward the centre
the zero of potential energy is at the lowest point and the top is one diameter above it; the string does no work because it always pulls perpendicular to the motion
Independent check on the tension by a different chain of reasoning: the weight of the ball is $mg = 1.47\ \mathrm{N}$, and the answer says the string carries exactly six weights. That factor can be got without any of the numbers above, since at the minimum condition $v_b^{2} = 5gr$, so $T = m(g + 5g) = 6mg$, and $6 \times 1.47 = 8.82\ \mathrm{N}$. Two routes, one number.
Three separate equations, and the sign of the weight term changed between the first and the third because the direction of the centre changed, not because anything physical changed.
The factor of six is worth carrying: a string swung in a vertical circle at the slowest speed that keeps it taut is pulled six times harder at the bottom than by the hanging ball alone.
Checkpoint
§10.2 — what belongs on a circular motion diagram●●○○○
Thirty seconds. A student draws a free body diagram for a car on a flat bend and labels three arrows: the weight down, the normal force up, and a centripetal force pointing at the centre of the bend.
Given
The car is going round a flat bend at steady speed
The three arrows drawn are the weight, the normal force, and an arrow labelled centripetal force
Find
(a) Is the diagram correct as drawn?
Hint 1/4
Ask what a free body diagram is a list of. Every arrow on one has to have an agent: something in the world that is doing the pushing or the pulling.
Hint 2/4
The word centripetal describes a direction, not a source. The radial equation says that the real forces, added along the radius, must come to $mv^{2}/r$; it does not add a force to the list.
Hint 3/4
For this car the real horizontal force is the friction of the road on the tyres. Substituting into the radial equation gives $f_s = mv^{2}/r$, and the friction arrow is the third arrow that should be there.
Hint 4/4
So the diagram is wrong: the third arrow should be labelled friction, and $mv^{2}/r$ belongs on the other side of the equals sign, not on the diagram.
Show solution
The test applied is the agent test, because it is mechanical and needs no judgement: name the object exerting the force.
Apply the agent test to each arrow
$$\text{weight}: \ \text{the Earth pulls the car}$$
an agent exists, so the arrow is legitimate
$$\text{normal force}: \ \text{the road pushes the car}$$
an agent exists, so the arrow is legitimate
$$\text{centripetal force}: \ \text{agent} = ?$$
no object in the problem is exerting it, because the word names a direction rather than an interaction
Answer $$\boxed{\text{False: the third arrow must be labelled friction}}$$
Check
Independent check by counting equations. With the wrong diagram the radial equation reads $f_s + mv^{2}/r = mv^{2}/r$, which forces the friction to be zero, and a car with no horizontal force cannot turn at all. The wrong diagram therefore contradicts the situation it was drawn for, which is a stronger objection than any appeal to terminology.
The same test disposes of the other invented arrow, the outward one students sometimes add for the feeling of being thrown outward in a turning car.
⚠ Adding a centripetal force to the diagram
the phrase centripetal force appears in the textbook, so it sounds like the name of a force rather than the name of the job a real force is doing
⚠ Keeping the same sign for the weight at the top and the bottom of a vertical circle
the weight really does point down at both places, so it feels as though its sign in the radial equation should not change either
wrong$$\text{top: } T + mg = \frac{mv^{2}}{r},\qquad \text{bottom: } T + mg = \frac{mv^{2}}{r}$$
right$$\text{top: } T + mg = \frac{mv^{2}}{r},\qquad \text{bottom: } T - mg = \frac{mv^{2}}{r}$$
10.3Gravitation read twice: as a force and as an energy
One inverse square law supplies the force that holds an orbit and the energy that decides whether a body ever comes back.
Gravitation is the one topic in this half of the course that has a full force form and a full energy form, so it is where the choice of route has to be made twice in the same question.
TheoremTheorem 10.3: the two faces of the inverse square law
Conditions
the distance $r$ is measured between the centres of the two bodies, so an altitude above a surface must have the planet radius added to it first
the energy form takes the zero of potential energy at infinite separation, which is what makes it negative everywhere else
the near surface form $U = mgy$ is a straight line approximation to the energy form, and is safe only while the height climbed is a small fraction of the planet radius
both forms treat a uniform sphere as though its whole mass sat at its centre, which is legitimate for any body outside the sphere
The first expression is the pull between two masses and it weakens with the square of the separation. The second is the energy stored in that pull, and it weakens only with the separation itself, which is why it falls off more slowly and why it is negative: a bound body sits in a well and has to be given energy to climb out. The third comes from setting the pull equal to the circular requirement, and the fourth from asking for just enough energy to reach the top of the well with nothing left over.
Getting the orbital and escape speeds from the two forms
For a circular orbit of radius $r$, gravity is the only force and it points at the centre, so the radial equation reads $GmM/r^{2} = mv^{2}/r$.
Cancel one power of $r$ and the orbiting mass: $v^{2} = GM/r$. The mass of the satellite has vanished, which is why a bolt and a space station in the same orbit travel at the same speed.
For escape, run the energy ledger from the surface, at radius $R$ with speed $v$, out to infinite separation with nothing left: $\tfrac{1}{2}mv^{2} - GmM/R = 0 + 0$.
Solve: $v^{2} = 2GM/R$. Notice this is twice the square of the orbital speed at the same radius, so the escape speed is $\sqrt{2}$ times the speed of a low circular orbit, a ratio worth carrying because it checks both formulas at once.
Gravitational potential energy per kilogram against distance from the centre of the Earth, with the zero placed at infinite separation. The curve is steep near the surface and almost flat far out, which is the whole reason mgy is safe for a ladder and useless for a satellite: mgy assumes this curve is a straight line.
Looks like this, but is not
This is the near surface form working: a 2.00 kg book is lifted 1.50 m onto a shelf. Over that climb the field strength changes by about one part in four million, so treating it as constant is not an approximation anybody can measure, and $\Delta U = mgh = 29.4\ \mathrm{J}$ is exact for every purpose in this course.
This looks like the same calculation and is not: lifting the same 2.00 kg from the ground to an altitude of one Earth radius, 6380 km up. The near surface form gives $mgh = 2.00 \times 9.80 \times 6.38\times 10^{6} = 125\ \mathrm{MJ}$. The honest energy form gives $GmM/(2R_E) = 62.5\ \mathrm{MJ}$ for the same climb. The near surface form is not slightly wrong here; it is wrong by exactly a factor of two, because it charges full price for every metre while the real field has halved by the halfway mark.
Speed and period of a satellite 400 km above the Earth
A satellite is in a circular orbit 400 km above the surface of the Earth. Find its orbital speed and the time it takes to go once around, and check the period against something you already know.
Given
Altitude $h = 4.00\times10^{5}\ \mathrm{m}$ above the surface
The orbit is circular and gravity is the only force acting
Find
The orbital speed and the period.
Solution
The force route is taken first because a circular orbit is a circular motion problem before it is an energy problem: the radial equation delivers the speed in one line.
Independent check from a different physical fact: the International Space Station orbits at roughly this altitude and is widely known to circle the Earth about sixteen times a day. Sixteen orbits in 1440 minutes is 90 minutes each, and this calculation gives 92.6 minutes without using that fact anywhere. A second check is the field strength: $GM/r^{2} = 8.67\ \mathrm{m/s^{2}}$ at this radius, sensibly a little below the surface value of 9.79.
Three short stages, and the only place the calculation could have gone badly wrong was the first line, where the altitude had to be turned into a radius.
Because the satellite mass cancels, everything in that orbit moves at the same speed: the station, a dropped spanner, and an astronaut outside it.
Escape speed from the surface of the Earth, and why mgh cannot produce it
Find the speed at which a projectile would have to leave the surface of the Earth, with no further propulsion, in order never to fall back. Then show what the near surface energy form would predict for the same question, and explain the difference.
$G = 6.674\times10^{-11}\ \mathrm{N\,m^{2}/kg^{2}}$, so $GM = 3.984\times10^{14}\ \mathrm{m^{3}/s^{2}}$
Air resistance and the rotation of the Earth are ignored
Never falling back means reaching infinite separation with zero speed left
Find
The escape speed, and what mgh would give instead.
Solution
The energy route is used because the question names two states, the launch and the far away limit, and asks nothing about time or about any force along the way.
Turn never comes back into an energy statement
$$K_i + U_i = K_f + U_f$$
the two instants are the launch and the far limit; only gravity acts, so there is no other work term
$$\tfrac{1}{2}mv^{2} - \frac{GmM}{R_E} = 0 + 0$$
at infinite separation the potential energy is zero by the chosen convention, and just barely escaping means arriving with no speed left
Independent check using the ratio proved in the box rather than repeating the arithmetic: the escape speed should be $\sqrt{2}$ times the speed of a circular orbit skimming the surface. That orbital speed is $\sqrt{GM/R_E} = 7.90\ \mathrm{km/s}$, and $1.414 \times 7.90 = 11.2\ \mathrm{km/s}$. The two agree, and neither used the other's arithmetic.
Two lines for the answer and three more for the comparison, which is the part that carries the lesson.
The finiteness of the well is the whole point. A constant field has infinite depth and nothing can ever leave; a real inverse square field has a finite depth, and that is why escape speed is a number rather than an impossibility.
Checkpoint
§10.3 — which potential energy form is legal●●○○○
Thirty seconds, no arithmetic. A problem asks for the energy needed to raise a 500 kg payload from the surface of the Earth to a circular orbit at an altitude of 3190 km, which is half an Earth radius.
Given
Payload mass 500 kg
Climb from the surface, at $6.38\times10^{6}\ \mathrm{m}$ from the centre, to $9.57\times10^{6}\ \mathrm{m}$ from the centre
The two candidate forms are $U = mgy$ and $U = -GmM/r$
Find
(a) Which form is legal here, and what is the test you applied?
Hint 1/4
The near surface form is a straight line drawn tangent to a curve. Ask over what range a tangent line stays close to the curve it touches.
Hint 2/4
The test is the size of the climb compared with the radius: $mgy$ is safe while $h \ll R$, because only then is the field strength effectively unchanged.
Hint 3/4
Here $h = 3.19\times10^{6}\ \mathrm{m}$ and $R_E = 6.38\times10^{6}\ \mathrm{m}$, so the climb is half a radius, and at the top the field is down to about four ninths of its surface value.
Hint 4/4
So only the general form is legal: use $U = -GmM/r$ at both ends and subtract.
Show solution
The ratio test is used rather than computing both answers, because the ratio can be done in your head at the start of a problem and the two full calculations cannot.
the field at the top is under half its surface value, so no single constant field strength describes the climb
$$\Delta U = GmM\left(\frac{1}{R_E} - \frac{1}{r_2}\right)$$
the general form, evaluated at both ends and subtracted, which is what must be used
Answer $$\boxed{U = -\frac{GmM}{r}\ \text{only; the test is } h/R \ll 1}$$
Check
Sanity check on the size of the error rather than on the choice: $mgh$ would give $500 \times 9.80 \times 3.19\times10^{6} = 1.56\times10^{10}\ \mathrm{J}$, while the general form gives $1.04\times10^{10}\ \mathrm{J}$. The near surface form is too large by about fifty per cent, which confirms it was the wrong tool rather than a slightly imprecise one.
A useful working rule: if the problem mentions an orbit, a satellite, or a height quoted in hundreds of kilometres, the general form is required.
⚠ Using the altitude in place of the orbital radius
the problem quotes an altitude because that is what is measured from the ground, and the symbol in the formula is also a distance, so the substitution feels dimensionally safe
wrong$$v = \sqrt{\frac{GM}{h}}$$
right$$v = \sqrt{\frac{GM}{R_E + h}}$$
⚠ Dropping the minus sign from the gravitational potential energy
every other energy in the course is positive, so a negative energy looks like a slip rather than a consequence of putting the zero at infinity
wrong$$E = \tfrac{1}{2}mv^{2} + \frac{GmM}{r}$$
right$$E = \tfrac{1}{2}mv^{2} - \frac{GmM}{r}$$
10.4The ledger: one line, two instants, and the friction term that ruins it
Write what the body had, add what was put in or taken out, set it equal to what it ends with.
The route chart sends most review problems here. This block is about writing the line so that nothing is counted twice and nothing is silently dropped.
RuleRule 10.4: the energy ledger between two instants
Conditions
the two instants have to be named in words before any symbol is written, because every term in the line is evaluated at one of them
a force may appear either as a potential energy or in the other work term, never in both; gravity and springs go in the potential energies, everything else goes in the work term
the friction contribution is $-f_k d$ with $d$ the distance actually slid along the surface, so a there and back journey pays twice
the zero of potential energy is declared once and used at both instants
Take a snapshot at the first instant and add up the two kinds of energy the body has: the energy of its motion and the energy of its position. Then add every joule that something outside pushed in, and subtract every joule that friction or a brake took out. What you are left with is exactly what the body has at the second instant, split between motion and position in whatever way the geometry demands. Nothing is created and nothing disappears; the friction term is not an exception but a bookkeeping entry for energy that has gone into warming the surfaces.
Where the potential energies come from and why friction cannot have one
Start from the work energy principle, which is exact: the total work of all forces equals the change in kinetic energy, $W_{\text{total}} = K_f - K_i$.
Split the total into the work done by gravity and springs, and the work done by everything else: $W_{\text{grav}} + W_{\text{spring}} + W_{\text{other}} = K_f - K_i$.
For gravity and for an ideal spring, the work between two points depends only on the two endpoints and not on the route taken. That fact, and only that fact, allows a function of position $U$ to be defined so that the work done is $-(U_f - U_i)$.
Substituting and rearranging gives the boxed line. Friction cannot be handled this way because the work it does depends on the length of the path: two routes between the same two points, one short and one long, remove different amounts. So friction has no potential energy and stays in the other work term forever.
The slide from the opening problem drawn as a ledger. The single column on the left is everything the child had at the top; the two columns on the right are what is left as motion and what the slide took. The two sides have the same total height, and that equality is the whole method.
Looks like this, but is not
This is the ledger working: a crate is pushed 4.00 m along a level floor and comes back to where it started. Gravity did zero net work, because the crate returned to the same height and $\Delta U = 0$.
This looks like the same statement about friction and is not: friction did not do zero net work on the round trip. It removed $f_k \times 4.00$ joules on the way out and another $f_k \times 4.00$ on the way back, because it always opposes the sliding and therefore always takes, never gives. Returning to the starting point resets the potential energy and resets nothing else, which is precisely the difference between a force that has a potential energy and one that does not.
How much energy a 2.50 m slide takes from a 28.0 kg child
The child from the opening problem, of mass 28.0 kg, starts from rest at the top of a slide whose top is 2.50 m above the bottom, and leaves the bottom at 4.20 m/s. The sliding surface is 5.00 m long. Find the energy the slide removed, and the average friction force along it.
Given
Mass $m = 28.0\ \mathrm{kg}$
Starts from rest, so $K_i = 0$
Vertical drop $h = 2.50\ \mathrm{m}$
Speed at the bottom $v_f = 4.20\ \mathrm{m/s}$
Length of the sliding surface $d = 5.00\ \mathrm{m}$
$g = 9.80\ \mathrm{m/s^{2}}$
Find
The energy removed by the slide, and the average friction force.
Solution
The ledger is used with friction as an unknown rather than as a given, which is the reverse of the usual direction.
Set up the ledger with the loss as the unknown
$$K_i + U_i + W_{\text{fric}} = K_f + U_f$$
the two instants are the release at the top and the exit at the bottom; the zero of potential energy is placed at the bottom
Independent plausibility check by a different quantity: 439 J out of 686 J is sixty four per cent of the drop, so the child arrives with about a third of the energy a smooth slide would have delivered. A smooth slide would give $v = \sqrt{2gh} = 7.00\ \mathrm{m/s}$, and $4.20/7.00 = 0.600$, so the speed ratio squared is 0.360, which is the same one third. Two different ways of expressing the loss agree.
One ledger line and two substitutions. The only decision that mattered was which distance to divide by at the end.
Note what a friction force of 87.8 N implies: with the surface at about thirty degrees the normal force is roughly 238 N, so the coefficient is about 0.37, which is high for a slide and is why children in dry cotton clothes have to push themselves along.
How high a spring launched block rises after crossing a rough patch
A spring of stiffness 480 N/m is compressed by 0.180 m and released, pushing a 0.250 kg block along a horizontal track. The first 0.900 m of the track is rough, with a coefficient of kinetic friction of 0.220; after that the track is smooth and curves upward. Find the speed of the block where the rough patch ends, and the greatest height it reaches on the smooth curve.
Independent check by the limiting case of a smooth track: with $\mu_k = 0$ the whole 7.776 J becomes height, giving $h = 7.776/2.45 = 3.17\ \mathrm{m}$. The rough patch costs 0.19 m of height, and $0.485/2.45 = 0.198\ \mathrm{m}$, which is the same number reached without touching either speed. A unit check also passes: newtons per metre times metres squared is newton metres, which is joules.
Two ledgers rather than one, and the split cost nothing because the second one had only two terms in it.
The highest point was found without ever computing a speed on the curve, and that is the habit worth taking: if the question asks for a height, aim the ledger at the height and leave the intermediate speed alone unless it is asked for.
Checkpoint
§10.4 — which distance multiplies the friction force●●○○○
Thirty seconds, no calculator. A block slides 3.00 m down a straight ramp inclined at 30.0 degrees, so it descends 1.50 m vertically.
Given
Distance along the ramp surface 3.00 m
Vertical drop 1.50 m
Kinetic friction force a steady 6.00 N
Find
(a) How many joules did friction remove?
Hint 1/4
Friction acts along the surface and is paid for by the surface. Ask which of the two lengths in the problem the block actually rubbed along.
Hint 2/4
The work done by a constant force is the force times the displacement in the direction of that force, so a force lying along the ramp is multiplied by the distance measured along the ramp.
Hint 3/4
Here the friction force is 6.00 N and the distance along the surface is 3.00 m, while the 1.50 m is the vertical component of the same displacement and belongs to gravity, not to friction.
Hint 4/4
So the energy removed is $6.00 \times 3.00 = 18.0\ \mathrm{J}$.
Show solution
Each force is paired with the displacement measured along its own direction, because that is what the definition of work says and it removes the temptation to pick whichever length is nearest.
Independent check by a limiting case: flatten the ramp toward horizontal. The drop goes to zero while the sliding distance stays 3.00 m, and friction plainly still removes 18.0 J from a block sliding 3.00 m on a level floor. Any rule that used the drop would predict zero loss on a level floor, which is absurd.
The general habit: before multiplying a force by a distance, say out loud which direction the force points, then take the part of the displacement that lies along it.
⚠ Multiplying the friction force by the height dropped
the height is the number the ledger has just used for gravity, so it is the one sitting in working memory when the friction line is written
10.5Where the zero of potential energy sits, and what it cannot change
Potential energies are only ever subtracted, so the level they are measured from is yours to choose.
The ledger needed a zero level declared before it could be written. This block is about why that declaration is free, and what happens when it is made twice.
DefinitionDefinition 10.5: potential energy is defined up to a constant
Conditions
the same zero level has to be used at both instants of one ledger, or the difference means nothing
the near surface form measures $y$ upward from the declared level, so a body below it has a negative potential energy and that is not an error
the general gravitational form has its zero fixed at infinite separation and cannot be moved, so it must not be mixed with the near surface form inside one problem
the spring form measures $x$ from the natural length of the spring, which is fixed by the spring rather than chosen by you
$$\boxed{\;U_g = mgy + C,\qquad \Delta U_g = mg\,\Delta y \ \text{is independent of } C,\qquad U_s = \tfrac{1}{2}kx^{2}\;}$$
You may add any fixed number you like to a potential energy without changing a single answer, because every answer contains a difference of two potential energies and the added number cancels. What you may not do is add one number at the start of the problem and a different one halfway through. The spring is the exception that proves the rule: its zero is not free, because the natural length is a property of the spring and not a choice of yours.
Why the choice of level cannot reach an answer
Suppose two students solve the same problem with zero levels a distance $c$ apart, so that where one writes $y$ the other writes $y + c$.
The first writes the ledger with $U_i = mgy_i$ and $U_f = mgy_f$; the second writes $U_i' = mg(y_i + c)$ and $U_f' = mg(y_f + c)$.
Every ledger contains the potential energies on opposite sides, so what actually enters the algebra is $U_i - U_f$, and for the second student that is $mg(y_i + c) - mg(y_f + c) = mg(y_i - y_f)$.
The constant $c$ has cancelled exactly, so both students get the same equation and the same answer. This is not an approximation and does not depend on the size of $c$; a level chosen a hundred kilometres underground works as well as one chosen at the floor.
The same swinging bob with two different choices for the zero of potential energy. The four potential energies printed are all different; the two drops between them are identical, and only the drop ever reaches an answer. Choosing the lowest point simply saves you from carrying a minus sign.
Looks like this, but is not
This is the freedom working: a swinging bob is analysed with the zero at the lowest point, giving 0.78 J at release and 0 J at the bottom, and then analysed again with the zero at the pivot, giving $-3.53\ \mathrm{J}$ and $-4.31\ \mathrm{J}$. The drop is 0.78 J both times and the speed at the bottom comes out identical.
This looks like the same freedom and is not: deciding to measure the spring compression from wherever the block first touches the spring rather than from the natural length. That is not a change of zero level, it is a change of the variable inside a squared term, and $\tfrac{1}{2}kx^{2}$ with the wrong $x$ is simply the wrong energy. A constant added to a potential energy cancels; a constant added inside the square does not.
Speed and string tension for a bob released at 35.0 degrees, done from two zero levels
A 0.400 kg bob hangs on a light string of length 1.10 m. It is pulled aside until the string makes 35.0 degrees with the vertical and released from rest. Find its speed at the lowest point and the tension there, working the potential energies once from the lowest point and once from the pivot.
Released from rest at 35.0 degrees from the vertical
The string is light and does not stretch, and air resistance is ignored
$g = 9.80\ \mathrm{m/s^{2}}$
Find
The speed at the lowest point and the tension in the string there.
Solution
The energy route gives the speed because the string tension varies all the way down and no force at an intermediate instant is wanted; the force route then converts that speed into the tension, which no ledger could have produced.
using the squared speed directly avoids rounding the speed and then squaring the rounded value
$$T = 5.34\ \mathrm{N}$$
against a weight of 3.92 N
Answer $$\boxed{v = 1.97\ \mathrm{m/s},\qquad T = 5.34\ \mathrm{N}}$$
Check
Independent check on the tension by a limiting case rather than by repeating it: if the bob were released from a very small angle, the speed at the bottom would tend to zero and the expression would collapse to $T = mg = 3.92\ \mathrm{N}$, the reading for a bob hanging still. Released from ninety degrees instead, $v^{2} = 2gL$ and $T = 3mg = 11.8\ \mathrm{N}$. The answer 5.34 N sits between those two bounds and nearer the lower one, as a thirty five degree release should.
Two extra lines were spent doing the potential energies a second way, purely to show the cancellation; in an exam only one of the two would be written.
The tension exceeding the weight even in this gentle swing is worth carrying: at the bottom of any swing the string is always pulled harder than the hanging weight, because it has to hold the bob up and bend its path at the same time.
Maximum compression when a 0.600 kg block is dropped onto a vertical spring
A 0.600 kg block is released from rest 0.250 m above the top of a vertical spring of stiffness 1200 N/m standing on the floor. Find how far the spring is compressed at the instant the block is momentarily at rest.
Given
Mass $m = 0.600\ \mathrm{kg}$, released from rest
Height of release above the free top of the spring $h = 0.250\ \mathrm{m}$
Spring stiffness $k = 1200\ \mathrm{N/m}$, standing vertically on the floor
At maximum compression the block is momentarily at rest
$g = 9.80\ \mathrm{m/s^{2}}$
Find
The maximum compression of the spring.
Solution
Both instants are chosen where the block is at rest, which deletes both kinetic energies and leaves an equation in the compression alone.
Choose the instants and the zero levels
$$i:\ \text{at rest, } 0.250\ \mathrm{m}\ \text{above the spring};\qquad f:\ \text{at rest, spring compressed by } x$$
picking two instants of rest is the whole trick, and it is available because the question asks for the maximum compression
$$U_g = 0 \ \text{at the final position};\qquad U_s = 0 \ \text{at the natural length}$$
the gravitational zero is chosen freely at the lowest point; the spring zero is not a choice, it is the natural length
Write the ledger
$$mg(h + x) = \tfrac{1}{2}kx^{2}$$
the block falls the gap plus the compression before stopping, and all of that potential energy ends up in the spring
$$(0.600)(9.80)(0.250 + x) = 600x^{2}$$
the commonest slip here is writing $mgh$ alone and forgetting that the block keeps descending while it squashes the spring
$$600x^{2} - 5.88x - 1.47 = 0$$
collected into a standard quadratic, with all quantities in SI so the roots come out in metres
the discriminant is dominated by the second term, which is the gravitational energy delivered from the drop
$$x = 0.0546\ \mathrm{m}\quad\text{or}\quad x = -0.0448\ \mathrm{m}$$
a negative compression would mean the spring stretched upward to catch a falling block, which nothing in the problem can do
$$x = 0.0546\ \mathrm{m} = 5.46\ \mathrm{cm}$$
three significant figures, matching the data
Answer $$\boxed{x = 0.0546\ \mathrm{m}}$$
Check
Independent check by putting the answer back into the two sides separately, which uses the number rather than the algebra: the left side is $mg(h+x) = (5.88)(0.3046) = 1.791\ \mathrm{J}$ and the right side is $\tfrac{1}{2}(1200)(0.0546)^{2} = 1.791\ \mathrm{J}$. A limiting check also passes: dropping the block from rest right on top of the spring, $h = 0$, gives $x = 2mg/k = 0.0098\ \mathrm{m}$, much smaller, as it should be.
One quadratic, which is the price of the extra $x$ inside the height. Forgetting that $x$ turns the problem into a linear one and produces a compression about ten per cent too small.
Whenever a body falls onto or through a spring, the height it falls is the gap plus the compression.
Checkpoint
§10.5 — moving the zero level of potential energy●●○○○
Thirty seconds. A student solves a ramp problem with the zero of gravitational potential energy at the top of the ramp instead of at the bottom, so every potential energy in the working comes out negative.
Given
The same physical problem is solved twice
The only difference is that the zero of gravitational potential energy has been moved to the top of the ramp
In the second solution both potential energies come out negative
Find
(a) Will the calculated final speed differ from the one obtained with the zero at the bottom?
Hint 1/4
Ask what the ledger actually does with the two potential energies. Are they used separately, or is only some combination of them used?
Hint 2/4
Both potential energies enter on opposite sides of the ledger, so what reaches the algebra is the difference $U_i - U_f$, and adding the same constant to both leaves that difference alone.
Hint 3/4
With the zero at the top the two values are $0$ and $-mgh$; with the zero at the bottom they are $+mgh$ and $0$. Either way the difference is $mgh$.
Hint 4/4
So the final speed is identical: the choice of level cannot reach the answer.
Show solution
The two solutions are written out side by side rather than argued about in words, because the cancellation is visible in two lines and an argument in words invites doubt.
Write both ledgers
$$\text{zero at the bottom:}\quad 0 + mgh = \tfrac{1}{2}mv^{2} + 0$$
the starting height is $h$ above the chosen level and the finish is on it
$$\text{zero at the top:}\quad 0 + 0 = \tfrac{1}{2}mv^{2} - mgh$$
the start is on the level and the finish is $h$ below it, so the final potential energy is negative
$$\text{both give}\quad v = \sqrt{2gh}$$
the two lines differ by moving one term across the equals sign, which changes nothing
Answer $$\boxed{\text{False: } v = \sqrt{2gh} \ \text{either way}}$$
Check
Independent check by choosing a third, deliberately absurd level: put the zero one kilometre underground. Then $U_i = mg(1000 + h)$ and $U_f = mg(1000)$, both enormous, and their difference is still $mgh$. If the answer depended on the level, an absurd level would produce an absurd speed, and it does not.
Because the level is free, choose it to make the arithmetic easy: the lowest point of the motion is almost always the cheapest choice, since it makes one of the two potential energies zero.
⚠ Changing the zero level halfway through a ledger
the initial state is often described from the ground and the final state from some other landmark, so each line is written from whichever level the sentence in the question suggested
wrong$$mgh_{\text{from the ground}} = \tfrac{1}{2}mv^{2} + mgh_{\text{from the table}}$$
right$$mgy_i = \tfrac{1}{2}mv^{2} + mgy_f\quad (y \text{ from one declared level})$$
⚠ Measuring a spring compression from the point of first contact
that is where the interesting part of the motion begins, so it feels like the natural origin, and the block really is at rest there in some problems
Power is how fast work is being done. Over an interval, take the joules and divide by the seconds. At an instant, take the part of the force that lies along the velocity and multiply it by the speed. The second form is the more useful one in mechanics questions because it needs no interval: a motor pulling a lift upward at a steady speed is delivering force times speed at every moment, and that number is what the electricity bill is charged for.
Why force times velocity is a power
Over a short time $\Delta t$ a body moves through a small displacement $\Delta \vec{s} = \vec{v}\,\Delta t$.
The work done by a force $\vec{F}$ over that displacement is $\Delta W = \vec{F}\cdot\Delta\vec{s} = \vec{F}\cdot\vec{v}\,\Delta t$.
Dividing by $\Delta t$ gives the rate at which that force is doing work, which is $\vec{F}\cdot\vec{v}$, and the angle in the scalar product is the angle between the force and the direction of motion at that moment.
A force perpendicular to the velocity therefore delivers zero power at every instant, which is the same statement as the earlier fact that a normal force or a string tension does no work.
The four checks, in the order that costs least. Units and limits take seconds and catch most of what goes wrong; the independent route is expensive and is kept for the answer that is worth marks. A number that survives all four is one you can defend without the marking scheme.
Looks like this, but is not
This is the power formula working: a cyclist holds a steady 8.00 m/s against a total resisting force of 32.0 N, so she is delivering $P = Fv = 256\ \mathrm{W}$ to the road, which is a plausible sustained output for a fit rider.
This looks like the same calculation and is not: the same cyclist accelerating from 8.00 m/s while still pushing against 32.0 N of resistance. Now $Fv$ with the resisting force gives only the power being wasted; the rider is delivering more than that, and the extra is going into kinetic energy. The formula $P = Fv$ is exact at every instant, but only if $F$ is the force whose power you actually want, and there are three different forces in this picture.
Power a cyclist delivers against 32.0 N of resistance at 8.00 m/s
A cyclist rides at a steady 8.00 m/s along level ground. The total resisting force from the air and the road is 32.0 N. Find the power she is delivering, and the energy she spends in a twenty minute ride at that speed.
Given
Steady speed $v = 8.00\ \mathrm{m/s}$ on level ground
Total resisting force $F = 32.0\ \mathrm{N}$, opposing the motion
Riding time 20.0 minutes at that speed
Steady speed means the kinetic energy is not changing
Find
The power delivered, and the energy spent in twenty minutes.
Solution
The instantaneous form is used rather than work over time, because the question gives a force and a speed and never mentions a distance.
Use the steady speed condition
$$\Delta K = 0 \;\Rightarrow\; P_{\text{rider}} = P_{\text{resistance}}$$
at constant speed nothing is accumulating as kinetic energy, so everything the rider puts in is being taken straight out again
$$P = Fv\cos 0^{\circ} = (32.0)(8.00)$$
the driving force is along the motion, so the cosine is one; the resisting force is opposite to it and removes at the same rate
$$P = 256\ \mathrm{W}$$
three significant figures, and the unit is watts because newtons times metres per second is joules per second
Turn the rate into a total
$$\Delta t = 20.0\ \mathrm{min} = 1200\ \mathrm{s}$$
converted to seconds first, because a watt is defined per second
$$W = P\,\Delta t = (256)(1200) = 3.07\times10^{5}\ \mathrm{J}$$
a rate times a time is a total, and the units confirm it: joules per second times seconds
Answer $$\boxed{P = 256\ \mathrm{W},\qquad W = 3.07\times10^{5}\ \mathrm{J}}$$
Check
Order of magnitude check against a known scale: a fit amateur cyclist sustains roughly 200 to 300 W for long periods and a professional a little over 400 W, so 256 W is squarely in range. A second check by a different route: in 1200 s at 8.00 m/s she covers 9600 m, and $W = Fd = (32.0)(9600) = 3.07\times10^{5}\ \mathrm{J}$, matching the first calculation without reusing the power.
Two lines. The only place to go wrong was leaving the time in minutes.
Three hundred thousand joules sounds enormous and is about the food energy in a small chocolate bar. That comparison is worth carrying, because it makes energy answers in the hundreds of kilojoules feel checkable rather than abstract.
Motor power for a 1150 kg lift, at steady speed and while accelerating
A lift together with its load has a total mass of 1150 kg. Find the power the motor delivers while raising it at a steady 2.40 m/s, and then the power at the instant when it is moving upward at 2.40 m/s while accelerating upward at 1.20 m/s squared. Ignore friction in the mechanism and the mass of the cable.
Given
Total mass $m = 1150\ \mathrm{kg}$
Case one: constant upward velocity $v = 2.40\ \mathrm{m/s}$
Case two: upward velocity 2.40 m/s at the instant of interest, with upward acceleration $a = 1.20\ \mathrm{m/s^{2}}$
Friction in the mechanism and the mass of the cable are ignored
$g = 9.80\ \mathrm{m/s^{2}}$
Find
The motor power in each case.
Solution
The force route supplies the cable tension first and the power formula converts it, because power needs a force and the only way to get the cable force is from the second law.
Case one: steady speed
$$\sum F_y = T - mg = 0 \;\Rightarrow\; T = mg = (1150)(9.80) = 11270\ \mathrm{N}$$
constant velocity means zero acceleration, so the cable pulls exactly as hard as gravity
$$P = Tv = (11270)(2.40) = 2.70\times10^{4}\ \mathrm{W}$$
the cable force is straight up and the motion is straight up, so the angle between them is zero
$$P = 27.0\ \mathrm{kW}$$
converted for readability only at the end
Case two: accelerating upward
$$\sum F_y = T - mg = ma \;\Rightarrow\; T = m(g + a) = (1150)(11.00)$$
the cable now has to hold the lift up and speed it up, so the tension exceeds the weight
$$T = 12650\ \mathrm{N}$$
about twelve per cent more than in the first case, which is the ratio 11.00 to 9.80
Independent check by splitting the second answer into two pieces that can be computed separately: at that instant the lift is gaining potential energy at $mgv = 27.0\ \mathrm{kW}$ and gaining kinetic energy at $mav = (1150)(1.20)(2.40) = 3.31\ \mathrm{kW}$. The two add to 30.3 kW, agreeing to rounding with the direct calculation, and the split explains where the extra power went. An order of magnitude check also passes: 27 kW is about thirty six horsepower, a reasonable lift motor.
Two short cases. The only new physics between them was one term on the right of the second law.
Power at an instant depends on the force at that instant, so a lift draws its largest power not at top speed but while accelerating near it.
Checkpoint
§10.6 — reading a power answer for unit errors●●○○○
Thirty seconds, no calculator. A student computes the power of a pump that raises 45.0 kg of water per minute through 12.0 m and writes 5290 W.
Given
Mass raised: 45.0 kg per minute
Height raised: 12.0 m
The student's answer: 5290 W
$g = 9.80\ \mathrm{m/s^{2}}$
Find
(a) Which check catches the error, and what is the correct power?
Hint 1/4
Run the cheapest check first. Look at what the student's number would be the answer to if the units were taken literally.
Hint 2/4
A watt is a joule per second. The quantity $mgh$ is an energy in joules, and turning it into a power means dividing by the number of seconds it took.
Hint 3/4
Here $mgh = (45.0)(9.80)(12.0) = 5290\ \mathrm{J}$, which is the energy per minute, and the time is 60.0 s, not 1 s.
Hint 4/4
So the unit check catches it, and the correct power is $5290/60.0 = 88.2\ \mathrm{W}$.
Show solution
The unit check is applied first because it costs seconds and needs no physics; only if it passes is it worth spending time on a second route.
a watt is a joule per second, so the minute has to be converted before the number can be called a power
Answer $$\boxed{P = 88.2\ \mathrm{W}}$$
Check
Order of magnitude check, independent of the arithmetic: 88 W is a bit more than a household filament bulb, which is the right size for a small pump lifting three quarters of a kilogram of water per second by about a storey. The rejected 5290 W would be a five kilowatt motor, the size of an industrial machine, for a job a garden pump does.
Whenever a rate is quoted per minute or per hour, convert it to per second before anything else.
⚠ Reporting an energy per minute as a power in watts
the data are given per minute, so the per minute rides silently through the calculation and the answer looks finished
⚠ Using the resisting force to find the power delivered while accelerating
at steady speed the two are equal, and that is the case every worked example starts with, so the equality gets remembered as a definition
wrong$$P_{\text{delivered}} = f v \quad\text{while } a \neq 0$$
right$$P_{\text{delivered}} = (f + ma)v \quad\text{while } a \neq 0$$
Writing an energy ledger that balances
Whenever the route chart sends you to the energy side, and always before substituting any numbers.
Name the two instants in words
Write, for example, released from rest at the top and passing the bottom. Every term afterwards is evaluated at one of these two moments, and a term that belongs to neither is a mistake.
Declare the zero level and the body
One body, one gravitational zero level, and if a spring is involved its natural length. Write them on the page; they cost one line and save the commonest sign error in the topic.
Write the full line before deleting anything
Put down the whole line with every term present, then cross out the ones that are zero and say why each is zero. Deleting on purpose is safe; never writing a term is not.
Handle every non conservative force explicitly
Friction contributes the friction force times the distance slid, with a minus sign. An applied pull contributes its force times the displacement along it. A force perpendicular to the motion contributes nothing, and saying so out loud is worth doing.
Solve, then check the size
Solve for the single unknown, put the unit on it, and compare it with something familiar before writing the box.
Where it goes wrong
Multiplying the friction force by the height dropped instead of the distance slid.
Counting gravity twice, once as a potential energy and once as a work done.
Using a different zero level in the initial and the final line of the same ledger.
Forgetting the second leg of friction on a there and back journey.
Setting up a circular motion problem without inventing a force
Whenever a body moves on a circular arc: a bend, a loop, a swing, an orbit, or a rotating ride.
Draw the body at the instant in question
Circular motion problems change from point to point, so the diagram is of one instant, not of the whole journey. Mark where the centre of the circle is from that point.
Draw only forces with an agent
Gravity from the Earth, the normal force from a surface, the tension from a string, friction from a road. If you cannot name the object exerting an arrow, do not draw it.
Point the radial axis at the centre and call it positive
Then read each force off the diagram with the sign its direction gives. At the top of a vertical circle the weight is positive; at the bottom it is negative. That is why the two cases look different.
Set the radial sum equal to the requirement
Write $\sum F_r = mv^{2}/r$. Nothing is added to the left because the path is curved; the curvature is entirely on the right.
Get the speed from wherever it lives
If the speed at that instant is not given, it usually comes from an energy ledger between that instant and one where the speed is known.
Where it goes wrong
Drawing an arrow labelled centripetal force, so that the requirement is counted twice.
Keeping the same sign for the weight at the top and the bottom of a vertical circle.
Using the length of a string as the radius when the circle drawn by the body is smaller, as in a conical swing.
Using the kinetic coefficient of friction for a rolling tyre, which is not sliding.
Scaffolding comes off
The common skeleton
Name the body and the two instants in words, and say which direction along the slope is positive.
Declare the zero of gravitational potential energy, normally the lower of the two positions.
Write the full ledger line with every term present, then delete the zero terms and say why each is zero.
Write the friction term as the friction force times the distance actually slid on that leg, with a minus sign.
Substitute numbers with units, solve for the single unknown, and box it with its unit.
Check the answer with a limiting case, normally by sending the coefficient of friction to zero.
1 · fully worked
How far a 1.60 kg block runs up a rough 25.0 degree slope at 6.00 m/s
A 1.60 kg block is launched up a 25.0 degree slope with an initial speed of 6.00 m/s. The coefficient of kinetic friction between block and slope is 0.180. Find how far along the slope it travels before stopping, and the height it gains.
Given
Mass $m = 1.60\ \mathrm{kg}$, initial speed $v_0 = 6.00\ \mathrm{m/s}$ up the slope
Slope angle $\theta = 25.0^{\circ}$ above the horizontal
Coefficient of kinetic friction $\mu_k = 0.180$
The block stops momentarily at the far point, so its kinetic energy there is zero
$g = 9.80\ \mathrm{m/s^{2}}$
Find
The distance travelled along the slope, and the height gained.
Solution
The energy route is chosen because two instants are described, the launch and the stop, and no time is asked for.
Name the instants and the zero level
$$i:\ \text{launch at } 6.00\ \mathrm{m/s};\qquad f:\ \text{momentarily at rest, a distance } d \text{ up the slope}$$
the far point is defined by the speed being zero there, which is what makes it a usable second instant
$$U = 0 \ \text{at the launch point};\qquad \text{positive direction: up the slope}$$
the lower of the two positions is the cheap choice, since it makes the initial potential energy zero
Write the ledger and delete the zeros
$$K_i + U_i + W_{\text{fric}} = K_f + U_f$$
the full line first, so that every deletion afterwards is a decision
the height is asked for separately because it is what the potential energy actually used
Answer $$\boxed{d = 3.14\ \mathrm{m},\qquad h = 1.33\ \mathrm{m}}$$
Check
Limiting check with a different value of one input: send $\mu_k \to 0$ and the formula gives $d = v_0^{2}/(2g\sin\theta) = 36.0/8.283 = 4.35\ \mathrm{m}$, further than the rough answer as it must be. A second, independent route is the force route: $a = g(\sin\theta + \mu_k\cos\theta) = 5.74\ \mathrm{m/s^{2}}$ and $d = v_0^{2}/2a = 3.14\ \mathrm{m}$, the same number from an equation the ledger never used.
One ledger line, one substitution and one division. The friction term needed the normal force, which cost the one extra line in the middle.
Gravity and friction both oppose the motion going up, so they add inside the bracket. On the way back down they oppose each other, and the bracket becomes a subtraction.
2 · you write the reasoning
Same launch, same block, but the slope is now smooth. A 1.60 kg block is launched up a 25.0 degree frictionless slope at 6.00 m/s. How far along the slope does it go before stopping? The steps are given; supply the reason for each one, then open the model reasons and compare.
reasoning
The full ledger is written before anything is deleted. Because the slope is smooth there is no friction work, so the other work term is absent from the start rather than deleted later; the string of the problem, the normal force, is perpendicular to the motion and does no work either.
reasoning
The block starts at the chosen zero level, so its initial potential energy is zero, and it stops at the far point, so its final kinetic energy is zero. What is left is the launch kinetic energy on one side and the height gained on the other, and that height is $d\sin\theta$ because the slope is at 25.0 degrees.
reasoning
The mass cancels on both sides, which is worth noticing rather than just doing: it means every block launched at 6.00 m/s runs the same distance up this slope, heavy or light.
reasoning
The number is larger than the 3.14 m of the rough case, which is the check: removing the only thing that was taking energy away has to let the block travel further, and the extra 1.21 m is what friction was costing.
3 · find the buried error
Back to the rough slope: the 1.60 kg block launched at 6.00 m/s up the 25.0 degree slope with $\mu_k = 0.180$, which we found runs 3.14 m up before stopping. It then slides back down. A student computes the speed with which it returns to the launch point. Two of the lines below are wrong. Find the first one.
the two buried errors (2)
⚠ step 2
The friction term keeps its sign from the upward leg. On the way down the block slides downward, so friction now acts up the slope and opposes gravity instead of joining it: the correct value is $a_{\text{down}} = g(\sin\theta - \mu_k\cos\theta) = 2.54\ \mathrm{m/s^{2}}$.
Friction is remembered as a property of the surface, which does not change, rather than as a force whose direction is set by the direction of sliding, which reverses at the turning point.
right
Redraw the diagram for the downward leg before writing the second acceleration. The weight component is unchanged, the friction arrow has flipped, so the two now subtract: $a_{\text{down}} = 9.80(0.4226 - 0.1632) = 2.54\ \mathrm{m/s^{2}}$, and $v = \sqrt{2(2.543)(3.136)} = 3.99\ \mathrm{m/s}$.
⚠ step 4
The friction loss is computed with the vertical height 1.33 m instead of the distance slid, and it counts only one leg. Friction is paid along the path, and the round trip covers $2d = 6.27\ \mathrm{m}$.
The height is the number the gravity term has just used, so it is the one in working memory; and the symmetry of the picture suggests that a round trip should be accounted for once.
right
Use the path length on each leg: $|W_{\text{fric}}| = \mu_k mg\cos\theta \times 2d = (2.559)(6.272) = 16.0\ \mathrm{J}$. That is over half of the launch energy of 28.8 J, which is exactly why the block comes back at 3.99 m/s rather than 6.00 m/s.
4 · the bare problem
§10.4 — round trip on a rough slope●●●●○
A block is launched up a rough slope, runs to a stop, and slides back to where it started. Nothing else acts on it.
Given
Mass $m = 2.20\ \mathrm{kg}$, launched up the slope at $v_0 = 4.50\ \mathrm{m/s}$
Slope angle $20.0^{\circ}$ above the horizontal
Coefficients of friction: $\mu_k = \mu_s = 0.250$, and $\tan 20.0^{\circ} = 0.364$ exceeds 0.250, so the block does slide back
$g = 9.80\ \mathrm{m/s^{2}}$
Find
(a) How far along the slope does the block travel before stopping?
(b) With what speed does it return to the launch point?
(c) How much energy has been converted to thermal energy over the round trip?
Hint 1/4
Treat the two legs separately. The geometry is the same on both, but one force on the block points a different way on the second leg than on the first.
Hint 2/4
For each leg write the ledger $K_i + U_i + W_{\text{fric}} = K_f + U_f$, with the friction term as the friction force times the distance slid on that leg, and remember that the normal force on a slope is $mg\cos\theta$.
Hint 3/4
The data again: $m = 2.20\ \mathrm{kg}$, $v_0 = 4.50\ \mathrm{m/s}$, $\theta = 20.0^{\circ}$, $\mu_k = 0.250$, $g = 9.80\ \mathrm{m/s^{2}}$. Up the slope the bracket is $\sin\theta + \mu_k\cos\theta$; down the slope it is $\sin\theta - \mu_k\cos\theta$.
Hint 4/4
The answers are 1.79 m, 1.94 m/s and 18.1 J.
Show solution
Each leg gets its own ledger because the friction term changes sign between them; trying to write one ledger for the whole trip works only if the friction distance is entered as the total path length, and that is exactly the step most people get wrong.
Independent check by a completely different bookkeeping: the block began with $\tfrac{1}{2}(2.20)(4.50)^{2} = 22.28\ \mathrm{J}$ and returned to the same height with $\tfrac{1}{2}(2.20)(1.9388)^{2} = 4.13\ \mathrm{J}$. The difference is 18.14 J, which must all be thermal since the potential energy came back exactly. That number was reached without using the friction force at all, and it agrees with the 18.14 J computed from the force, since the two are the same calculation seen from opposite ends.
The ratio of the return speed to the launch speed is the square root of the ratio of the two brackets, $\sqrt{0.1071/0.5769} = 0.431$, and it does not depend on the launch speed at all.
Full exam-style question
Energy needed to move a 620 kg satellite from a 7.20 Mm orbit to a 9.60 Mm orbitexam format
A 620 kg satellite is in a circular orbit of radius $7.20\times10^{6}\ \mathrm{m}$ about the Earth. (a) Find its orbital speed. (b) Find its kinetic energy, its gravitational potential energy and its total energy. (c) Find the energy that must be supplied to place it in a circular orbit of radius $9.60\times10^{6}\ \mathrm{m}$, and state whether it is then moving faster or slower.
Final circular orbit radius $r_2 = 9.60\times10^{6}\ \mathrm{m}$
Earth mass $M = 5.97\times10^{24}\ \mathrm{kg}$, $G = 6.674\times10^{-11}\ \mathrm{N\,m^{2}/kg^{2}}$, so $GM = 3.984\times10^{14}\ \mathrm{m^{3}/s^{2}}$
The zero of gravitational potential energy is at infinite separation
Find
The orbital speed, the three energies, the energy required for the transfer, and whether the satellite speeds up or slows down.
Solution
The force route supplies the speed, because a circular orbit is a radial equation. The energy route then handles the transfer, because it names two states and asks for an energy, and no ledger could have produced the speed while no radial equation could have produced the transfer energy.
Independent check on the transfer energy by a route that never uses the total energy formula: the kinetic energy falls from $1.715\times10^{10}$ to $\tfrac{1}{2}(620)(4.150\times10^{7}) = 1.287\times10^{10}\ \mathrm{J}$, a drop of $4.29\times10^{9}\ \mathrm{J}$, while the potential energy rises from $-3.431\times10^{10}$ to $-2.573\times10^{10}$, a gain of $8.58\times10^{9}\ \mathrm{J}$. The net change is $8.58 - 4.29 = 4.29\times10^{9}\ \mathrm{J}$, which is the boxed value reached without the total energy formula. A structural check also passes: $E_1 = -K_1$ exactly, as the circular orbit relation demands.
Four stages, but only two ideas: one radial equation and one pair of energies. The rest is substitution.
The counterintuitive part is worth stating plainly: you burn fuel to go higher and you end up slower.
Practice
A · concept 3 questions
1§10.2 — work done by the inward force on a circular path●●○○○
A two mark opener of the kind that starts a paper. Decide the verdict, and be ready to justify it in one sentence, because the sentence is where the marks are.
Given
The claim: in uniform circular motion the inward force does work on the body, because it continually changes the direction of the velocity.
Find
(a) True or false, and why?
Hint 1/4
Work is not about changing something; it is about a specific geometric relationship between a force and a displacement. Recall what that relationship is.
Hint 2/4
The work done by a force over a small displacement is $Fd\cos\theta$, and the power it delivers is $Fv\cos\theta$, where the angle is measured between the force and the direction of motion.
Hint 3/4
In uniform circular motion the net inward force points at the centre while the velocity is tangent to the circle, so the angle between them is 90 degrees at every instant and the cosine is zero.
Hint 4/4
So the claim is false: the inward force changes the direction of the velocity without changing its size, and only changes of size cost energy.
Show solution
The power form is used rather than the work form, because it settles the question at every instant rather than over an interval and so leaves no room for a claim that the works cancel.
the radius and the tangent are perpendicular for a circle, which is a geometric fact rather than a physical assumption
$$\Delta K = \int P\,dt = 0 \;\Rightarrow\; v = \text{constant}$$
zero power at every instant is stronger than zero net work over a lap, and it is what the phrase uniform circular motion means
Answer $$\boxed{\text{False: } P = Fv\cos 90^{\circ} = 0}$$
Check
Independent check by the converse: if the inward force did work, the kinetic energy would change and the speed would not be constant, contradicting the word uniform in the premise. The claim therefore destroys its own hypothesis, which is the strongest kind of refutation available.
The same argument disposes of the tension in a swinging string, the normal force on a car in a bend, and gravity on a circular orbit: all three are perpendicular to the motion and all three deliver zero power.
2§10.4 — what is conserved when a block slides down a rough ramp●●●○○
A block slides from rest down a straight rough ramp and reaches the bottom moving. A paper asks which single statement about the energies is correct.
Given
A block slides from rest down a rough straight ramp
It arrives at the bottom with a measurable speed
The only forces are gravity, the normal force and kinetic friction
Find
(a) Which statement is correct?
Hint 1/4
Write the ledger for this motion before reading the options. Which terms are present, and which of them is not zero?
Hint 2/4
The line is $K_i + U_i + W_{\text{fric}} = K_f + U_f$, and $W_{\text{fric}}$ is negative and not zero because the ramp is rough.
Hint 3/4
With $K_i = 0$ and the zero level at the bottom, the line reads $mgh - f_kd = \tfrac{1}{2}mv_f^{2}$, so the kinetic energy gained is less than the potential energy lost.
Hint 4/4
So the mechanical energy falls by exactly the friction loss, and the total energy including the thermal share is what stays constant.
Show solution
The ledger is written before the options are read, because reading four plausible sentences first makes each one sound like a possible answer.
Write the line and read it
$$0 + mgh - f_kd = \tfrac{1}{2}mv_f^{2} + 0$$
the friction term is negative and non zero, so the two sides cannot be the mechanical energy alone
$$\Delta(K + U) = -f_kd < 0$$
the mechanical energy falls, and the amount it falls by is exactly the friction loss
$$\Delta(K + U) + E_{\text{thermal}} = 0$$
nothing is destroyed; the ledger balances once the thermal share is written down
Independent check by a limiting case: make the ramp smooth, so $f_k = 0$, and the statement collapses to conservation of mechanical energy, which is the case already known to be true. Any correct statement about the rough case must reduce to the smooth case in that limit, and this one does.
Conserved and constant are not loose words here. Total energy is conserved in every problem in this course; mechanical energy is conserved only when the non conservative forces do no work.
3§10.3 — comparing two circular orbits●●●○○
Two identical satellites are in circular orbits about the Earth, one at radius $r$ and one at radius $2r$. Decide which comparison is right, without computing anything.
Given
Two identical satellites, each in its own circular orbit about the Earth
Orbit radii are $r$ and $2r$ measured from the centre of the Earth
The relevant relations are $v = \sqrt{GM/r}$ and $E = -GmM/(2r)$
Find
(a) Which comparison between the outer and the inner satellite is correct?
Hint 1/4
Write down the two relations that depend on the radius before looking at the options, and notice that one of them is a negative quantity.
Hint 2/4
The orbital speed goes as $r^{-1/2}$ and the total energy goes as $-1/(2r)$, so doubling the radius divides the speed by $\sqrt{2}$ and halves the size of the total energy.
Hint 3/4
With radii $r$ and $2r$: $v_{\text{outer}} = v_{\text{inner}}/\sqrt{2} = 0.707\,v_{\text{inner}}$, and $E_{\text{outer}} = E_{\text{inner}}/2$, which for a negative number means closer to zero.
Hint 4/4
So the outer satellite is slower and yet has the greater total energy, since halving a negative number raises it.
Show solution
Ratios are formed rather than numbers computed, because the question compares two cases and every constant cancels in a ratio.
Independent check with the potential and kinetic energies taken separately: the outer satellite has half the kinetic energy, because $K = GmM/(2r)$, and a potential energy half as negative, a gain of $GmM/(2r)$. The net change is a gain of $GmM/(4r)$, positive, agreeing with the ratio argument by a completely different accounting.
Slower and higher energy at the same time is the standard trap in orbit questions, and it has a one line reason: the potential energy gains twice what the kinetic energy loses.
B · computation 6 questions
1§10.4 — energy a rough slope removes from a skier●●●○○
A skier of total mass 62.0 kg starts from rest at the top of a straight run whose bottom is 8.50 m lower, and reaches the bottom at 11.0 m/s.
Given
Mass $m = 62.0\ \mathrm{kg}$, starting from rest
Vertical drop $h = 8.50\ \mathrm{m}$
Speed at the bottom $v_f = 11.0\ \mathrm{m/s}$
Length of the run along the surface $d = 22.0\ \mathrm{m}$
$g = 9.80\ \mathrm{m/s^{2}}$
Find
(a) How much energy did friction and air resistance remove?
(b) What was the average resisting force along the run?
Hint 1/4
You are not being asked to model the friction; you are being asked what is missing from a ledger. Set up the ledger with the loss as the unknown.
Hint 2/4
Write $K_i + U_i + W_{\text{other}} = K_f + U_f$ with the zero of potential energy at the bottom, and solve for $W_{\text{other}}$.
Hint 3/4
The data again: $m = 62.0\ \mathrm{kg}$, $h = 8.50\ \mathrm{m}$, $v_f = 11.0\ \mathrm{m/s}$, $d = 22.0\ \mathrm{m}$. Compute $mgh$ and $\tfrac{1}{2}mv_f^{2}$ and subtract; then divide the difference by the distance along the surface.
Hint 4/4
The loss is $1.41\times10^{3}\ \mathrm{J}$ and the average resisting force is 64.3 N.
Show solution
The loss is left as the unknown in the ledger rather than modelled from a coefficient, because no coefficient is given and the run is not of uniform steepness.
Independent check on the size of the force: the skier weighs 608 N, so the resisting force is about a tenth of the weight, which is what a coefficient of roughly 0.1 on snow would give and is far too small to be a mistake in an order of magnitude. A second check on the energy: the skier keeps 3751 of 5165 joules, that is 73 per cent, so a smooth run would have delivered $\sqrt{5165/3751} = 1.17$ times the speed, about 12.9 m/s, which is only slightly faster and matches a run that is nearly frictionless.
Whenever a problem gives you both a drop and an arrival speed and no coefficient, it is asking you to run the ledger backwards.
2§10.2 — banking angle for a bend taken without friction●●●○○
A road is being designed so that a car can round a bend of radius 150 m at 22.0 m/s with no reliance on friction at all, which is what makes the bend safe in the wet.
Given
Bend radius $r = 150\ \mathrm{m}$
Design speed $v = 22.0\ \mathrm{m/s}$
Friction is to play no part, so the road surface must be banked
$g = 9.80\ \mathrm{m/s^{2}}$
Find
(a) At what angle to the horizontal must the road be banked?
(b) For a 1400 kg car at the design speed, what is the normal force from the road?
Hint 1/4
With no friction there are only two forces on the car, and one of them is vertical. Draw them and ask which direction their sum has to point.
Hint 2/4
Take a vertical axis and a horizontal radial axis. The vertical equation is $F_N\cos\theta - mg = 0$ and the radial one is $F_N\sin\theta = mv^{2}/r$; dividing one by the other eliminates the normal force.
Hint 3/4
With $v = 22.0\ \mathrm{m/s}$, $r = 150\ \mathrm{m}$ and $g = 9.80\ \mathrm{m/s^{2}}$, the division gives $\tan\theta = v^{2}/(rg) = 484/1470$.
Hint 4/4
So $\theta = 18.2^{\circ}$, and then $F_N = mg/\cos\theta = 13720/0.9500 = 1.44\times10^{4}\ \mathrm{N}$.
Show solution
The two axes are kept horizontal and vertical rather than tilted with the road, because the acceleration of the car is horizontal and choosing axes along the acceleration keeps one equation free of it.
Write the two axis equations
$$\sum F_y = F_N\cos\theta - mg = 0$$
the car goes round a level circle, so it has no vertical acceleration however steeply the road is banked
$$\sum F_r = F_N\sin\theta = m\frac{v^{2}}{r}$$
the horizontal component of the normal force is the entire inward force, since friction has been excluded by design
Limiting check on the formula: as the design speed goes to zero the required angle goes to zero, a flat road, which is right because a stationary car needs no inward force. As the radius grows the angle also falls, which is why motorway curves are barely banked while a velodrome is nearly vertical. A unit check passes too: metres squared per second squared divided by metres times metres per second squared is dimensionless, as a tangent must be.
A banked bend has exactly one speed at which friction is not needed. Above it the car needs friction pointing down the slope, below it friction pointing up, which is why an icy banked bend is dangerous at both ends of the speed range.
3§10.5 — speed a spring gives a block on a smooth track●●○○○
A spring of stiffness 260 N/m is compressed 0.140 m and used to launch a 0.320 kg block along a smooth horizontal track.
Block mass $m = 0.320\ \mathrm{kg}$, released from rest
The track is horizontal and smooth, and the spring is ideal
Find
(a) How much energy is stored in the compressed spring?
(b) What speed does the block have once it has left the spring?
Hint 1/4
The spring is the only thing that had energy at the start and the block is the only thing that has it at the end. Nothing else changes.
Hint 2/4
The stored energy is $U_s = \tfrac{1}{2}kx^{2}$, measured from the natural length, and the ledger on a level smooth track is simply $U_s = \tfrac{1}{2}mv^{2}$.
Hint 3/4
With $k = 260\ \mathrm{N/m}$, $x = 0.140\ \mathrm{m}$ and $m = 0.320\ \mathrm{kg}$: compute the stored energy first, then solve the ledger for the speed.
Hint 4/4
The stored energy is 2.55 J and the speed is 3.99 m/s.
Show solution
The energy route is the only sensible one because the spring force varies through the launch, so no constant acceleration exists and no kinematic formula applies.
using the unrounded stored energy so that the rounding happens only once, in the box
Answer $$\boxed{U_s = 2.55\ \mathrm{J},\qquad v = 3.99\ \mathrm{m/s}}$$
Check
Independent check by the average force route: the spring force falls linearly from $kx = 36.4\ \mathrm{N}$ to zero, so the average force is 18.2 N over 0.140 m, giving $W = 2.55\ \mathrm{J}$ without using the one half formula at all. The speed of about 4 m/s is also the right size for a hand compressed spring, roughly a brisk walking pace.
Quadrupling the stored energy only doubles the speed, because the energy goes as the square of the speed.
4§10.4 — crate dragged by a rope at an angle over a rough floor●●●●○
A 24.0 kg crate is dragged 6.00 m across a level floor by a rope pulling with 95.0 N at 30.0 degrees above the horizontal.
Given
Mass $m = 24.0\ \mathrm{kg}$, starting from rest
Rope force 95.0 N at 30.0 degrees above the horizontal, constant
Displacement 6.00 m horizontally along the floor
Coefficient of kinetic friction $\mu_k = 0.180$
$g = 9.80\ \mathrm{m/s^{2}}$
Find
(a) How much work does the rope do?
(b) How much work does friction do?
(c) What is the speed of the crate after 6.00 m?
Hint 1/4
Three forces do work or fail to do work here, and one of them changes the size of another. Sort out the vertical direction before touching the horizontal one.
Hint 2/4
Work by a constant force is $Fd\cos\theta$. The normal force follows from the vertical equation $F_N + F\sin\theta - mg = 0$, because the upward tilt of the rope helps hold the crate up, and then $f_k = \mu_k F_N$.
Hint 3/4
With $F = 95.0\ \mathrm{N}$ at 30.0 degrees, $m = 24.0\ \mathrm{kg}$, $d = 6.00\ \mathrm{m}$ and $\mu_k = 0.180$: the vertical component of the rope is 47.5 N, so the normal force is $235.2 - 47.5 = 187.7\ \mathrm{N}$.
Hint 4/4
The rope does 494 J, friction does $-203\ \mathrm{J}$, and the crate reaches 4.92 m/s.
Show solution
The vertical equation is solved before any work is computed, because the friction force depends on the normal force and the normal force depends on the tilt of the rope.
The vertical direction first
$$\sum F_y = F_N + F\sin 30.0^{\circ} - mg = 0$$
the crate slides along the floor and does not lift off it, so the vertical acceleration is zero
Independent check by the force route: the net horizontal force is $F\cos 30.0^{\circ} - f_k = 82.3 - 33.8 = 48.5\ \mathrm{N}$, so $a = 48.5/24.0 = 2.02\ \mathrm{m/s^{2}}$ and $v = \sqrt{2(2.02)(6.00)} = 4.92\ \mathrm{m/s}$. The same number from a chain of reasoning that never computed a work.
Tilting the rope upward is a genuine trade: it wastes part of the pull, since only the cosine does work, but it also lightens the crate on the floor and cuts the friction.
5§10.2 — least release height for a loop the loop●●●●○
A small block is released from rest on a smooth track and runs down into a vertical circular loop of radius 6.50 m.
The block is released from rest at height $h$ above the bottom of the loop
At the top of the loop the block must still be touching the track
$g = 9.80\ \mathrm{m/s^{2}}$
Find
(a) What is the least speed the block can have at the top of the loop?
(b) From what least height must it be released?
Hint 1/4
Two different pieces of physics are needed, one at the top of the loop and one between the release point and the top. Decide which is which before starting.
Hint 2/4
At the top, the radial equation with the track on the point of losing contact: $F_N \to 0$, so $mg = mv_{\text{top}}^{2}/R$. Between release and the top, the ledger on a smooth track: $mgh = \tfrac{1}{2}mv_{\text{top}}^{2} + mg(2R)$.
Hint 3/4
With $R = 6.50\ \mathrm{m}$ and $g = 9.80\ \mathrm{m/s^{2}}$: get $v_{\text{top}}$ from the first relation, then feed its square into the second, remembering that the top of the loop is a height $2R$ above the bottom.
Hint 4/4
The least speed at the top is 7.98 m/s and the least release height is 16.3 m, which is two and a half loop radii.
Show solution
The radial equation is applied at the top and the ledger between the two heights, because the condition for staying on the track is a statement about forces while the link between the two heights is a statement about energy.
Independent check on the general result rather than on this loop: for any smooth loop, $v_{\text{top}}^{2} = gR$ gives $h = R/2 + 2R = 2.5R$, and $2.5 \times 6.50 = 16.3\ \mathrm{m}$. A physical check also passes: the release point has to be above the top of the loop, which is at 13.0 m, and it is, by a quarter of a radius, which is the height needed to buy the speed.
The commonest wrong answer here is $2R$, obtained by requiring the block merely to reach the top. Reaching the top with zero speed is not enough: it would have left the track long before, because contact requires a minimum speed rather than a minimum height.
6§10.6 — engine power for a car climbing a hill●●●○○
A 1200 kg car climbs a straight hill inclined at 5.00 degrees at a steady 20.0 m/s. Resistance from the air and the road totals 420 N, opposing the motion.
(a) What driving force must the engine supply at the wheels?
(b) What power is that, in kilowatts?
Hint 1/4
At a steady speed on a slope nothing is accumulating as kinetic energy, so the driving force is fighting two separate things at once. Name them before adding.
Hint 2/4
Along the slope: $F_{\text{drive}} - mg\sin\theta - F_{\text{resist}} = 0$ at constant speed, and then $P = F_{\text{drive}}v$ because the driving force lies along the motion.
Hint 3/4
With $m = 1200\ \mathrm{kg}$, $\theta = 5.00^{\circ}$, $F_{\text{resist}} = 420\ \mathrm{N}$ and $v = 20.0\ \mathrm{m/s}$: the weight component along the slope is $11760\sin 5.00^{\circ} = 1025\ \mathrm{N}$.
Hint 4/4
So the driving force is 1445 N and the power is 28.9 kW.
Show solution
The force route supplies the driving force and the power formula converts it, because the question gives a speed and wants a rate.
newtons times metres per second is joules per second, which is watts
Answer $$\boxed{F_{\text{drive}} = 1.44\times10^{3}\ \mathrm{N},\qquad P = 28.9\ \mathrm{kW}}$$
Check
Independent check by splitting the power into its two jobs: the car gains height at $v\sin\theta = 1.743\ \mathrm{m/s}$, so it is gaining potential energy at $mgv\sin\theta = 20.5\ \mathrm{kW}$, while it wastes $(420)(20.0) = 8.4\ \mathrm{kW}$ against resistance. The two add to 28.9 kW. An order of magnitude check also passes: a small family car develops perhaps 70 kW at full power, so using 29 kW to hold 72 km per hour up a gentle hill is sensible.
Most of the power here goes into the hill, not into the air. On level ground the same car at the same speed would need only 8.4 kW, which is why a modest engine is enough on the flat and strains on a climb.
C · exam level 4 questions
1§10.4 — spring, rough patch and smooth incline in one problem●●●●○
A three stage problem of the kind that carries ten marks. A 0.400 kg block is held against a spring of stiffness 750 N/m, compressed 0.120 m, on a horizontal surface.
Given
Block mass $m = 0.400\ \mathrm{kg}$, released from rest
Rough horizontal patch of length $d = 1.50\ \mathrm{m}$ with $\mu_k = 0.300$
Beyond it, a smooth incline at $35.0^{\circ}$ above the horizontal
$g = 9.80\ \mathrm{m/s^{2}}$
Find
(a) What speed does the block have as it leaves the spring?
(b) What speed does it have at the foot of the incline?
(c) How far along the incline does it travel before stopping?
Hint 1/4
Three stages with different physics in each: a spring that does work, a rough level patch that removes energy, and a smooth climb that converts it. Handle them as three ledgers rather than one.
Hint 2/4
The stored energy is $\tfrac{1}{2}kx^{2}$; the friction loss is $\mu_k mgd$ on the level; and on the smooth incline the remaining kinetic energy buys a height, with the distance along the slope being that height divided by $\sin\theta$.
Hint 3/4
The data again: $m = 0.400\ \mathrm{kg}$, $k = 750\ \mathrm{N/m}$, $x = 0.120\ \mathrm{m}$, $d = 1.50\ \mathrm{m}$, $\mu_k = 0.300$, $\theta = 35.0^{\circ}$. The stored energy is 5.40 J and the friction loss is 1.76 J.
Hint 4/4
The answers are 5.20 m/s, 4.26 m/s and 1.62 m along the incline.
Show solution
The journey is split at each change of surface, because the friction term must be multiplied by the length of the rough patch alone.
Independent check on the last stage by the force route, which the ledger never used: on a smooth 35.0 degree incline the deceleration is $g\sin\theta = 5.62\ \mathrm{m/s^{2}}$, and $L = v_1^{2}/(2a) = 18.18/11.24 = 1.62\ \mathrm{m}$. A structural check also passes: the friction removed 1.76 J of 5.40 J, that is 33 per cent of the energy, and the speed fell from 5.20 to 4.26 m/s, a ratio of 0.820 whose square is 0.673, so 33 per cent of the energy has indeed gone.
Each stage handed exactly one number to the next, and that number was an energy rather than a speed.
2§10.2 — conical pendulum with the radius that is not the string●●●●○
A 0.250 kg ball on a light string 0.850 m long is swung so that it travels in a horizontal circle with the string making a constant angle of 30.0 degrees with the vertical.
Given
Ball mass $m = 0.250\ \mathrm{kg}$
String length $L = 0.850\ \mathrm{m}$, making $30.0^{\circ}$ with the vertical
The ball travels in a horizontal circle at constant speed
$g = 9.80\ \mathrm{m/s^{2}}$
Find
(a) What is the tension in the string?
(b) What is the speed of the ball?
(c) How long does one revolution take?
Hint 1/4
Draw the ball at one instant and mark the centre of the circle it is travelling on. That centre is not at the top of the string, and the radius is not the string length.
Hint 2/4
Two axes: vertically $T\cos\theta - mg = 0$, since the ball stays at the same height; radially $T\sin\theta = mv^{2}/r$, with $r = L\sin\theta$ the radius of the horizontal circle.
Hint 3/4
With $m = 0.250\ \mathrm{kg}$, $L = 0.850\ \mathrm{m}$ and $\theta = 30.0^{\circ}$: the radius is $0.850\sin 30.0^{\circ} = 0.425\ \mathrm{m}$, and $mg = 2.45\ \mathrm{N}$.
Hint 4/4
The tension is 2.83 N, the speed is 1.55 m/s, and one revolution takes 1.72 s.
Show solution
The vertical equation is solved first because it contains only the tension, so it delivers that unknown immediately; the radial equation then has only the speed left in it.
Vertical: the ball does not rise or fall
$$T\cos\theta - mg = 0 \;\Rightarrow\; T = \frac{mg}{\cos 30.0^{\circ}}$$
the circle is horizontal, so the vertical acceleration is zero even though the ball is accelerating
Independent check by eliminating the tension algebraically before substituting, which is a different path through the same two equations: dividing the radial equation by the vertical one gives $\tan\theta = v^{2}/(rg)$, so $v^{2} = rg\tan\theta = (0.425)(9.80)(0.5774) = 2.405\ \mathrm{m^{2}/s^{2}}$, matching. A limiting check also passes: as the angle goes to zero the tension tends to the weight and the speed tends to zero, which is a ball hanging still.
The same trap appears in the rotor ride, the banked bend and the swinging seat of a fairground ride: the radius of the circle is the horizontal distance from the axis, and it is almost never the length of the thing holding the body up.
3§10.3 — escape speed from a moon of known surface gravity●●●○○
A probe is to be launched from the surface of a moon whose surface free fall acceleration is $1.62\ \mathrm{m/s^{2}}$ and whose radius is $1.74\times10^{6}\ \mathrm{m}$.
Given
Surface free fall acceleration $g_m = 1.62\ \mathrm{m/s^{2}}$
Moon radius $R = 1.74\times10^{6}\ \mathrm{m}$
The probe is to leave with no further propulsion and never return
Find
(a) What is the escape speed from the surface?
Hint 1/4
You are not given the mass, so the general formula in terms of $GM$ is not directly usable. Look for a relation that lets the surface gravity stand in for $GM$.
Hint 2/4
At the surface $g_m = GM/R^{2}$, so $GM = g_mR^{2}$. Substituting into $v_{\text{esc}} = \sqrt{2GM/R}$ gives $v_{\text{esc}} = \sqrt{2g_mR}$.
Hint 3/4
With $g_m = 1.62\ \mathrm{m/s^{2}}$ and $R = 1.74\times10^{6}\ \mathrm{m}$: the product $2g_mR$ is $5.64\times10^{6}\ \mathrm{m^{2}/s^{2}}$, and one square root finishes it.
Hint 4/4
So the escape speed is $2.37\times10^{3}\ \mathrm{m/s}$, a little over 2.3 km per second.
Show solution
The surface gravity is used to replace $GM$ rather than to compute the mass and then put it back, because the substitution is one line while the detour is three and introduces two chances to slip an exponent.
Independent check by the ratio to a circular orbit skimming the surface: that orbit has $v = \sqrt{g_mR} = 1.68\times10^{3}\ \mathrm{m/s}$, and the escape speed should be $\sqrt{2}$ times it, which is $2.37\times10^{3}\ \mathrm{m/s}$. The ratio was proved from the general forms and never used these numbers.
The resemblance between $\sqrt{2g_mR}$ and the familiar $\sqrt{2gh}$ is a coincidence of algebra rather than of physics: the field is not constant over the escape, and the radius here is not a height climbed.
4§10.2 — finding the fault in a loop the loop solution●●●●○
A student is asked for the least height from which a block must be released on a smooth track so that it completes a vertical loop of radius 2.00 m.
Given
The problem: smooth track, vertical loop of radius $R = 2.00\ \mathrm{m}$, block released from rest at height $h$ above the bottom of the loop, $g = 9.80\ \mathrm{m/s^{2}}$
Step 1: at the top the track push is on the point of vanishing, so $mg = mv_{\text{top}}^{2}/R$ and $v_{\text{top}}^{2} = gR = 19.6\ \mathrm{m^{2}/s^{2}}$
Step 2: energy from the release point to the top gives $mgh = \tfrac{1}{2}mv_{\text{top}}^{2}$
Step 4: this is below the top of the loop at 4.00 m, but the block picks up speed on the way down, so it can still get round
Find
(a) Which is the first wrong line, and what is the correct release height?
Hint 1/4
Take each line and ask what it is claiming, then test the claim on its own. Two of the four claims cannot both be right about the same block.
Hint 2/4
The ledger between two instants has to include every energy at both of them. At the top of the loop the block is at a height $2R$ above the bottom, so the ledger reads $mgh = \tfrac{1}{2}mv_{\text{top}}^{2} + mg(2R)$.
Hint 3/4
With $R = 2.00\ \mathrm{m}$ and $v_{\text{top}}^{2} = 19.6\ \mathrm{m^{2}/s^{2}}$: the correct height is $h = R/2 + 2R = 2.5R$.
Hint 4/4
So Step 2 is the first wrong line, and the correct release height is 5.00 m.
Show solution
Each line is tested against a fact that does not come from the other lines, because a fake solution is internally consistent by construction and can only be broken from outside.
the first term is $R/2$, so the general result is $h = 2.5R$ for any smooth loop
$$\text{Step 4 is also wrong}: \ h = 1.00\ \mathrm{m} < 2R = 4.00\ \mathrm{m}$$
on a smooth track a block released below the top of the loop cannot reach it at all, so the student's excuse contradicts conservation of energy rather than rescuing it
Answer $$\boxed{\text{Step 2 is the first fault};\qquad h_{\min} = 5.00\ \mathrm{m} = 2.5R}$$
Check
Independent check on the repaired answer by a limit that needs none of the algebra: the release height must exceed the height of the top of the loop, $2R = 4.00\ \mathrm{m}$, since a smooth track never gives back more than it was given. The corrected 5.00 m clears that bar, the student's 1.00 m does not, and this test alone would have flagged the solution.
A solution that ends below an obvious floor is wrong even before the error is located. Building that floor first, in one line, is faster than checking the algebra and catches exactly this class of missing term.
D · interleaved 4 questions
1§10.4 — a ball thrown from a roof●●●○○
No hint is given about which method this needs. A ball is thrown from the edge of a flat roof 12.0 m above the ground, leaving the hand at 14.0 m/s at 40.0 degrees above the horizontal.
Given
Launch speed 14.0 m/s at 40.0 degrees above the horizontal
Launch point 12.0 m above the ground
Air resistance is negligible
$g = 9.80\ \mathrm{m/s^{2}}$
Find
(a) With what speed does the ball strike the ground?
(b) Would the answer change if the ball were thrown at 40.0 degrees below the horizontal at the same speed?
Hint 1/4
Read what is asked for. A speed at a named instant, with a known launch instant and no time mentioned, and the launch angle appearing nowhere in the question.
Hint 2/4
The ledger $\tfrac{1}{2}mv_i^{2} + mgh = \tfrac{1}{2}mv_f^{2}$, with the zero of potential energy at the ground. Gravity is the only force doing work, and its work depends only on the drop.
Hint 3/4
With $v_i = 14.0\ \mathrm{m/s}$ and $h = 12.0\ \mathrm{m}$: the ledger gives $v_f^{2} = v_i^{2} + 2gh = 196 + 235.2$, and the launch angle never enters.
Hint 4/4
So $v_f = 20.8\ \mathrm{m/s}$, and the answer is the same at any launch angle.
Show solution
The energy route is used although this is a projectile problem, because the unknown is a speed and not a position or a time.
the mass cancels, and so does every trace of the launch direction
$$v_f = \sqrt{431.2} = 20.8\ \mathrm{m/s}$$
three significant figures, matching the data
Why the angle is absent
$$W_{\text{grav}} = mg\,\Delta y$$
the work done by gravity depends only on the vertical drop, not on the path taken to achieve it
$$v_f = \sqrt{v_i^{2} + 2gh} \ \text{for every launch angle}$$
so a ball thrown up, down or horizontally at 14.0 m/s all land at 20.8 m/s from this roof
Answer $$\boxed{v_f = 20.8\ \mathrm{m/s},\ \text{independent of the launch angle}}$$
Check
Independent check by the component route, which uses the angle everywhere: the launch components are $v_x = 14.0\cos 40.0^{\circ} = 10.72\ \mathrm{m/s}$ and $v_y = 14.0\sin 40.0^{\circ} = 9.00\ \mathrm{m/s}$. The horizontal component is unchanged at landing, and the vertical satisfies $v_{y,f}^{2} = 9.00^{2} + 2(9.80)(12.0) = 316.2$. Then $v_f^{2} = 10.72^{2} + 316.2 = 431.1$, giving 20.8 m/s. Two entirely different calculations, one number.
This is the shape of an interleaved question: it looks like a projectile problem from an earlier week and is answered fastest with a tool from a later one.
2§10.4 — two blocks, a table and a string●●●●○
No method is suggested. A 3.00 kg block on a horizontal table is joined by a light string over a light frictionless pulley at the table edge to a 2.00 kg block hanging freely.
Given
Block on the table $m_1 = 3.00\ \mathrm{kg}$, hanging block $m_2 = 2.00\ \mathrm{kg}$
Coefficient of kinetic friction between table and block $\mu_k = 0.200$
The string is light and inextensible, the pulley light and frictionless
Released from rest; the hanging block descends 0.800 m
$g = 9.80\ \mathrm{m/s^{2}}$
Find
(a) What is the acceleration of the system?
(b) How fast are the blocks moving after the hanging block has descended 0.800 m?
(c) Confirm the speed by an energy ledger for the two blocks together.
Hint 1/4
Two bodies joined by a taut string move with the same size of acceleration and the same speed. Decide whether you want the acceleration at all before you compute it.
Hint 2/4
Force route: treat the pair as one system of mass $m_1 + m_2$, driven by $m_2g$ and resisted by $\mu_km_1g$, so $a = (m_2g - \mu_km_1g)/(m_1+m_2)$. Energy route: $m_2gd - \mu_km_1gd = \tfrac{1}{2}(m_1+m_2)v^{2}$.
Hint 3/4
The data again: $m_1 = 3.00\ \mathrm{kg}$ on the table, $m_2 = 2.00\ \mathrm{kg}$ hanging, $\mu_k = 0.200$, $d = 0.800\ \mathrm{m}$. The driving force is 19.6 N and the friction is 5.88 N.
Hint 4/4
The acceleration is $2.74\ \mathrm{m/s^{2}}$ and the speed after 0.800 m is 2.10 m/s.
Show solution
The two blocks are treated as one system, because the string tension is internal to that system and never has to be computed.
both blocks move at the same speed, so one kinetic energy term with the total mass is correct
Answer $$\boxed{a = 2.74\ \mathrm{m/s^{2}},\qquad v = 2.10\ \mathrm{m/s}}$$
Check
The energy line in part (c) is itself the independent check, and it is genuinely independent: it never used the acceleration and never used a kinematic formula. A limiting check also passes: with $\mu_k = 0$ the acceleration would be $19.6/5.00 = 3.92\ \mathrm{m/s^{2}}$, larger as it must be, and with $\mu_k$ raised to 0.667 the driving and resisting forces would balance and the system would not start at all.
The tension never appeared. Whenever a question asks only about motion and not about the string, the pair can be treated as one body and one equation replaces two.
3§10.2 — a string breaks and the stone becomes a projectile●●●●○
The type is not announced. A 0.240 kg stone is whirled in a horizontal circle of radius 0.900 m at a steady 6.00 m/s, at a height of 1.40 m above level ground.
Given
Stone mass $m = 0.240\ \mathrm{kg}$
Circle radius $r = 0.900\ \mathrm{m}$, speed $v = 6.00\ \mathrm{m/s}$, horizontal circle
Height above the ground when the string breaks $1.40\ \mathrm{m}$
Air resistance is negligible; $g = 9.80\ \mathrm{m/s^{2}}$
Treat the circle as horizontal, so the string is very nearly horizontal too
Find
(a) What was the tension in the string just before it broke?
(b) How far horizontally from the break point does the stone land?
(c) With what speed does it hit the ground?
Hint 1/4
Three different pieces of the course are needed here, one for each part, and the parts do not use the same method. Sort out which is which before starting any of them.
Hint 2/4
Part (a) is the radial equation $T = mv^{2}/r$. Part (b) is projectile motion with a horizontal launch: the fall time comes from $h = \tfrac{1}{2}gt^{2}$ and the range from $x = vt$. Part (c) is fastest as a ledger, $v_f^{2} = v^{2} + 2gh$.
Hint 3/4
The data again: $m = 0.240\ \mathrm{kg}$, $r = 0.900\ \mathrm{m}$, $v = 6.00\ \mathrm{m/s}$, $h = 1.40\ \mathrm{m}$. The fall time is $\sqrt{2h/g}$.
Hint 4/4
The tension is 9.60 N, the horizontal distance is 3.21 m, and the landing speed is 7.96 m/s.
Show solution
Three different tools are used deliberately, one per part, because each part asks for a different kind of quantity: a force at an instant, a position after a time, and a speed between two states.
Part (a): the radial equation
$$\sum F_r = T = m\frac{v^{2}}{r} = (0.240)\frac{36.0}{0.900}$$
the circle is horizontal and the string very nearly so, so the tension supplies essentially the whole inward force
$$T = 9.60\ \mathrm{N}$$
against a weight of 2.35 N, so the string is pulled about four times as hard as by the hanging stone
Independent check on part (c) by components, which never touches energy: the horizontal speed stays 6.00 m/s and the vertical speed at landing is $gt = (9.80)(0.5345) = 5.24\ \mathrm{m/s}$, so the landing speed is $\sqrt{36.0 + 27.4} = 7.96\ \mathrm{m/s}$. A limiting check also passes on part (a): a stone whirled slowly would need almost no tension, and the formula gives zero at zero speed.
Notice that the radius mattered only in part (a). Once the string breaks the circle is irrelevant, and the stone remembers nothing about it except the speed and the direction it happened to have at that instant.
4§10.4 — stopping distance and what it scales with●●●○○
The type is hidden on purpose. A 1200 kg car travelling at 100 km/h brakes hard on a level road with locked wheels; the coefficient of kinetic friction between tyres and road is 0.800.
Given
Car mass $m = 1200\ \mathrm{kg}$
Initial speed 100 km/h, to be converted to SI
Coefficient of kinetic friction with locked wheels $\mu_k = 0.800$, level road
$g = 9.80\ \mathrm{m/s^{2}}$
Find
(a) Convert the speed to metres per second and find the kinetic energy of the car.
(b) How far does the car slide before stopping?
(c) By what factor does that distance change if the same car does this at 50 km/h instead?
Hint 1/4
Start by converting the speed, because every later quantity depends on it and a wrong conversion here poisons all three parts.
Hint 2/4
One hundred kilometres in one hour is $10^{5}$ metres in 3600 seconds. Then $K = \tfrac{1}{2}mv^{2}$, and the sliding distance follows from $\mu_kmgd = K$, in which the mass cancels.
Hint 3/4
With $m = 1200\ \mathrm{kg}$ and $\mu_k = 0.800$: the speed is 27.8 m/s, and the friction force is $\mu_kmg = 9408\ \mathrm{N}$.
Hint 4/4
The kinetic energy is $4.63\times10^{5}\ \mathrm{J}$, the distance is 49.2 m, and halving the speed divides the distance by four.
Show solution
The distance is derived symbolically before any number is put in, because the symbolic form shows at once that the mass cancels and that the speed enters squared, which is what part (c) is really asking about.
Independent check by the force route: the deceleration with locked wheels is $\mu_kg = 7.84\ \mathrm{m/s^{2}}$, and $d = v^{2}/(2a) = 771.6/15.68 = 49.2\ \mathrm{m}$. An order of magnitude check also passes: published stopping distances for a car at 100 km/h are around 45 to 55 m on dry road, so this is right without being suspiciously exact.
The mass cancelling is worth stating plainly, because it contradicts what most people expect: a heavy car does not need a longer distance on the same road, since the extra weight buys exactly as much extra friction as it costs in energy.
Mistake ledger (16 entries)
⚠ Using a slope formula on a curved surface
the result $a = g\sin\theta$ is memorised as a fact about ramps rather than as a consequence of one fixed angle, so a curve looks like a ramp with a slightly awkward angle
wrong$$v = \sqrt{2\,(g\sin\theta)\,L}\quad\text{on a curved chute}$$
right$$v = \sqrt{2gh}\quad\text{on any smooth track of drop } h$$
⚠ Trying to get a time out of an energy ledger
the ledger answers so many questions cheaply that it feels universal, and the missing clock is invisible because no symbol for it is absent from the page
⚠ Keeping the same sign for the weight at the top and the bottom of a vertical circle
the weight really does point down at both places, so it feels as though its sign in the radial equation should not change either
wrong$$\text{top: } T + mg = \frac{mv^{2}}{r},\qquad \text{bottom: } T + mg = \frac{mv^{2}}{r}$$
right$$\text{top: } T + mg = \frac{mv^{2}}{r},\qquad \text{bottom: } T - mg = \frac{mv^{2}}{r}$$
⚠ Using the altitude in place of the orbital radius
the problem quotes an altitude because that is what is measured from the ground, and the symbol in the formula is also a distance, so the substitution feels dimensionally safe
wrong$$v = \sqrt{\frac{GM}{h}}$$
right$$v = \sqrt{\frac{GM}{R_E + h}}$$
⚠ Dropping the minus sign from the gravitational potential energy
every other energy in the course is positive, so a negative energy looks like a slip rather than a consequence of putting the zero at infinity
wrong$$E = \tfrac{1}{2}mv^{2} + \frac{GmM}{r}$$
right$$E = \tfrac{1}{2}mv^{2} - \frac{GmM}{r}$$
⚠ Multiplying the friction force by the height dropped
the height is the number the ledger has just used for gravity, so it is the one sitting in working memory when the friction line is written
⚠ Changing the zero level halfway through a ledger
the initial state is often described from the ground and the final state from some other landmark, so each line is written from whichever level the sentence in the question suggested
wrong$$mgh_{\text{from the ground}} = \tfrac{1}{2}mv^{2} + mgh_{\text{from the table}}$$
right$$mgy_i = \tfrac{1}{2}mv^{2} + mgy_f\quad (y \text{ from one declared level})$$
⚠ Measuring a spring compression from the point of first contact
that is where the interesting part of the motion begins, so it feels like the natural origin, and the block really is at rest there in some problems
the force route needs a constant acceleration to be followed by a kinematic formula; the energy route needs two named instants and cannot return a time
The radial equation
$$\boxed{\;\sum F_r = m\,\frac{v^{2}}{r}\quad\text{(taking toward the centre as positive)},\qquad \sum F_t = m\,a_t\;}$$
the radius is that of the path actually followed, the speed is the speed at that instant, and only forces with an agent appear on the left
straight slope, sliding, and the second expression is meaningful only if the body does slide back, which needs the tangent of the angle to exceed the static coefficient
smooth track or light inextensible string, and the limiting case where the track push or the tension is on the point of vanishing at the top
Check yourself
Close the page and write, from memory and on one side of paper: the two routes and the one sentence that decides between them; the radial equation and what may not appear in it; the energy ledger with its friction term and the distance that friction is multiplied by; the two forms of gravitational potential energy and the test for which one is legal; and the four checks in the order they cost. Then reopen and mark what you missed, because the gaps are the reading list.
Given an unseen problem, write one line naming the route and the word in the question that decided it, before doing any algebra?
c-route-choice
Write the radial equation for a car on a bend, a ball at the top of a vertical circle and a conical pendulum, getting the sign of the weight right in each and inventing no force?
c-radial-axis
Compute an orbital speed and period from an altitude, and say in one line why the near surface potential energy form is illegal for that height?
c-gravitation-two-forms
Write a ledger for a body that slides over a rough patch and then up a smooth slope, with the friction force multiplied by the right distance on each leg?
c-energy-ledger
Solve the same problem twice with two different zero levels and show that the answer is unchanged, and handle a block falling onto a spring without losing the compression from the height?
c-reference-levels
Compute the power of a motor at a given speed and then check a finished answer with a unit test, a limiting case and an order of magnitude comparison?
c-power-and-checks
Glossary (15 terms)
mechanical energymekanik enerji
The sum of the kinetic energy and the potential energy of a body or a system at one instant. It stays constant only while every force doing work has a potential energy, and it falls by exactly the friction loss when it does not.
conservative forcekorunumlu kuvvet
A force whose work between two points is the same along every route, so that the work around any closed path is zero.
korunumsuz kuvvet
A force whose work depends on the route taken, so no potential energy can be defined for it. Sliding friction is the standard example, since a longer path between the same two points costs more.
potential energypotansiyel enerji
Energy a body has because of where it is rather than how fast it is moving, defined so that the work done by the matching conservative force is minus its change.
referans düzeyi
The height chosen as the zero of gravitational potential energy in a particular problem. The choice is free and must be used at both ends of the same energy line, because only differences of potential energy reach an answer.
The energy stored in the pull between two masses. Near a planet surface it is written as the weight times a height above a chosen level, and in general it is the negative quantity that goes to zero at infinite separation.
The energy stored in a stretched or compressed spring, equal to half the stiffness times the square of the displacement from the natural length.
thermal energyısıl enerji
The energy that appears in two rubbing surfaces when friction removes mechanical energy from a body. It is what makes the total balance even though the mechanical part does not.
energy ledger
The single line used throughout this section: what the body had at the first instant, plus whatever was put in or taken out, equals what it has at the second instant.
radial equation
Newton's second law written along the direction pointing at the centre of a circular path, with the real forces on the left and the mass times the speed squared over the radius on the right.
escape speedkaçış hızı
The least launch speed at which a projectile leaving a planet surface never falls back, obtained from the energy line by requiring it to reach infinite separation with nothing left over.
The sum of the kinetic and gravitational potential energies of a body in orbit. For a circular orbit it is negative and equal to minus the kinetic energy, so a higher orbit has a greater total energy and a smaller speed.
powergüç
The rate at which work is done, measured in watts. Over an interval it is the work divided by the time; at an instant it is the component of the force along the velocity multiplied by the speed.
wattvat
The SI unit of power, equal to one joule per second. A quantity reported in watts that was computed per minute or per hour has not been converted and is not yet a power.
limiting casesınır durum
A special value of one input, such as a coefficient of friction set to zero, for which the answer is already known.
What comes next
§11 · Linear Momentum
Every problem in this section had one body, or two bodies joined so tightly that they could be treated as one. The next section asks what happens when two bodies meet for a fraction of a second and then separate, where the forces between them are enormous, unknown and short lived. Energy will not be enough on its own there, because some of it disappears into deformation, and a second conserved quantity has to be built to take its place.
Sources
D. Giancoli, Physics for Scientists and Engineers with Modern Physics, 5th edition The set textbook for the course. Its chapters on applying Newton's laws, on gravitation and on energy cover the same ground that this review pulls together; the end of chapter problems are harder than the ones here, deliberately, and are the right next step once this set feels comfortable.
Course syllabus, week 10 line and assessment table The scope of this section comes from the week line, which reads catch up and review and names no chapter numbers, so no chapter number is quoted anywhere in this section.
SI values used throughout: g = 9.80 m/s squared, G = 6.674e-11 N m squared per kg squared, Earth mass 5.97e24 kg, Earth radius 6.38e6 m These four numbers are fixed for the whole section and are mutually consistent: the last three fed into the inverse square law return 9.79 m/s squared at the surface, which is the first to three figures.