7 concepts20 worked examples32 exercises5 exam-level7 figures
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08Work and Energy
Two identical cars brake as hard as their tyres allow on the same dry road. The first is doing 50 km/h when the driver hits the pedal and it stops in about 16 m. The second is doing 100 km/h, twice as fast, and it needs about 66 m, which is four times the room and not twice. Nothing about the car, the road or the driver has changed, so the factor of four has to come from somewhere.
By the end of this section you can produce that 16 m and that 66 m in two lines each, say exactly which quantity is doing the squaring, and check both answers by a route that never mentions time.
In 60 seconds
Work is the part of a force that points along the motion, multiplied by the distance moved, and the total work done on a body is exactly the change in the quantity $\tfrac12 mv^{2}$ that the body carries; that one sentence replaces a page of kinematics whenever the question asks about speeds and distances rather than about times.
Work done by a constant force
$$W = F d \cos\theta$$
the force keeps the same size and direction and the body moves in a straight line; $\theta$ is the angle between the force and the displacement
Work as a scalar product
$$W = \vec F \cdot \vec d = F_x d_x + F_y d_y$$
the force and the displacement are given in components, so no angle has to be found first
more than one force acts; add the works one by one, or work with the net force, and check that the two agree
Kinetic energy
$$KE = \tfrac12 m v^{2}$$
any body of mass $m$ moving at speed $v$; never negative, and it does not care about direction
Work-energy principle
$$W_{\rm net} = \Delta KE = \tfrac12 m v_2^{2} - \tfrac12 m v_1^{2}$$
a question links a force and a distance to a change of speed and never mentions time
Work done by a varying force
$$W = \int_{x_1}^{x_2} F(x)\,dx$$
the force changes as the body moves; the integral is the area under the force against position graph
Work to stretch or compress a spring
$$W = \tfrac12 k x^{2}$$
an of stiffness $k$ taken from its to a stretch or squash of $x$
Three most common mistakes
Writing $W = Fd$ when the force is at an angle to the motion. The cosine is not decoration: a rope at $60^{\circ}$ delivers half the work of the same rope pulled along the floor, and a force at $90^{\circ}$ delivers none at all.
Setting the work done by one force equal to the change in kinetic energy. The principle uses the net work. In the dragging example in this section the rope does 786 J and the kinetic energy rises by only 42 J, because friction takes the rest.
Using $W = Fd$ for a force that changes as the body moves, usually by picking the starting value of the force. For a spring, or for any curved force graph, the work is the area under the graph and the starting value overstates it.
The assessment table gives 20% to each midterm, 25% to the final, 10% to quizzes in total, 5% to homework and 20% to lab work. It says nothing about how the topics are spread across those papers, so no claim is made here about where work and questions turn up; the safe assumption is everywhere.
How much time do you have?
10 minutes
You leave with the two formulas that carry most of the marks, $W = Fd\cos\theta$ and $W_{\rm net} = \Delta KE$, and with the single sentence that decides whether you use them correctly: it is the net work, not the work of your favourite force.
The 60 second card · Formula card · Work done by a constant force: only the part along the motion counts · The work-energy principle: net work is the change in kinetic energy · Mistake ledger
45 minutes
You add the parts that turn the formulas into marks: the that keeps the signs straight, kinetic energy and why doubling the speed is not doubling anything, and the area rule that handles a force which refuses to stay constant.
The 60 second card · Work done by a constant force: only the part along the motion counts · Adding the works up: the ledger for one body · Kinetic energy: the number a moving body carries · The work-energy principle: net work is the change in kinetic energy · Work done by a varying force: the area under the graph · Method boxes · Fading ladder · Practice B (computation) · Check yourself
full read
Everything above plus the scalar product, which is how work arrives in component form on an exam paper, the spring, which is the varying force you are most likely to meet, and the interleaved set that forces you to decide which tool a question wants before you reach for it.
The opening pages · What you should already have · Notation · All seven concept blocks · Method boxes · Contrast pairs · Fading ladder · Exam level example · Practice A to D · Mistake ledger · Formula card · Glossary · Check yourself
By the end of this section
Compute the work done by a constant force on a body that moves in a straight line, including the cases where the force is at an angle, perpendicular, or opposing the motion.
Evaluate the work as a scalar product when the force and the displacement are given in components, and get the angle between two vectors from the same product.
Assemble a work ledger for every force acting on one body, get the net work two independent ways, and read the signs correctly.
Calculate the kinetic energy of a body, and the work needed to take it from one speed to another, keeping the units and the order of magnitude honest.
Apply the work-energy principle to link a force and a distance to a change of speed, including and forces that bring a body to rest.
Integrate a varying force over a displacement, or read the same work off the area under a force against position graph, and say why an average of the endpoints can fail.
Solve spring problems: find the stiffness from a measurement, find the work stored in a stretch or a , and use it to launch a body.
Syllabus coverage
Work
What work is, how a constant force at an angle delivers it, the scalar product form, the ledger of several forces, a force that varies with position, and the spring
The week line carries no chapter numbers, so no chapter number is quoted anywhere in this section. The scope is the standard content of that topic in the set textbook, split across five of the seven blocks.
covered
Energy
Kinetic energy as the number a moving body carries, and the work-energy principle that ties it to the net work
Energy in this section means kinetic energy and nothing else, which is stated in the conventions block so that the word cannot quietly widen.
covered
Potential energy and the conservation of mechanical energy
Energy stored by position rather than by motion, and the bookkeeping that keeps a total constant
Deferred to the next section, whose title is exactly that subject. Nothing here needs it: every gravity or spring problem in this section is handled by computing the work that force does, which is a calculation you can do with what is on this page.
deferred
Power
The rate at which work is done, measured in watts
Not named in this week's line, and it belongs with the energy chapter that follows. Every question here asks how much work, never how fast the work was delivered, so no time appears in any energy answer in this section.
deferred
Work along a curved path
Adding up the work along a path that bends, where the angle between force and motion changes from point to point
Named in three sentences so that the straight-line restriction on the main formula means something, then dropped. Every path in this section is straight, or is split into straight pieces, so treat the curved case as background rather than examinable.
off_syllabus
Recall first
Newton's second law in components
$\sum F_x = m a_x$ and $\sum F_y = m a_y$, written for one named body after every force on it has been drawn.
The work-energy principle is derived from it in this section, and every ledger here starts from the same free-body diagram you were already drawing.
The normal force comes from an equation, not from memory
$N$ is the perpendicular push of a surface. On a level floor with nothing else vertical, $N = mg$; in general it is whatever the perpendicular component equation gives, for example $N = mg - F\sin\theta$ under a rope pulling upward at $\theta$.
Friction is $\mu_k N$, and friction is the force whose work you will be subtracting all section. An error in $N$ becomes an error in the final speed.
Kinetic friction
$F_{fr} = \mu_k N$ while the surfaces slide, directed against the sliding.
It is the standard second entry in every work ledger on this page, and it is the only force here that is guaranteed to take energy away.
Components of the weight on a slope
With axes tilted so that $x$ runs along a slope of angle $\theta$ and $y$ runs out of it, the weight splits into $mg\sin\theta$ down the slope and $mg\cos\theta$ into it, so that $N = mg\cos\theta$.
Half the harder problems in this section put the body on a ramp, and the resolving was done in an earlier section; it is reused here unchanged.
The constant acceleration formula that links speed to distance
$v^{2} = v_0^{2} + 2 a d$, valid only while the acceleration keeps the same value.
It is the tool the work-energy principle is derived from, and it is also the tool that fails the moment the force varies, which is the reason the new machinery is worth building.
Components of a vector
A vector of magnitude $A$ at angle $\theta$ to the $x$ axis has $A_x = A\cos\theta$ and $A_y = A\sin\theta$; conversely $A = \sqrt{A_x^{2}+A_y^{2}}$.
The scalar product block is written entirely in components, and the angle between two vectors is recovered from them.
The definite integral as an area
$\int_a^b f(x)\,dx$ is the signed area between the curve $y = f(x)$ and the $x$ axis, counted positive above the axis and negative below it.
Work done by a varying force is exactly that area, with force on the vertical axis and position on the horizontal one. If the integral is new, the areas used here are triangles and rectangles and can be read off geometrically.
Try it yourself first (3 questions)
1§08.0 — splitting a force into components●○○○○
Three warm-up questions before the new material, on the pieces this section leans on. Getting one wrong is not a problem; it just tells you which recall above to read first. A rope pulls on a sledge with a force of 90.0 N directed 25.0 degrees above the horizontal floor.
Given
$F = 90.0\ \mathrm{N}$
The rope makes $25.0^{\circ}$ with the horizontal floor
The floor is level
Find
(a) Find the horizontal component of the rope's pull.
(b) Find the vertical component of the rope's pull.
Hint 1/4
You are not being asked for anything to do with motion here, only for the two numbers the single arrow is equivalent to.
Hint 2/4
For a vector at angle $\theta$ to the $x$ axis, $A_x = A\cos\theta$ and $A_y = A\sin\theta$.
Hint 3/4
With $F = 90.0$ N and $\theta = 25.0^{\circ}$: $\cos 25.0^{\circ} = 0.9063$ and $\sin 25.0^{\circ} = 0.4226$.
Hint 4/4
The components are 81.6 N horizontally and 38.0 N vertically.
Rebuild the vector from the parts: $\sqrt{81.6^{2}+38.0^{2}} = 90.0$ N, and $\tan^{-1}(38.0/81.6) = 25.0^{\circ}$, so nothing was lost.
The horizontal part is the one that will do work when the sledge slides along the floor, and the vertical part is the one that will change the normal force. Both matter in this section, for different reasons.
2§08.0 — when the constant acceleration formulas are allowed●●○○○
The second warm-up is the one worth getting wrong now rather than in the exam. A block moves along a straight line under a single force that pushes it forward the whole way, but the size of that force falls off as the block advances.
Given
The motion is along a straight line
One force acts along the direction of motion
The size of that force changes with position
Find
(a) True or false: because the motion is in a straight line, $v^{2} = v_0^{2} + 2ad$ can be used to find the final speed. Give your reason in one line.
Hint 1/4
Look at what the formula needs, not at what the motion looks like. Straightness is one condition; ask whether it is the only one.
Hint 2/4
The formula $v^{2} = v_0^{2} + 2ad$ comes from integrating a constant acceleration. Every symbol in it assumes $a$ never changes.
Hint 3/4
Here the force changes as the block advances, and $a = F/m$ with $m$ fixed, so $a$ changes too.
Hint 4/4
False: the acceleration is not constant, so that formula does not apply.
Show solutionTrace the assumption back
$$a = \frac{F(x)}{m}$$
the second law is still true point by point, so a varying force gives a varying acceleration
that relation is what you get by integrating a constant acceleration once; with $a$ changing there is no single value to put in
Answer $$\boxed{\;\text{False: } a \text{ is not constant here}\;}$$
Check
Test it on a case you can settle another way: a force that is 60 N at the start and 20 N at the end cannot give the same answer as a steady 60 N, yet the formula with the starting value would say it does.
Whenever a problem tells you the force changes with position, the kinematic shortcut is gone and something else has to take its place. That something is the area rule built later in this section.
3§08.0 — acceleration with friction on a level floor●●○○○
The third warm-up is a plain second-law question of the kind this section will soon replace with a shorter route. A 12.0 kg block on a level floor is pushed by a horizontal force of 60.0 N, and the coefficient of kinetic friction between block and floor is 0.300.
Given
$m = 12.0\ \mathrm{kg}$
Horizontal push $F = 60.0\ \mathrm{N}$
$\mu_k = 0.300$
Level floor
$g = 9.80\ \mathrm{m/s^{2}}$
Find
(a) Find the normal force on the block.
(b) Find its acceleration while it slides.
Hint 1/4
Two equations, one for each axis. Decide which one produces the normal force before touching the horizontal direction.
the leftover force divided by the mass, which is the second law read forwards
Answer $$\boxed{\;N = 118\ \mathrm{N},\qquad a = 2.06\ \mathrm{m/s^{2}}\;}$$
Check
Order of magnitude: a couple of metres per second squared is a brisk walk turning into a jog over a second or two, which is what a 60 N push on a 12 kg block ought to feel like.
Hold on to this answer. Later in this section the same block appears again and its speed after a given distance comes out in one line instead of three, without the acceleration ever being computed.
Notation
symbol
reads as
means
watch out
$W$
capital W
the work done by one named force on one named body, in
always say which force did it; a bare $W$ in a problem with three forces is the commonest way of losing track of a sign
$W_{\rm net}$
W net
the sum of the works done by every force acting on the body
this, and only this, is the quantity the work-energy principle uses; the work of the force you happen to be interested in is not it
$\theta$
theta
the angle between the force arrow and the displacement arrow, drawn tail to tail
not the angle to the horizontal, and not the slope angle, though on level ground the three often agree
$\vec F \cdot \vec d$
F dot d
the scalar product of the force and the displacement, which is the work
the result is a number, not a vector; writing an arrow over the answer is a sign that the product has been confused with something else
$KE$
K E
the kinetic energy $\tfrac12 mv^{2}$ of a body, in joules
it is built from the speed, so it never carries a direction and never comes out negative
$\Delta KE$
delta K E
the change in kinetic energy, final value minus initial value
the order is final minus initial, so a body slowing down has a negative $\Delta KE$; reversing the order flips every answer's sign
$J$
joule
the unit of work and of energy, equal to one newton metre
a joule is small on a human scale: lifting an apple by one metre is about one joule, and a braking car sheds hundreds of thousands of them
$k$
k
the stiffness of a spring, in newtons per metre, from $F = kx$
it is a property of the spring alone; the same $k$ appears whether the spring is stretched or squashed, and it is not a force
$x$
x
in the spring blocks, the stretch or compression measured from the spring's natural length
measured from the natural length, not from the floor or from any other origin; a spring already stretched 3 cm and pulled to 8 cm has moved from $x = 0.03$ m to $x = 0.08$ m
Conventions used here
The angle that goes into a work calculation
In $W = Fd\cos\theta$ the angle $\theta$ is measured between the force arrow and the displacement arrow, with both drawn from the same point. It is not the angle to the horizontal and it is not the angle to the surface, although on a level floor those often coincide. Draw the two arrows tail to tail before reading the angle off.
What a means on this page
Work carries a sign and the sign is not a direction. means the force is helping the motion and the body is gaining kinetic energy from it; negative work means the force opposes the motion and is taking kinetic energy away. A friction force on a sliding body almost always does negative work, and writing it without the minus sign is the fastest way to a wrong final speed.
Energy in this section means kinetic energy only
The only energy in these seven blocks is $\tfrac12 mv^{2}$. There is no stored energy, no total that has to stay constant and therefore no zero level to declare anywhere. When gravity or a spring appears, its effect is computed as work done by that force and entered in the ledger like any other force. Keeping to that discipline is why nothing in this section needs a result from the next one.
The unit that every answer in this section carries
Work and kinetic energy are both in joules, and one joule is one newton metre. A bare number is not an answer: $42$ is wrong where $42\ \mathrm{J}$ is right. Forces stay in newtons, distances in metres, masses in kilograms, speeds in metres per second, and the spring stiffness in newtons per metre.
The value taken for g in these blocks, and its sign
$g = 9.80\ \mathrm{m/s^{2}}$ everywhere in this section, and it is a positive number. Whether the work done by gravity comes out positive or negative is decided by whether the body moved down or up, never by hiding a minus sign inside $g$.
How many digits survive into an answer here
Three significant figures, unless the data given is coarser, in which case the answer follows the coarsest datum. Intermediate values are carried at full precision and only the last line is rounded, so a middle step recomputed from the printed numbers can differ in the final digit.
What the idealised words mean in the work blocks
Smooth or frictionless means no friction force at all, so that surface contributes nothing to the ledger. A light rope has no mass, does not stretch, and transmits the same tension at both ends. An ideal spring obeys $F = kx$ for every stretch used in the question and has no mass of its own. Air resistance is ignored unless a question says otherwise.
8.1Work done by a constant force: only the part along the motion counts
Work measures how much a force helps or hinders motion: force times distance, times the cosine of the angle between them.
You can already turn forces into an acceleration and that into a speed; this block builds the number that skips the middle step.
Solvable with what we have
Find a body's acceleration from the forces on it, friction included.
Turn it into a final speed with $v^{2} = v_0^{2} + 2ad$.
Handle a rope at an angle, a slope, or a circular path.
Not solvable yet
A 6.0 kg sledge is pulled from rest along 8.0 m of smooth track by a rope whose tension falls steadily from 60 N to 20 N. How fast is it going at the end?
What does it cost to stretch a spring twice as far as before?
The rope pulls with 60 N, so $a = 60/6.0 = 10\ \mathrm{m/s^{2}}$, and then $v = \sqrt{2(10)(8.0)} = 12.6\ \mathrm{m/s}$.
Why it fails
The second step needs a constant acceleration, and here the force, so the acceleration, falls to a third of its starting value on the way. The 12.6 m/s is what a rope that never weakened would deliver; the true answer is 10.3 m/s. What is missing is a way of adding a force up over a distance rather than at an instant.
DefinitionDefinition 8.1: work done by a constant force
Conditions
The force keeps the same size and direction throughout
The body moves in a straight line through a displacement $d$
$\theta$ is the angle between the force and the displacement, drawn tail to tail
A scalar in joules, with $1\ \mathrm{J} = 1\ \mathrm{N\,m}$
$$\boxed{\;W = F\,d\cos\theta\;}$$
Take the size of the force, keep only the fraction that points the way the body went, and multiply by how far it went. Leaning forward gives a positive answer; leaning backward a negative one; square on to the motion, zero, however large the force.
A 120 N rope at $35^{\circ}$ drags a 45 kg trunk 8.0 m along a level floor. Only the $\textcolor{#cf222e}{\text{horizontal part, } 98.3\ \mathrm{N}}$, lies along the $\textcolor{#1f6feb}{\text{displacement}}$ and does work; the $\textcolor{#8250df}{\text{vertical part, } 68.8\ \mathrm{N}}$, has nothing to move along and contributes nothing, although it does lighten the load on the floor.
Looks like this, but is not
Work is force times distance. It is the sentence everyone arrives with, and it is right whenever the force points exactly the way the body goes.
Carry a 20 kg suitcase 30 m along a level corridor. Your hand pushes up with 196 N, the case moves sideways, the angle between them is a right angle, and the work on the case is zero, not the 5880 J that force times distance would claim. Your arm aches all the same, which is a fact about muscle chemistry, not about the suitcase.
angle between force and motion
cos of that angle
work done (J)
what it is doing
0°
1.000
+500
pushing the body along, full effect
30°
0.866
+433
helping, but a seventh of it is wasted sideways
60°
0.500
+250
half of it wasted; the same force delivers half the work
90°
0.000
0
carried along for the ride, contributing nothing
120°
−0.500
−250
fighting the motion, taking energy out
Read down the third column: the force never changes and the distance never changes, yet the work runs from +500 J to −250 J. Everything in that column is the angle doing its work, which is why the angle is the first thing to look for in a work question and the last thing to guess.
The work a 120 N rope does dragging a trunk 8.0 m
A 45 kg trunk is dragged 8.0 m along a level floor by a rope held at $35^{\circ}$ above the horizontal with a steady tension of 120 N. Find the work done by the rope, by gravity, and by the floor's normal push.
Given
$m = 45\ \mathrm{kg}$
$T = 120\ \mathrm{N}$ at $35^{\circ}$ above the horizontal
$d = 8.0\ \mathrm{m}$ along the floor
$g = 9.80\ \mathrm{m/s^{2}}$
Find
the work done by each of the three named forces
SolutionThe rope: keep the part that lies along the floor
$$W_T = T d\cos\theta = (120)(8.0)\cos 35^{\circ}$$
the displacement is horizontal, so the angle to use is the rope's angle to the horizontal
$$= (120)(8.0)(0.8192) = 786\ \mathrm{J}$$
equivalently, the horizontal part of the pull is 98.3 N and it acts through the whole 8.0 m
Gravity and the floor: read the angle before reaching for a calculator
$$W_G = mg\,d\cos 90^{\circ} = 0$$
the weight points straight down while the trunk moves horizontally, so the two are square on
$$W_N = N d\cos 90^{\circ} = 0$$
the normal push is perpendicular to the surface, and the surface is what the trunk slides along
Bound the answer without repeating the multiplication: the largest work this rope could ever do over 8.0 m is $120 \times 8.0 = 960$ J, delivered only if the rope lay flat along the floor. Since $\cos 35^{\circ}$ is a little over four fifths, an answer a little over four fifths of 960 J is what we should have, and 786 J is.
Two of the three forces cost nothing to handle once the angle was read off the diagram rather than off the numbers in the question.
The two zeros are worth more than the 786 J. In every level-floor problem in this section, gravity and the normal force will drop out of the ledger for exactly this reason, which is why the ledger is usually shorter than the free-body diagram.
Lowering a 12 kg box: who does the negative work
You lower a 12 kg box from a shelf to the floor, a drop of 1.5 m, moving it at a steady slow speed the whole way. Find the work done by your hand and the work done by gravity, and then say what changes if you simply hold the box still for a minute.
Given
$m = 12\ \mathrm{kg}$
$d = 1.5\ \mathrm{m}$ downward
The box moves at constant speed, so the forces on it balance
$g = 9.80\ \mathrm{m/s^{2}}$
Find
the work done by the hand and by gravity, and the work done while holding the box still
SolutionGet the hand's force from the fact that the speed is steady
The two works cancel, and they have to: the box left the shelf at rest and reached the floor at rest, so nothing about its motion changed and the total work on it must be zero. That cancellation is an independent check on both signs at once.
Negative work is not a smaller kind of work; it is work in the other direction. A force doing negative work is taking motion out of the body, which is exactly what your hand is for when you lower something.
Checkpoint
§08.1 — work done by a handle at an angle●●○○○
Thirty seconds, on the one formula in this block. A 25 kg suitcase is pulled 15 m along a level airport floor by a handle held at $40^{\circ}$ above the horizontal, with a steady pull of 65 N along the handle.
Given
$m = 25\ \mathrm{kg}$
$F = 65\ \mathrm{N}$ along the handle
The handle is at $40^{\circ}$ above the horizontal floor
$d = 15\ \mathrm{m}$
Find
(a) Find the work done on the case by the pull along the handle.
(b) Find the work done on the case by gravity.
Hint 1/4
Two forces, two angles. The only decision to make is what angle each force makes with the direction the case actually travels.
Hint 2/4
$W = Fd\cos\theta$, with $\theta$ measured between the force and the displacement.
Hint 3/4
Here $F = 65$ N, $d = 15$ m and the handle is at $40^{\circ}$, so $\cos 40^{\circ} = 0.766$; gravity is at $90^{\circ}$ to the floor.
Hint 4/4
The handle does 747 J of work and gravity does none.
Show solutionApply the definition twice
$$W = (65)(15)\cos 40^{\circ} = 747\ \mathrm{J}$$
only the horizontal part of the pull, 49.8 N, travels with the case
$$W_G = (245)(15)\cos 90^{\circ} = 0$$
the weight is square on to the motion, so its size never enters
Sanity bound: $65 \times 15 = 975$ J is the most this pull could do, and $\cos 40^{\circ}$ is about three quarters, so an answer near 750 J is the right size.
Notice the mass was never used. Work done by a given force does not care how heavy the body is; the mass will only matter when we ask what that work does to the motion.
⚠ Dropping the cosine when the force is at an angle
the numbers for the force and the distance are both printed in the question and the angle is in a picture, so the two numbers get multiplied and the picture never gets used
8.2The scalar product: work when the vectors arrive in components
Multiply matching components and add: it gives the same work as the cosine formula, without ever finding an angle.
The last block needed an angle, and exam questions often hand you components instead; this block is the bridge between the two, and it is one line long.
DefinitionDefinition 8.2: the scalar product, and work written with it
Conditions
$\vec A$ and $\vec B$ are any two vectors and $\theta$ is the angle between them
The result is a scalar: a number with a sign and a unit, never an arrow
The component form needs both vectors written on the same set of axes
For work, $\vec A$ is the constant force and $\vec B$ is the straight-line displacement
$$\boxed{\;\vec A \cdot \vec B = AB\cos\theta = A_xB_x + A_yB_y + A_zB_z, \qquad W = \vec F\cdot\vec d\;}$$
Line the two arrows up tail to tail, shorten the first one to just the part that lies along the second, and multiply the two lengths. Written in components the same number comes out of multiplying the two x parts, multiplying the two y parts, and adding, which is why no angle is needed when the components are given.
The force $\textcolor{#1f6feb}{\vec F = (12\,\hat\imath + 9\,\hat\jmath)\ \mathrm{N}}$ projected onto the line of the displacement. What survives the projection is $\textcolor{#cf222e}{7.80\ \mathrm{N}}$, and multiplying that by the length of the displacement gives the same 42 J that the component sum gives in one line.
Looks like this, but is not
Two vectors multiplied together give a vector. It is the natural guess, and it is what the word product suggests.
The scalar product hands back a plain number, and that is the point of it: work is not a direction, it is an amount. There is a second product of two vectors that does return a vector, but it is not this one and it is not what appears in $W = \vec F\cdot\vec d$. A quick test that the arrow does not belong: the two vectors $(8.0, -6.0)\ \mathrm{N}$ and $(3.0, 4.0)\ \mathrm{m}$ have a scalar product of $24 - 24 = 0$, and zero has no direction to point in even though neither vector is zero.
Work from components, then the same work from the angle
A constant force $\vec F = (12\,\hat\imath + 9\,\hat\jmath)\ \mathrm{N}$ acts on a crate while the crate undergoes the displacement $\vec d = (5.0\,\hat\imath - 2.0\,\hat\jmath)\ \mathrm{m}$. Find the work done by this force, and then find the angle between the force and the displacement.
Given
$\vec F = (12\,\hat\imath + 9\,\hat\jmath)\ \mathrm{N}$
$\vec d = (5.0\,\hat\imath - 2.0\,\hat\jmath)\ \mathrm{m}$
The force is constant over the whole displacement
Find
the work done, and the angle between the two vectors
SolutionMultiply matching components and add
$$W = F_xd_x + F_yd_y = (12)(5.0) + (9)(-2.0)$$
the component form needs no angle, which is why it is the cheaper route when components are given
$$W = 60 - 18 = 42\ \mathrm{J}$$
the negative second term is the part of the force fighting the sideways motion, and it is subtracted, not ignored
Recover the angle from the other form of the same product
Check the number by the other formula: $Fd\cos\theta = (15)(5.39)(0.520) = 42$ J, which is the value the component route gave without ever mentioning an angle. Two independent routes, one answer.
The component route took one line; the angle route took three and was only needed because the angle itself was asked for.
If a question gives you components, do not convert to magnitudes and angles first. The conversion is where the arithmetic errors live, and the component form is exact and short.
A force that does no work without being zero
A crate is pushed by a constant force $\vec F = (8.0\,\hat\imath - 6.0\,\hat\jmath)\ \mathrm{N}$ while it undergoes the displacement $\vec d = (3.0\,\hat\imath + 4.0\,\hat\jmath)\ \mathrm{m}$. Find the work done, and explain the result geometrically.
Given
$\vec F = (8.0\,\hat\imath - 6.0\,\hat\jmath)\ \mathrm{N}$, of size 10 N
$\vec d = (3.0\,\hat\imath + 4.0\,\hat\jmath)\ \mathrm{m}$, of length 5.0 m
Find
the work done by this force, and why it takes that value
SolutionDo the component sum first, before deciding what it means
Test the perpendicularity independently by slopes: the force direction has slope $-6/8 = -0.75$ and the displacement has slope $4/3 = 1.33$, and $(-0.75)(1.33) = -1$, which is the condition for two lines to cross at a right angle.
A 10 N force acting through 5.0 m did nothing at all to the crate's motion. This is the same fact as the normal force doing no work, dressed in components, and it is the quickest test for perpendicularity you will ever have.
Checkpoint
§08.2 — work as a component sum●●○○○
Thirty seconds, one line of arithmetic. A constant force acts on a trolley while the trolley moves through a given displacement, and both vectors are written on the same axes.
Given
$\vec F = (6.0\,\hat\imath + 8.0\,\hat\jmath)\ \mathrm{N}$
$\vec d = (3.0\,\hat\imath - 1.0\,\hat\jmath)\ \mathrm{m}$
Find
(a) Find the work done by this force on the trolley.
Hint 1/4
You are not being asked for an angle or a magnitude, so do not compute either. Ask which single operation turns two component vectors into a work.
Hint 2/4
$W = \vec F\cdot\vec d = F_xd_x + F_yd_y$.
Hint 3/4
Here $F_x = 6.0$ N, $F_y = 8.0$ N, $d_x = 3.0$ m and $d_y = -1.0$ m.
matching components multiply; the mismatched pairs never meet each other
Answer $$\boxed{\\;W = 10\ \mathrm{J}\\;}$$
Check
Cross-check with magnitudes: $F = 10$ N, $d = 3.16$ m, so $\cos\theta = 10/31.6 = 0.316$ and $\theta = 71.6^{\circ}$, an acute angle, which agrees with the positive sign.
Whenever a work comes out positive but small compared with $Fd$, the reason is always the same: a large slice of the force is pointing somewhere the body did not go.
⚠ Adding the components instead of multiplying them in pairs
the formula contains both a multiplication and an addition and the two get swapped under time pressure
the minus sign belongs to the displacement, not to the force, so it looks like it is not part of this force's business
wrong$$W = (12)(5) + (9)(2) = 78\ \mathrm{J}$$
right$$W = (12)(5) + (9)(-2) = 42\ \mathrm{J}$$
8.3Adding the works up: the ledger for one body
List every force on the body, work out what each one contributes, and add with signs; that total is the only one that matters.
One force at a time is arithmetic; the reason the last two blocks were worth building is that a real problem has four forces and only their total does anything.
RuleRule 8.3: net work, computed two ways
Conditions
One named body, with every force on it listed exactly once
All the forces constant and the displacement a straight line, so each work is $F_id\cos\theta_i$
The same displacement $\vec d$ is used for every entry, because it is the same body moving
The two routes must agree; if they do not, a force has been missed or a sign has been dropped
Either work out what each force contributes and add the contributions, or add the forces first and let the total force do the work; both give the same number. The first route tells you where the energy went, the second is faster, and doing both is the cheapest check available in this section.
Proof
Work is built from a scalar product, and scalar products distribute over addition in the same way ordinary multiplication does: $(\vec A + \vec B)\cdot\vec C = \vec A\cdot\vec C + \vec B\cdot\vec C$.
Every force on the body acts through the same displacement $\vec d$, because there is only one body and it made one move.
So $\sum_i \vec F_i\cdot\vec d = \left(\sum_i \vec F_i\right)\cdot\vec d$, which is the statement in the box.
Nothing here is special to work; it is the distributive law, and it is the reason the two routes are guaranteed to agree rather than merely observed to.
The full force list for the trunk of the first block, with each force's contribution written next to it. The $\textcolor{#cf222e}{\text{rope and the friction}}$ are the only two entries that are not zero, and they nearly cancel; the $\textcolor{#8250df}{\text{weight and the normal force}}$ are perpendicular to the $\textcolor{#1f6feb}{\text{displacement}}$ and contribute nothing at all.
Looks like this, but is not
The net work is the work done by the force you applied. Nobody says this out loud, but it is what happens whenever a solution computes one work and puts it straight into an energy equation.
Drag the trunk and the rope contributes 786 J while friction contributes $-744$ J, so the net work is 42 J. Treating the rope's 786 J as the net work overstates the trunk's gain by a factor of nearly nineteen, and predicts a final speed of 5.9 m/s instead of 1.4 m/s. The rope's work is a real number and it is correctly computed; it is simply not the number the next block will need.
force
size (N)
angle to the motion
work (J)
rope tension
120
35°
+786
kinetic friction
93.0
180°
−744
weight
441
90°
0
normal force from the floor
372
90°
0
net
5.26 along the motion
0°
+42
The largest force in the table, the 441 N weight, contributes nothing, and the smallest entry in the last column is the only one that will matter in the next block. Size does not decide relevance here; the angle does. Notice too that the last row was computed independently, by adding the forces first, and it agrees with the sum of the column above it.
The full ledger for the dragged trunk, checked two ways
The 45 kg trunk of the first block is dragged 8.0 m along a level floor by a rope at $35^{\circ}$ with a tension of 120 N. The coefficient of kinetic friction between trunk and floor is 0.250. Build the complete work ledger and find the net work, then confirm it with the net force.
Given
$m = 45\ \mathrm{kg}$
$T = 120\ \mathrm{N}$ at $35^{\circ}$ above the horizontal
$\mu_k = 0.250$
$d = 8.0\ \mathrm{m}$ along a level floor
$g = 9.80\ \mathrm{m/s^{2}}$
Find
the work done by each force, the net work, and a check by an independent route
SolutionThe normal force is not the weight here, so get it from the vertical equation
The two routes were built from different intermediate numbers, 786 and 744 in one and 5.26 in the other, and they land on the same 42 J. A missed force or a flipped sign would show up as a disagreement, so this is a genuine check rather than a repetition.
Five lines, of which the first two were about the normal force. In this section the normal force is where most of the arithmetic risk sits, and it is never the last thing you should compute.
Two forces of 786 J and 744 J leaving a net of 42 J is the signature of a heavy object being dragged: nearly everything the rope delivers is scraped off by the floor. That is also why dragging furniture is exhausting and slow.
Lowering a 350 kg lift 12.0 m at a steady speed
A 350 kg service lift is lowered 12.0 m at a constant speed by a single cable. Find the work done by gravity, the work done by the cable, and the net work on the lift.
Given
$m = 350\ \mathrm{kg}$
$d = 12.0\ \mathrm{m}$ downward
Constant speed throughout
$g = 9.80\ \mathrm{m/s^{2}}$
Find
the work done by gravity and by the cable, and the net work
SolutionUse the constant speed to get the cable tension
Independent check on the sign pattern rather than the arithmetic: raise the same lift 12.0 m at a steady speed instead, and every angle flips, so gravity does $-4.12\times10^{4}$ J and the cable $+4.12\times10^{4}$ J. The net work is zero either way, which is the only outcome consistent with a constant speed.
A net work of zero does not mean nothing happened. Two large works passed through the lift in opposite directions; the motor's bill was real. It means only that the lift's motion was left unchanged, and that is all net work reports on.
Checkpoint
§08.3 — the ledger when the speed is steady●●○○○
Thirty seconds, and it is a reasoning question dressed as a calculation. A 20 kg crate is pushed 4.0 m along a level floor by a horizontal force of 50 N, and it travels at a constant speed the whole way.
Given
$m = 20\ \mathrm{kg}$
Horizontal push $F = 50\ \mathrm{N}$
$d = 4.0\ \mathrm{m}$
The crate moves at constant speed
Find
(a) Find the net work done on the crate.
(b) Find the work done on it by friction.
Hint 1/4
One of the two answers can be written down before any arithmetic, from a single phrase in the question. Find that phrase first.
Hint 2/4
Constant speed means zero acceleration, so the forces balance and the net force is zero; and $W_{\rm net} = (\sum F)d$.
Hint 3/4
Here the push does $(50)(4.0) = 200$ J, gravity and the normal force do nothing, and the four works must add to the net work.
Hint 4/4
The net work is zero and friction does $-200$ J.
Show solutionThe state of motion gives the total for free
Check by computing the friction force directly: a balance gives $F_{fr} = 50$ N, and $(50)(4.0)\cos 180^{\circ} = -200$ J, which is the same entry reached without ever using the coefficient.
Constant speed is a gift in a work problem, exactly as it was in a force problem: it hands you one equation before you have done anything.
⚠ Leaving friction out of the ledger because it was not mentioned in the last sentence
the question asks about the rope, so the rope is what gets computed, and the ledger is never actually written down
8.4Kinetic energy: the number a moving body carries
Half the mass times the speed squared: one number that says how much motion a body has, regardless of direction.
Work is what a force delivers to a body; this block builds the account it is delivered into, and the squared speed in it is the reason the hook's four appeared.
DefinitionDefinition 8.4: translational kinetic energy
Conditions
$m$ is the mass in kilograms and $v$ the speed in metres per second
$v$ is the speed, so the direction of travel never enters
The result is in joules, the same unit as work, which is not a coincidence
Valid for a body moving as a whole; a spinning body carries more, which is not treated here
$$\boxed{\;KE = \tfrac12 m v^{2}\;}$$
Take the mass, halve it, and multiply by the speed multiplied by itself. Because the speed is squared, going twice as fast does not carry twice as much, it carries four times as much, and because a square is never negative, no body ever has less than zero of it.
Kinetic energy of a 1200 kg car against its speed. The curve is a parabola, so the $\textcolor{#1f6feb}{\text{step from 10 to 20 m/s}}$ costs three times as much as getting to 10 m/s did, and the $\textcolor{#cf222e}{\text{whole curve}}$ steepens as the car speeds up.
Looks like this, but is not
Kinetic energy is a vector, because velocity is. The formula does contain the velocity, so the guess is reasonable.
Two identical 1200 kg cars pass each other on a road at 25 m/s in opposite directions. Their velocities are $+25$ and $-25\ \mathrm{m/s}$, but $(-25)^{2}$ and $(+25)^{2}$ are the same number, so both carry $3.75\times10^{5}$ J and neither carries a negative amount. The squaring destroys the direction on purpose: kinetic energy answers how much, not which way. This is also why a body's kinetic energy can never be negative, while the work done on it very often is.
moving thing
mass (kg)
speed (m/s)
kinetic energy (J)
a mosquito
0.0000025
0.50
0.0000003
a thrown baseball
0.145
40
116
a sprinting person
70
10
3500
a car in town
1200
14
1.2 × 10⁵
the same car on a motorway
1200
33
6.5 × 10⁵
Two rows are worth staring at. The car in town and the car on the motorway are the same object with the same engine, and the second carries five and a half times as much, because the speed ratio 33 to 14 is squared into 5.5. And the person at 3500 J is a useful yardstick: a joule is a small unit, so any answer in this section that comes out in single digits describes something gentle, like a spring toy, while anything past $10^{5}$ J describes a vehicle.
A car at 25 m/s against a baseball at 40 m/s
Find the kinetic energy of a 1200 kg car travelling at 25 m/s, and of a 0.145 kg baseball thrown at 40 m/s. Then find the car's kinetic energy at 50 m/s and say what the comparison means for a driver.
Given
car: $m = 1200\ \mathrm{kg}$, $v = 25\ \mathrm{m/s}$ and later $50\ \mathrm{m/s}$
Order of magnitude check on the baseball: 116 J is roughly what it takes to lift a 12 kg suitcase one metre, which is a believable amount of effort for one hard throw. The car's $3.75\times10^{5}$ J is three thousand times that, and a car is far heavier and moving faster, so the gap is the right size.
The factor of four is the whole hook. A driver who doubles speed has not doubled anything about the crash; four times as much has to be taken out of the car before it stops, and the road can only take it out at a fixed rate per metre.
A lorry at walking pace matching a car at speed
A 12000 kg lorry moves at 7.90 m/s and a 1200 kg car at 25.0 m/s. Compare their kinetic energies, and then find the speed at which the car would carry twice the lorry's kinetic energy.
Check the last number by ratio instead of by formula: $35.3/25.0 = 1.41$, and $1.41^{2} = 2.00$, so the kinetic energy has indeed doubled. The square root of two turning up as a speed ratio is the signature of a doubled energy.
Two applications of one definition, plus one rearrangement. No forces were needed anywhere, because kinetic energy is a property of the body's motion alone.
Mass and speed are not interchangeable. To match a lorry you can be light and quick, but the price of speed is paid twice over, which is why the car needed 25 m/s to match a lorry doing less than 8.
Checkpoint
§08.4 — what doubling the speed costs●●○○○
Thirty seconds. A 1500 kg car speeds up from 15.0 m/s to 30.0 m/s on a motorway slip road, and you are asked only about the number the car carries, not about the forces that changed it.
Given
$m = 1500\ \mathrm{kg}$
$v_1 = 15.0\ \mathrm{m/s}$
$v_2 = 30.0\ \mathrm{m/s}$
Find
(a) Find the increase in the car's kinetic energy.
Hint 1/4
You need a difference of two values, not one value. Decide which two before computing anything.
Hint 2/4
$KE = \tfrac12 mv^{2}$, so the increase is $\tfrac12 m(v_2^{2} - v_1^{2})$.
Hint 3/4
Here $m = 1500$ kg, $v_1 = 15.0$ m/s and $v_2 = 30.0$ m/s, so $v_2^{2} - v_1^{2} = 900 - 225 = 675$.
Hint 4/4
The kinetic energy rises by $5.06\times10^{5}$ J.
Show solutionSubtract the values, not the speeds
$$\Delta KE = \tfrac12(1500)\left(30.0^{2} - 15.0^{2}\right)$$
the squares must be taken first; $(30.0-15.0)^{2}$ is a different and wrong quantity
arithmetic, with the units in joules because kilograms times metres squared per second squared is a joule
Answer $$\boxed{\;\Delta KE = 5.06\times10^{5}\ \mathrm{J}\;}$$
Check
Cross-check by the factor rule: doubling the speed multiplies the kinetic energy by four, so the increase must be three times the starting value. Three times $\tfrac12(1500)(225) = 1.6875\times10^{5}$ J gives $5.06\times10^{5}$ J, from a route with no subtraction in it.
The wrong route, $\tfrac12 m(v_2-v_1)^{2}$, gives $1.69\times10^{5}$ J, which is a third of the truth and looks perfectly reasonable on a page. Squares do not distribute over subtraction, and this is the place that fact costs marks.
⚠ Squaring the change in speed instead of changing the squares
the phrase change in kinetic energy invites subtracting the speeds first, and the resulting expression looks tidier
8.5The work-energy principle: net work is the change in kinetic energy
Whatever the net work adds up to, that exact number is what the body's kinetic energy gains or loses.
We now have a delivery, the net work, and an account, the kinetic energy; this block is the single sentence that says they are the same number.
TheoremTheorem 8.5: the work-energy principle
Conditions
The work on the left is the net work: every force, added with signs
$v_1$ is the speed at the start and $v_2$ the speed at the end
The mass does not change during the move
Nothing here needs a constant force, although the derivation below starts with one
$$\boxed{\;W_{\rm net} = \Delta KE = \tfrac12 m v_2^{2} - \tfrac12 m v_1^{2}\;}$$
Add up the work every force does on the body over a move. That total, sign included, is exactly how much the quantity half m v squared changes over the same move. Positive total, the body ends faster; negative total, it ends slower; zero total, it ends at the speed it started with, however violent the journey in between.
Proof
Take one body of mass $m$ under a constant net force $F$ along its straight-line motion, so the net work over a distance $d$ is $W_{\rm net} = Fd$.
Newton's second law replaces the force: $F = ma$, so $W_{\rm net} = mad$.
The acceleration is constant, so $v_2^{2} = v_1^{2} + 2ad$, which rearranges to $ad = \tfrac12\left(v_2^{2}-v_1^{2}\right)$.
Substituting: $W_{\rm net} = m\cdot\tfrac12\left(v_2^{2}-v_1^{2}\right) = \tfrac12 mv_2^{2} - \tfrac12 mv_1^{2}$, which is the statement.
The derivation used a constant force, but the result outlives it: split a varying force into short steps, apply the argument to each, and add. The middle terms cancel in pairs and only the first and last speeds survive.
Stopping distance against the speed the brakes went on at, for a coefficient of friction of 0.60. The mass cancels, so this one curve serves every car. The $\textcolor{#1f6feb}{\text{two marked points}}$ are the hook: 50 km/h needs 16.4 m and 100 km/h needs 65.6 m, which is four times as far for twice the speed.
Looks like this, but is not
The work done by the force I applied equals the change in kinetic energy. This is the principle with one word quietly deleted, and the word is net.
The rope did 786 J on the trunk. If that were the change in kinetic energy, a 45 kg trunk starting from rest would leave the 8.0 m stretch at $\sqrt{2(786)/45} = 5.9\ \mathrm{m/s}$, which is a fast jog, for a trunk being dragged across a floor. The real net work is 42 J and the real speed is 1.4 m/s, a slow walk, because friction removed 744 J of the 786 J on the way. The principle is exact; it is exact about the total, and there is no version of it that works with one force picked out of the list.
Why 100 km/h needs four times the room of 50 km/h
A car travelling at 50.0 km/h brakes with all four wheels locked on a road where the coefficient of kinetic friction is 0.600, and stops. Find the stopping distance. Then repeat for 100 km/h, and say what happened to the car's mass along the way.
Given
$v_1 = 50.0\ \mathrm{km/h} = 13.9\ \mathrm{m/s}$, and later $100\ \mathrm{km/h} = 27.8\ \mathrm{m/s}$
$v_2 = 0$
$\mu_k = 0.600$ on a level road
The only horizontal force while braking is friction
$g = 9.80\ \mathrm{m/s^{2}}$
Find
the stopping distance at each speed, and the role of the mass
SolutionWrite the ledger in symbols before putting any number in
$$W_{\rm net} = -\mu_k m g\, d$$
friction is the only force with a component along the motion, and it opposes it, hence the minus
$$-\mu_k m g\, d = 0 - \tfrac12 m v_1^{2}$$
the work-energy principle, with a final speed of zero because the car stops
Cancel the mass and solve for the distance
$$d = \frac{v_1^{2}}{2\mu_k g}$$
the mass appears on both sides and disappears; a loaded car and an empty one stop in the same distance
Independent check with the old machinery: $a = \mu_k g = 5.88\ \mathrm{m/s^{2}}$, and $v^{2} = v_1^{2} - 2ad$ with $v = 0$ gives $d = 192.9/11.76 = 16.4$ m. The two routes are genuinely different, one through forces and times and one through energies, and they agree.
The energy route needed no acceleration, no time, and no direction; the force route needed all three. On a question that asks only for a distance, that is three chances to slip that never arise.
The car does not have twice as much to get rid of at twice the speed, it has four times as much, and the road removes it at a fixed number of joules per metre. That is the whole hook, and it is also why speed limits fall much faster than crash energies rise.
The dragged trunk finally moves: its speed after 8.0 m
The 45 kg trunk is dragged from rest through 8.0 m by the 120 N rope at $35^{\circ}$, against friction with $\mu_k = 0.250$. The ledger for this move was built in an earlier block and came to a net work of 42.0 J. Find the trunk's speed at the end of the 8.0 m.
Given
$m = 45\ \mathrm{kg}$
$v_1 = 0$ (it starts from rest)
$W_{\rm net} = +42.0\ \mathrm{J}$ over the 8.0 m
The ledger entries were $+786$ J from the rope and $-744$ J from friction
Find
the speed of the trunk after the 8.0 m drag
SolutionPut the net work into the principle
$$W_{\rm net} = \tfrac12 m v_2^{2} - \tfrac12 m v_1^{2} = \tfrac12 m v_2^{2}$$
the second term vanishes because the trunk started at rest, which is the only thing that fact is used for
$$42.0 = \tfrac12 (45) v_2^{2}$$
the net work, not the rope's 786 J; the rope's work is already inside this number
the square root keeps the answer small even though 42 J sounds like a lot for a trunk
Answer $$\boxed{\;v_2 = 1.37\ \mathrm{m/s}\;}$$
Check
Check through forces: the net force along the motion was 5.26 N, so $a = 5.26/45 = 0.117\ \mathrm{m/s^{2}}$, and $v^{2} = 2(0.117)(8.0)$ gives $v = 1.37\ \mathrm{m/s}$. Same answer, different machinery.
A slow walking pace after eight metres of heaving on a rope, which matches anyone's experience of moving furniture. If the answer had come out at 5.9 m/s, the friction entry would have been missing from the ledger, and this is the sort of implausible number that should send you back to the list of forces.
Checkpoint
§08.5 — the force needed to catch a ball●●●○○
Thirty seconds, and it is the everyday version of the braking car. A 0.300 kg ball arrives at your hands at 8.00 m/s and you bring it to rest, letting your hands travel back 0.250 m while you do it.
Given
$m = 0.300\ \mathrm{kg}$
$v_1 = 8.00\ \mathrm{m/s}$, $v_2 = 0$
the hands move back $d = 0.250\ \mathrm{m}$ during the catch
Find
(a) Find the your hands exert on the ball.
Hint 1/4
You are given two speeds and a distance and asked for a force. Ask which single statement in this block connects exactly those four things.
Hint 2/4
$W_{\rm net} = \Delta KE$, and for a single force opposing the motion, $W_{\rm net} = -Fd$.
Hint 3/4
Here $m = 0.300$ kg, $v_1 = 8.00$ m/s, $v_2 = 0$ and $d = 0.250$ m, so $\Delta KE = 0 - \tfrac12(0.300)(64.0)$.
Hint 4/4
The average force is 38.4 N.
Show solutionFind how much has to be removed
$$\Delta KE = 0 - \tfrac12(0.300)(8.00)^{2} = -9.60\ \mathrm{J}$$
negative because the ball is losing motion, and the sign will carry through to the force
Turn it into a force over the given distance
$$-F d = -9.60 \;\Rightarrow\; F = \frac{9.60}{0.250} = 38.4\ \mathrm{N}$$
the hands push against the motion, so their work is negative and the two minus signs cancel
Answer $$\boxed{\;F = 38.4\ \mathrm{N}\;}$$
Check
Plausibility: 38.4 N is the weight of about a 3.9 kg object, a firm but comfortable push, which is what catching a ball feels like. Halving the stopping distance would double it, which is also what catching badly feels like.
Every impact question in this section has this shape. The energy to be removed is fixed by the motion; the force is whatever the distance you allow makes it, which is the entire design principle behind crumple zones and crash mats.
⚠ Using the work of one force as if it were the net work
that force is the one the question talks about, and the principle looks like it is about work rather than about a total
wrong$$786 = \tfrac12(45)v^{2} \;\Rightarrow\; v = 5.9\ \mathrm{m/s}$$
right$$42.0 = \tfrac12(45)v^{2} \;\Rightarrow\; v = 1.37\ \mathrm{m/s}$$
⚠ Writing the change as the initial value minus the final one
the body is slowing down, so the instinct is to arrange the subtraction to make the answer positive
8.6Work done by a varying force: the area under the graph
When the force changes as the body moves, the work is the area under the force against position graph.
Every formula so far has assumed the force never changed; this block removes that assumption and settles the sledge problem the section opened with.
RuleRule 8.6: work done by a force that varies with position
Conditions
The motion is along a straight line, taken as the $x$ axis
$F(x)$ is the component of the force along that line at each position
Areas below the axis count as negative work, exactly as signed areas always do
For a constant force the rule collapses to $W = F\Delta x$, so nothing earlier is lost
$$\boxed{\;W = \int_{x_1}^{x_2} F(x)\,dx \;=\; \text{the signed area under the } F\text{--}x \text{ graph}\;}$$
Cut the journey into slices so short that the force does not change across any one of them, work out force times slice width for each, and add them all up. In the limit that is the integral, and on a graph of force against position it is just the area between the curve and the axis, counted positive above and negative below.
Proof
Over a slice of width $\Delta x$ so short that $F$ barely changes, the constant-force rule applies and the work is about $F(x)\Delta x$, which is the area of a thin rectangle under the graph.
Adding the slices gives $W \approx \sum F(x_i)\Delta x$, a sum of rectangle areas, which is already usable as a numerical estimate.
Letting the slices shrink turns the sum into $\int F(x)\,dx$ and the staircase of rectangles into the exact .
The only assumption used was that a short enough piece looks constant, which is why the rule needs no smoothness beyond the force having a value at each point.
The sledge from the opening of this section: a rope tension falling in a straight line from 60 N to 20 N over 8.0 m. The $\textcolor{#cf222e}{\text{shaded trapezium}}$ has an area of 320 J and that is the work done. The $\textcolor{#8250df}{\text{dashed rectangle}}$ is what a solution that used the starting tension would be claiming, 480 J, which is half as much again.
Looks like this, but is not
For a varying force, use the average of the starting and finishing values. It sounds safe, and for a force that changes in a straight line it is even correct.
Take $F(x) = 3.0x^{2}$ newtons acting from $x = 0$ to $x = 2.0\ \mathrm{m}$. The endpoint average is $(0 + 12)/2 = 6.0\ \mathrm{N}$, which over 2.0 m suggests 12 J. The true work is the area, $\int_0^2 3.0x^{2}dx = 8.0\ \mathrm{J}$, so the shortcut overstates it by half. The rule works for a straight-line force because the trapezium's area really is the mean height times the width, and it fails the moment the graph bends, which no exam question will warn you about.
The sledge with the weakening rope, settled at last
A 6.0 kg sledge starts from rest and is pulled 8.0 m along a smooth level track by a rope whose tension falls steadily from 60 N at the start to 20 N at the end. Find the work done by the rope and the sledge's final speed, and compare with what the constant-acceleration route claimed at the start of this section.
Given
$m = 6.0\ \mathrm{kg}$
$v_1 = 0$
the tension falls linearly from $60\ \mathrm{N}$ at $x = 0$ to $20\ \mathrm{N}$ at $x = 8.0\ \mathrm{m}$
the track is smooth, so friction contributes nothing
Bracket the answer without integrating: a rope that never fell below 20 N does at least $20\times8.0 = 160$ J, and one that never rose above 60 N does at most 480 J, so the work lies between 160 J and 480 J, and the speed between 7.3 and 12.6 m/s. The answers 320 J and 10.3 m/s sit inside both brackets.
One trapezium area replaced an integral, and the work-energy principle replaced a kinematic route that was not available at all here.
This closes the problem the section opened with. The naive answer was not a small slip: it was 22 percent too fast, because it charged the rope for a pull it stopped delivering after the first metre.
Work done by a force that grows as the square of the distance
A force directed along the motion has magnitude $F(x) = 3.0x^{2}$ newtons when the body is at position $x$ metres. Find the work it does between $x = 0$ and $x = 2.0\ \mathrm{m}$, and compare it with the estimate you would get from averaging the first and last values of the force.
Given
$F(x) = 3.0x^{2}\ \mathrm{N}$, directed along the motion
the body moves from $x_1 = 0$ to $x_2 = 2.0\ \mathrm{m}$
$F(0) = 0$ and $F(2.0) = 12\ \mathrm{N}$
Find
the work done, and how far the endpoint average is out
SolutionIntegrate, because the graph is curved and has no elementary area
this is the trapezium under the straight line joining the endpoints, not under the curve
$$\frac{12}{8.0} = 1.5$$
fifty percent too high, because the real curve sags below the straight line joining its ends
Answer $$\boxed{\;W = 8.0\ \mathrm{J},\ \text{while the endpoint average claims } 12\ \mathrm{J}\;}$$
Check
Check the integral by counting squares instead: split the run into four strips of width 0.5 m and take the force at each strip's midpoint, giving $3.0(0.0625 + 0.5625 + 1.5625 + 3.0625)(0.5) = 7.875$ J, which agrees with 8.0 J to the accuracy of such a coarse count.
One integral, one comparison. The comparison is the part worth keeping: it shows what a plausible shortcut costs.
One boundary of this section is worth naming here. Everything above assumed the path is a straight line, so that a single $x$ locates the body. When the path bends, the angle between force and motion changes from point to point and the work becomes a sum along the curve. That case is not examined in this section, and every path in the problems here is straight or is cut into straight pieces.
Checkpoint
§08.6 — reading a work off two simple shapes●●○○○
Thirty seconds with no calculator needed. A force acts along the direction of motion and its size depends on position: it is a steady 20 N from $x = 0$ to $x = 2.0\ \mathrm{m}$, and from there it falls in a straight line to zero at $x = 5.0\ \mathrm{m}$.
Given
$F = 20\ \mathrm{N}$ for $0 \le x \le 2.0\ \mathrm{m}$
$F$ falls linearly from $20\ \mathrm{N}$ at $x = 2.0\ \mathrm{m}$ to $0$ at $x = 5.0\ \mathrm{m}$
the force is along the motion throughout
Find
(a) Find the total work done by this force between $x = 0$ and $x = 5.0\ \mathrm{m}$.
Hint 1/4
Sketch the graph before computing anything; the region under it is made of two shapes you already know the areas of.
Hint 2/4
The work is the area under the force against position graph, and areas add.
Hint 3/4
Here the first piece is a rectangle 20 N high and 2.0 m wide, and the second is a triangle 20 N high and 3.0 m wide.
Hint 4/4
The total work is 70 J.
Show solutionSplit at the corner
$$W_1 = (20)(2.0) = 40\ \mathrm{J}$$
a constant force means a rectangle, and the old formula is the area of that rectangle
$$W_2 = \tfrac12(3.0)(20) = 30\ \mathrm{J}$$
a straight fall to zero means a triangle, whose area is half the base times the height
$$W = 40 + 30 = 70\ \mathrm{J}$$
the areas add because the works add, which is the same statement as before about the ledger
Answer $$\boxed{\;W = 70\ \mathrm{J}\;}$$
Check
Bracket it: the force never exceeded 20 N over 5.0 m, so the work cannot exceed 100 J, and the constant part alone already gives 40 J. The answer sits sensibly between the two bounds.
Nearly every varying-force question on an exam paper is built from rectangles and triangles for exactly this reason: the areas are meant to be read, not integrated.
⚠ Multiplying the starting value of a varying force by the distance
the starting value is the number printed first in the question, and $W = Fd$ is the formula that comes to hand
8.7The spring: the varying force you will actually be asked about
A spring pulls back in proportion to its stretch, so the work to stretch it is a triangle of area one half k x squared.
Areas under graphs are general; one particular graph turns up in more exam questions than all the others together, and it is a straight line through the origin.
RuleRule 8.7: the spring force and the work it takes
Conditions
$x$ is measured from the natural length, so $x = 0$ where the spring is relaxed
$k$ is the stiffness in newtons per metre, a property of that spring alone
$F = kx$ holds only within the spring's working range; stretch it far enough and it stops obeying
The formula below is the work done on the spring; the spring does the negative of it on the body
$$\boxed{\;F_{\rm spring} = -kx, \qquad W_{\rm on\ spring} = \int_0^{x} k s\,ds = \tfrac12 k x^{2}\;}$$
The spring pushes or pulls back in proportion to how far it has been moved from its resting length, and always in the direction that undoes the move, which is what the minus sign records. Because the force you must apply grows in a straight line from zero, the work to reach a stretch x is the area of a triangle: half the final force times the distance, or half k x squared.
Proof
To hold the spring at stretch $s$ you must pull with $ks$, so the graph of your force against stretch is a straight line through the origin with slope $k$.
The work you do reaching a stretch $x$ is the area under that line, which is a triangle of base $x$ and height $kx$.
That area is $\tfrac12(x)(kx) = \tfrac12 kx^{2}$, and the integral $\int_0^x ks\,ds$ gives the same thing.
Since the spring's own force is $-ks$ at every point, the work the spring does on the body is exactly the negative of this, $-\tfrac12kx^{2}$, while the body stretches it.
The pull needed against the stretch, for a spring with $k = 400\ \mathrm{N/m}$. The $\textcolor{#1f6feb}{\text{first 12 cm}}$ costs 2.88 J and the $\textcolor{#cf222e}{\text{second 12 cm}}$ costs 8.64 J, three times as much, because the strip is taller even though it is no wider.
Looks like this, but is not
Stretching twice as far takes twice as much work. Distance doubled, effort doubled; it is how nearly everything else in this section behaves.
With $k = 400\ \mathrm{N/m}$, reaching 0.12 m takes 2.88 J and reaching 0.24 m takes 11.5 J, which is four times as much, not twice. The second twelve centimetres alone costs 8.64 J, three times what the first twelve cost, because by then you are pulling against 48 N rather than against nothing. Anything with a squared quantity in it behaves this way, which is the same reason the braking car needed four times the road.
Finding a spring's stiffness, then paying for two equal stretches
A spring stretches 0.045 m when a steady 18 N pull is applied to it. Find its stiffness, then find the work needed to stretch it from its natural length to 0.12 m, and the further work needed to go from 0.12 m to 0.24 m.
Given
a pull of $18\ \mathrm{N}$ produces a stretch of $0.045\ \mathrm{m}$
the spring is ideal over the whole range used
stretches are measured from the natural length
Find
the stiffness, and the work for each of the two stretches
Check the second answer as an area instead of a difference: the strip is a trapezium with parallel sides $48\ \mathrm{N}$ and $96\ \mathrm{N}$ and width 0.12 m, so its area is $\tfrac12(48+96)(0.12) = 8.64$ J. Two shapes, one number.
One division and three squarings. The only place to go wrong is treating the second stretch as $\tfrac12k(0.12)^{2}$ again, which would give 2.88 J and miss the point of the block.
Ratios are the fast way to see this: the works for stretches 1 and 2 units are as $1^{2}$ to $2^{2}$, so the first quarter of the total range costs a sixteenth of the total work. Springs get expensive at the far end.
A spring launching a cart, and how fast it leaves
A 0.50 kg cart is pressed against a spring of stiffness $250\ \mathrm{N/m}$, compressing it 0.20 m, and is then released on a smooth horizontal track. Find the work the spring does on the cart and the speed at which the cart leaves the spring.
Given
$m = 0.50\ \mathrm{kg}$
$k = 250\ \mathrm{N/m}$
compression $x = 0.20\ \mathrm{m}$ from the natural length
the track is smooth and horizontal
the cart starts from rest
Find
the work done on the cart by the spring, and its launch speed
SolutionThe spring gives back exactly what was stored in squashing it
$$W_{\rm spring\ on\ cart} = \tfrac12 k x^{2} = \tfrac12(250)(0.20)^{2}$$
the spring now pushes the way the cart moves, so its work on the cart is positive here
$$= \tfrac12(250)(0.040) = 5.0\ \mathrm{J}$$
the same triangle as before, read in the other direction along the same graph
Feed it into the work-energy principle
$$5.0 = \tfrac12(0.50)v^{2} - 0$$
the spring is the only force with a component along the motion, so 5.0 J is the net work
the cart is light, so a modest 5.0 J buys a respectable speed
Answer $$\boxed{\;W = 5.0\ \mathrm{J},\qquad v = 4.5\ \mathrm{m/s}\;}$$
Check
Independent check on the size of the answer: the largest force the spring ever applied was $kx = 50\ \mathrm{N}$, and if it had somehow held that all the way it would have delivered 10 J and a speed of 6.3 m/s. The real answer must be below that, and 4.5 m/s is.
Two formulas and no forces resolved. Doing this with the second law would need an acceleration that changes at every instant, which is not a calculation available with the tools of the earlier sections.
Notice what the spring's own force never had to be: the answer used only the area, never the value of $F$ at any particular moment. That is the practical payoff of the area rule, and it is why spring questions are quick once the shape is recognised.
Checkpoint
§08.7 — the work stored in one stretch●●○○○
Thirty seconds, one substitution. A spring with a stiffness of $320\ \mathrm{N/m}$ is stretched from its natural length by 0.15 m and held there.
Given
$k = 320\ \mathrm{N/m}$
stretch $x = 0.15\ \mathrm{m}$ from the natural length
Find
(a) Find the work done in stretching the spring from its natural length to that point.
Hint 1/4
The force is not constant along this stretch, so decide which of the two work rules in this section applies before writing anything.
Hint 2/4
For a spring taken from its natural length to a stretch $x$, the work is $\tfrac12 kx^{2}$.
Hint 3/4
Here $k = 320\ \mathrm{N/m}$ and $x = 0.15\ \mathrm{m}$, so $x^{2} = 0.0225\ \mathrm{m^{2}}$.
Hint 4/4
The work is 3.6 J.
Show solutionUse the triangle
$$W = \tfrac12 k x^{2} = \tfrac12(320)(0.15)^{2} = 3.6\ \mathrm{J}$$
the pull rose in a straight line from zero to 48 N, so its average value over the stretch is 24 N
Answer $$\boxed{\;W = 3.6\ \mathrm{J}\;}$$
Check
Same number from the average force: $(24)(0.15) = 3.6$ J. The endpoint average is legitimate here because the graph is a straight line, which is the one case where that shortcut is exact.
Three and a half joules is about what it takes to lift a bag of sugar 40 cm, which is a fair description of pulling a stiff spring 15 cm. Spring answers in this section should stay in the single digits of joules.
⚠ Using force times distance for a spring
$F = kx$ hands you a force and the question hands you a distance, so the two get multiplied out of habit
the question often gives a total length or a position on a bench, and the formula silently expects the distance from the natural length
wrong$$W = \tfrac12 k (0.24 - 0.12)^{2} = 2.88\ \mathrm{J}$$
right$$W = \tfrac12 k (0.24)^{2} - \tfrac12 k (0.12)^{2} = 8.64\ \mathrm{J}$$
Working out the work done by one named force
Any time a question asks how much work a particular force does, and as the inner loop of every ledger you will build. Five steps, of which the first two are decisions and only the last is arithmetic.
Name the body and write down its displacement
One body, one move. Write the displacement as a length and a direction, because every angle in the calculation is measured against it. If the body's path bends, cut it into straight pieces and do them one at a time.
Draw the force and read the angle between it and that displacement
Draw the two arrows from the same point. The angle you want is between them, not between the force and the horizontal and not between the force and the surface. On a slope these differ, and that difference is where most lost marks live.
Decide whether the force is constant over the whole move
Constant means the same size and the same direction throughout. A rope with a fixed tension is constant; a spring never is; a rope described as weakening is not. This decision picks the formula, so make it explicitly rather than by habit.
Apply the matching rule
Constant force: $W = Fd\cos\theta$, or $W = F_xd_x + F_yd_y$ if components are given, which needs no angle. Varying force: the work is the area under the force against position graph, computed as an integral or as triangles and rectangles.
Attach the sign and the unit before moving on
A force that leans forward gives a positive work, one that leans backward a negative one, and one at a right angle gives zero. Write the joules down. An unsigned or unlabelled number will be added into a ledger later and there will be no way to tell then what it meant.
Where it goes wrong
Reading the slope angle off the diagram when the force and the displacement make a different angle with each other.
Using the constant-force rule on a spring, which gives an answer exactly twice too big.
Writing the friction contribution as a positive number because the friction force itself was quoted as positive.
Solving a problem with the work-energy principle
When a question links forces and a distance to a change of speed, and never mentions time. If the question asks for a time or an acceleration, this is the wrong tool and the second law is the right one.
Fix the two states and write the two speeds
State 1 is where the body starts and state 2 where it finishes. Write $v_1$ and $v_2$ even if one of them is zero, and especially if one of them is the unknown. Everything else in the solution refers to this pair.
Draw the free-body diagram and mark the displacement
Every force on the body, drawn once, plus the displacement between the two states. This is the same diagram you drew for force problems; nothing new is needed and nothing may be left off.
Get the normal force from the perpendicular equation, if friction is involved
Never write $N = mg$ from memory. On a level floor with nothing else vertical it comes out that way; on a slope it is $mg\cos\theta$; with a rope pulling upward at an angle it is smaller than $mg$. Friction is $\mu_k N$, so this step decides the friction entry.
Write one ledger line per force and add them
Force, angle to the displacement, work, with its sign. Perpendicular forces give zeros and are worth writing down as zeros so that you can see the list is complete. The sum is $W_{\rm net}$.
Set the total equal to the change in kinetic energy and solve
$W_{\rm net} = \tfrac12 mv_2^{2} - \tfrac12 mv_1^{2}$, in that order, final minus initial. Solve for whichever symbol the question left blank, in symbols first if you can, and substitute at the end.
Test the answer at a limit before writing it down
Set the friction coefficient to zero, or the angle to zero, or the mass to something absurd, and see whether the formula does what it obviously should. A stopping distance that does not grow with speed, or a final speed that depends on the mass when the mass ought to cancel, is telling you something.
Where it goes wrong
Putting the work of one force into the principle instead of the net work.
Writing $\Delta KE$ as initial minus final, which flips the sign of every answer.
Reaching for this tool on a question that asks how long the motion took, which it cannot answer.
The same rope pulled flat along the floor
A 30 kg crate is dragged 5.0 m across a level floor by a rope with a tension of 150 N held horizontally. The coefficient of kinetic friction is 0.300 and the crate starts from rest. Find the net work and the final speed.
Given
$m = 30\ \mathrm{kg}$
$T = 150\ \mathrm{N}$, horizontal
$\mu_k = 0.300$
$d = 5.0\ \mathrm{m}$
$g = 9.80\ \mathrm{m/s^{2}}$
Find
the net work and the final speed
SolutionNormal force, then friction
$$N = mg = (30)(9.80) = 294\ \mathrm{N}$$
the rope is horizontal, so it takes no share of the weight
$$F_{fr} = (0.300)(294) = 88.2\ \mathrm{N}$$
sliding, so the kinetic coefficient applies
Ledger and total
$$W_T = (150)(5.0) = +750\ \mathrm{J}$$
the pull is along the motion, so the cosine is one
$$W_{fr} = -(88.2)(5.0) = -441\ \mathrm{J}$$
friction opposes the motion for the whole 5.0 m
$$W_{\rm net} = 750 - 441 = 309\ \mathrm{J}$$
the vertical forces are perpendicular and contribute nothing
starting from rest, so the whole net work goes into the final kinetic energy
Answer $$\boxed{\;W_{\rm net} = 309\ \mathrm{J},\qquad v = 4.5\ \mathrm{m/s}\;}$$
Check
Force route: net force $150 - 88.2 = 61.8$ N, so $a = 2.06\ \mathrm{m/s^{2}}$ and $v = \sqrt{2(2.06)(5.0)} = 4.5\ \mathrm{m/s}$.
The same rope tilted 40 degrees upward
Everything as before, but the 150 N rope is now held at $40^{\circ}$ above the horizontal. A 30 kg crate, 5.0 m across a level floor, $\mu_k = 0.300$, starting from rest. Find the net work and the final speed.
Given
$m = 30\ \mathrm{kg}$
$T = 150\ \mathrm{N}$ at $40^{\circ}$ above the horizontal
a little slower than the flat rope, and only a little, because the two effects nearly cancel
Answer $$\boxed{\;W_{\rm net} = 278\ \mathrm{J},\qquad v = 4.3\ \mathrm{m/s}\;}$$
Check
Check the trade-off arithmetic on its own: pull lost is $Td(1-\cos 40^{\circ}) = 175$ J, friction saved is $\mu_k T\sin 40^{\circ}d = 145$ J, and $309 - 175 + 145 = 279$ J, which is the second net work back again to rounding.
Same rope, same tension, same distance, same floor: tilting the rope up by $40^{\circ}$ throws away 175 J of pull and saves 145 J of friction, so here it loses 31 J, but raise the coefficient to 0.500 and the same tilt gains 66 J instead.
How to tell them apart
Tilting the rope upward pays only when the friction it saves beats the pull it wastes, that is when $\mu_k\sin\theta > 1-\cos\theta$. At $40^{\circ}$ that means $\mu_k > 0.364$, so on a slick floor pull flat and on a sticky one pull up.
A constant 48 N pull through 0.12 m
A steady horizontal force of 48 N drags a block 0.12 m along a smooth track. Find the work done.
Given
$F = 48\ \mathrm{N}$, constant and along the motion
$d = 0.12\ \mathrm{m}$
Find
the work done by the force
SolutionConstant force, so a rectangle
$$W = Fd = (48)(0.12) = 5.76\ \mathrm{J}$$
the force had this value at every point of the move, so the area under its graph is a rectangle
Answer $$\boxed{\;W = 5.76\ \mathrm{J}\;}$$
Check
Dimension check: newtons times metres are joules, and 48 N through 12 cm is about the effort of lifting a 4 kg bag by 15 cm, which is small and believable.
A spring that reaches 48 N at 0.12 m
A spring of stiffness $400\ \mathrm{N/m}$ is stretched 0.12 m from its natural length, at which point the pull needed is exactly 48 N. Find the work done in stretching it.
the pull started at zero and only reached 48 N at the very end, so the area is half the rectangle
Answer $$\boxed{\;W = 2.88\ \mathrm{J}\;}$$
Check
Average force route: the pull rose in a straight line from 0 to 48 N, so its average is 24 N, and $(24)(0.12) = 2.88$ J.
Both moves end with the same 48 N being applied and both cover the same 0.12 m, and the spring takes exactly half the work, because it only reached 48 N at the last instant while the constant force had it from the start.
How to tell them apart
Ask what the force was doing in the middle of the move, not at the end of it. If the number quoted is the force throughout, the work is $Fd$; if it is the force only at the finish and the force grew from zero, the work is $\tfrac12 Fd$.
Scaffolding comes off
The common skeleton
Name the one body and mark its two states, with a speed written at each
Draw every force on it, and mark the displacement between the two states
Choose axes along and across the motion, and get the normal force from the across equation
Turn the normal force into a friction force if the surfaces are rough
Write one work line per force, with its angle and its sign, and add them to get the net work
Set the net work equal to the change in kinetic energy, solve, then test the result at a limit
1 · fully worked
A skier towed 40 m up a 15 degree slope at constant speed
A 65 kg skier is towed 40 m up a $15.0^{\circ}$ slope at a constant speed by a rope lying along the slope. The coefficient of kinetic friction between skis and snow is 0.100. Find the work done by the rope, and confirm the ledger adds up.
Given
$m = 65\ \mathrm{kg}$
slope angle $15.0^{\circ}$
$\mu_k = 0.100$
$d = 40\ \mathrm{m}$ up the slope
constant speed throughout
$g = 9.80\ \mathrm{m/s^{2}}$
Find
the work done by the tow rope, and a check that the whole ledger balances
The ledger closing to zero is an independent check on the tension: had $T$ been wrong by even a newton, the three works would have failed to cancel by 40 J. Order of magnitude: nine kilojoules is roughly what it takes to lift the skier 14 m straight up, and she has in fact risen $40\sin 15.0^{\circ} = 10.4$ m while dragging against friction, so the size is right.
Four steps, and the only one with any risk in it was the first. Every rung below reuses this same skeleton.
Constant speed did two jobs here: it gave the tension for free, and it told us in advance what the ledger had to add up to. Look for that phrase before you start computing.
2 · you write the reasoning
Easier than rung 1, because the slope is gone. A 65 kg skier is towed 40 m across level snow at constant speed by a horizontal rope, with $\mu_k = 0.100$. The three lines below are correct and the answer is right. Your job is to say why each line is allowed, in your own words, before opening the model reasons.
reasoning
The rope is horizontal, so it has no vertical component and the only two vertical forces are the normal force and the weight. The skier stays on the snow, so the vertical acceleration is zero and that axis is a balance. This is one of the few arrangements in which $N = mg$ is actually true, and notice that it came out of an equation rather than being assumed; tilt the rope and this line changes.
reasoning
The skier is sliding, so the friction is kinetic and is an equality rather than an inequality. The tension equals it because the speed is constant, which makes the along-motion axis a balance too. On the slope of rung 1 this step had a second term, the along-slope weight, and here that term is exactly zero.
reasoning
The rope lies along the motion, so the angle in $W = Td\cos\theta$ is zero and the cosine is one. The answer is a little over a quarter of the slope answer. Almost all of the difference is the 6.59 kJ that went into raising the skier 10.4 m, work that on level snow does not have to be done at all; the small remainder is friction, which is a touch smaller on a slope because the snow is pressed less hard.
3 · find the buried error
Harder than rung 2, because the rope is back at an angle and the crate is speeding up. A 12.0 kg crate on a level floor is pulled 6.00 m from rest by a rope at $30.0^{\circ}$ above the horizontal with a tension of 55.0 N, against a coefficient of kinetic friction of 0.200. A student's solution is written out below and reaches a final speed of 5.61 m/s. Exactly two of its four steps are faulty. Find them.
the two buried errors (2)
⚠ step 1
The normal force is taken as the whole weight. The rope pulls partly upward, so it unloads the floor: $N = mg - T\sin 30.0^{\circ} = 117.6 - 27.5 = 90.1\ \mathrm{N}$, and the friction is $(0.200)(90.1) = 18.0\ \mathrm{N}$, not 23.5 N.
$N = mg$ is true on every level floor with nothing else vertical, and the rope's angle looks like a fact about the pull rather than a fact about the floor. Nothing on the page looks wrong afterwards, because a normal force of 118 N is a perfectly reasonable number.
right
Write the vertical equation out every time before reaching for $\mu_k$. A quick test: anything pulling upward must make $N$ smaller than $mg$, so an $N$ equal to the weight when a rope is tilted up is wrong before the arithmetic is checked.
⚠ step 2
The cosine is missing. Only the horizontal part of the tension travels with the crate, so $W_{\rm rope} = (55.0)(6.00)\cos 30.0^{\circ} = 286\ \mathrm{J}$, not 330 J.
Both numbers needed for $Fd$ are printed in the question and the angle is in the picture, so the multiplication happens before the picture is consulted. It is the single commonest slip in this section.
right
Before multiplying a force by a distance, say out loud what angle the two arrows make. If the answer is not zero, the cosine belongs in the line.
4 · the bare problem
§08.5 — a box pushed up a rough incline●●●●○
No scaffolding this time; the same six-step skeleton, on a slope, with the push along the slope. A 8.00 kg box is pushed 4.00 m up a $20.0^{\circ}$ incline by a force of 90.0 N directed along the incline, starting from rest, with a coefficient of kinetic friction of 0.250 between box and incline.
Given
$m = 8.00\ \mathrm{kg}$
incline angle $20.0^{\circ}$
$F = 90.0\ \mathrm{N}$ directed up along the incline
$\mu_k = 0.250$
$d = 4.00\ \mathrm{m}$ up the incline
starts from rest
$g = 9.80\ \mathrm{m/s^{2}}$
Find
(a) Find the work done by each of the four forces on the box.
(b) Find the box's speed after the 4.00 m.
Hint 1/4
Four forces, four ledger lines, then one equation. Decide which of the four can be written down as zero without any arithmetic.
Hint 2/4
$N = mg\cos\theta$ on a slope, $F_{fr} = \mu_k N$, each work is $Fd\cos(\text{angle to the motion})$, and $W_{\rm net} = \tfrac12 mv^{2}$ from rest.
Hint 3/4
Here $m = 8.00$ kg, $\theta = 20.0^{\circ}$, $F = 90.0$ N, $\mu_k = 0.250$ and $d = 4.00$ m, with $\sin 20.0^{\circ} = 0.3420$ and $\cos 20.0^{\circ} = 0.9397$.
Hint 4/4
The push does 360 J, gravity $-107$ J, friction $-73.7$ J, the normal force nothing, and the box leaves at 6.69 m/s.
Limit test: set $\mu_k = 0$ and the net work becomes 253 J, giving 7.95 m/s, which must be faster than the rough answer and is. Set the slope to zero as well and the net work becomes the full 360 J with $v = 9.49$ m/s, the fastest of the three, exactly as it should be.
Two obstacles, gravity and friction, took 181 J of the push's 360 J between them. On a slope, gravity is usually the larger of the two, and it is the one that does not depend on how rough the surface is.
Full exam-style question
Exam format: a spring launch across a rough patch, in three partsexam format
A 2.50 kg block is pressed against a spring of stiffness $620\ \mathrm{N/m}$, compressing it 0.180 m, and is then released. The surface under the spring is smooth, but beyond the point where the block leaves the spring there is a rough patch 1.20 m long with a coefficient of kinetic friction of 0.320. (a) Find the work the spring does on the block. (b) Find the speed at which the block leaves the spring. (c) Find its speed at the far end of the rough patch.
Given
$m = 2.50\ \mathrm{kg}$
$k = 620\ \mathrm{N/m}$
compression $x = 0.180\ \mathrm{m}$
rough patch length $L = 1.20\ \mathrm{m}$
$\mu_k = 0.320$ on the rough patch, smooth elsewhere
block starts from rest, surface horizontal throughout
$g = 9.80\ \mathrm{m/s^{2}}$
Find
the spring's work, the launch speed, and the speed after the rough patch
Solution(a) The spring's contribution is an area, not a product
$$W_S = \tfrac12 k x^{2} = \tfrac12(620)(0.180)^{2}$$
the spring force fell from $kx = 112$ N to zero as the block moved out, so the constant-force rule does not apply
Independent check on part (c): find the patch length that would just stop the block. It is $10.04/7.84 = 1.28\ \mathrm{m}$, only 8 cm longer than the patch it actually crossed, so a very small surviving speed is exactly what should come out. A final answer near 2 m/s would have been inconsistent with that margin.
Three parts, but only two ideas: an area for the spring and a ledger for the patch. Part (c) was done in one equation over the whole journey rather than in two stages, which halves the number of intermediate numbers that can go wrong.
The last check is the transferable part. Whenever a body only just makes it, compute the distance that would exactly stop it and compare; if the two are close, a small final speed is confirmed, and if they are not, something upstream is wrong.
Practice
A · concept 4 questions
1§08.1 — when a force does no work at all●●○○○
A one-mark statement of the kind that opens a paper, and one that most people accept on the first reading. A porter walks 30 m along a level corridor carrying a heavy case at a steady height and a steady speed.
Given
the case is carried at constant height
the corridor is level
the porter walks 30 m at a steady speed
Find
(a) True or false: a force acting on a body while the body moves must do some work on it. Give your reason in one sentence.
Hint 1/4
The statement claims a link between two things. Name the two things precisely and ask what actually connects them.
Hint 2/4
$W = Fd\cos\theta$, and the cosine is zero when the force is at a right angle to the displacement.
Hint 3/4
Here the porter's force on the case is vertical, holding it up, while the case's displacement is horizontal.
Hint 4/4
False: a perpendicular force does no work however long the journey.
Show solutionApply the definition to the geometry given
the angle is between the force and the displacement, and here they are square on
$$W = Fd(0) = 0$$
no size of force and no length of corridor can rescue a factor of zero
Answer $$\boxed{\;\text{False: } W = 0 \text{ when } \vec F \perp \vec d\;}$$
Check
Same conclusion by components: with the force $(0, F)$ and the displacement $(d, 0)$, the scalar product is $0\cdot d + F\cdot 0 = 0$, reached without any mention of an angle.
This is the same fact as the normal force and the centripetal force doing no work. Any force that stays perpendicular to the motion is a passenger in the work ledger.
2§08.1 — work done by the hand that carries a case●●○○○
The same corridor, now with numbers, because the numerical version is where the marks are. A 20.0 kg case is carried 30.0 m along a level corridor at constant height and constant speed.
Given
$m = 20.0\ \mathrm{kg}$
$d = 30.0\ \mathrm{m}$, horizontal
the case is held at constant height and constant speed
$g = 9.80\ \mathrm{m/s^{2}}$
Find
(a) How much work does the carrying force do on the case?
Hint 1/4
Before computing anything, decide which way the carrying force points and which way the case goes.
Hint 2/4
$W = Fd\cos\theta$, with $\theta$ the angle between the force and the displacement.
Hint 3/4
Here the force is vertical with size $mg = 196\ \mathrm{N}$, the displacement is horizontal with length 30.0 m, so the angle between them is a right angle.
Hint 4/4
The work is zero.
Show solutionSize the force, then check the angle
$$F = mg = (20.0)(9.80) = 196\ \mathrm{N}$$
constant speed and constant height mean the carrying force exactly balances the weight
$$W = (196)(30.0)\cos 90^{\circ} = 0$$
the angle does all the work here, and it makes the answer independent of both other numbers
Answer $$\boxed{\;W = 0\;}$$
Check
Component check: force $(0, 196)$ N, displacement $(30.0, 0)$ m, and the scalar product is $0 + 0 = 0$.
The distractors in this question are all real numbers from real calculations, which is why they are tempting. Each of them answers a question that was not asked.
3§08.4 — two bodies carrying the same kinetic energy●●●○○
A comparison question, which is the shape most conceptual energy items take. Two blocks slide along a bench and a measurement shows that they carry exactly the same kinetic energy, but one of them has four times the mass of the other.
Given
both blocks have the same kinetic energy
$m_{\rm heavy} = 4 m_{\rm light}$
Find
(a) How do their speeds compare?
Hint 1/4
Write the equality the question gives you as an equation before trying to picture the answer.
Hint 2/4
$KE = \tfrac12 mv^{2}$, so equal kinetic energies means $m_1v_1^{2} = m_2v_2^{2}$.
Hint 3/4
Here $m_{\rm heavy} = 4m_{\rm light}$, so $4m_{\rm light}v_{\rm heavy}^{2} = m_{\rm light}v_{\rm light}^{2}$.
Hint 4/4
The lighter block is moving twice as fast.
Show solutionSet the two energies equal and cancel
Test with numbers: 1.0 kg at 4.0 m/s carries 8.0 J, and 4.0 kg at 2.0 m/s carries 8.0 J. Equal energies, speed ratio two.
Every time the energies are equal, mass ratios turn into speed ratios through a square root. That single fact answers most comparison questions in this section without any arithmetic.
4§08.3 — the sign of the work done by friction●●●●○
A statement that is true in every worked example you have seen so far, which is exactly what makes it dangerous. Consider a crate sitting on the flat bed of a lorry, not sliding, while the lorry pulls away from a junction and speeds up.
Given
the crate does not slide on the bed
the lorry, and with it the crate, is speeding up
the only horizontal force on the crate is the friction from the bed
Find
(a) True or false: the work done by a friction force on a body is always negative. Give your reason in one sentence.
Hint 1/4
Ask what friction is doing to this particular body, and which way the body is going while it does it.
Hint 2/4
Work is negative only when the force opposes the displacement; the sign comes from the angle, not from the name of the force.
Hint 3/4
Here the crate accelerates forward, and the only forward force acting on it is the friction from the lorry bed, so friction points the way the crate travels.
Hint 4/4
False: the friction on the crate does positive work.
Show solutionFind the direction of the friction from the acceleration
$$\textstyle\sum F_x = ma_x > 0$$
the crate is speeding up, so something must be pushing it forward
$$F_{fr} = ma_x \ \text{forward}$$
friction from the bed is the only horizontal force on the crate, so it is that something
Consistency check with the previous section: this crate has zero relative sliding, so the friction acting is static, and static friction is precisely the force that was described there as supplying whatever the motion demands.
Rules of thumb about signs are worth having, but only if you can say which condition they rest on. This one rests on the body sliding backwards relative to the surface, and it fails whenever the surface is what drives the body.
B · computation 8 questions
1§08.1 — a mower pushed with a downward slanting force●●○○○
The standard first computation of this section, with the angle below the horizontal rather than above it, which changes nothing in the formula and quite a lot in the diagram. A 24.0 kg lawnmower is pushed 12.0 m across level ground by a force of 55.0 N directed along the handle, which points $32.0^{\circ}$ below the horizontal.
Given
$m = 24.0\ \mathrm{kg}$
$F = 55.0\ \mathrm{N}$ along the handle
the handle is at $32.0^{\circ}$ below the horizontal
$d = 12.0\ \mathrm{m}$ across level ground
$g = 9.80\ \mathrm{m/s^{2}}$
Find
(a) Find the work done on the mower by the push along the handle.
(b) Find the work done on the mower by gravity and by the ground's normal force.
Hint 1/4
Two arrows, one angle. Decide what angle the push makes with the direction the mower actually travels before writing any formula.
Hint 2/4
$W = Fd\cos\theta$ with $\theta$ measured between the force and the displacement; a force tilted below the horizontal makes the same angle with a horizontal displacement as one tilted above it.
Hint 3/4
Here $F = 55.0$ N, $d = 12.0$ m and $\theta = 32.0^{\circ}$, with $\cos 32.0^{\circ} = 0.8480$.
Hint 4/4
The push does 560 J and the two vertical forces do nothing.
Bound: the most this push could do over 12.0 m is $55.0 \times 12.0 = 660$ J, and $\cos 32.0^{\circ}$ is about 0.85, so an answer near 560 J is the right size.
Downward slanting is not the same as unhelpful. The vertical part still does no work, but unlike an upward pull it presses the mower harder into the ground, which raises the normal force and so the friction. The work it does is zero; its effect on the ledger is not.
2§08.2 — work from a pair of position vectors●●●○○
Component questions usually arrive with the displacement disguised as a pair of positions, and the first job is to undisguise it. A constant force $\vec F = (18\,\hat\imath - 7.0\,\hat\jmath)\ \mathrm{N}$ acts on a particle that moves from the point $(2.0, 1.0)\ \mathrm{m}$ to the point $(6.0, 4.0)\ \mathrm{m}$.
Given
$\vec F = (18\,\hat\imath - 7.0\,\hat\jmath)\ \mathrm{N}$, constant
start point $(2.0, 1.0)\ \mathrm{m}$
end point $(6.0, 4.0)\ \mathrm{m}$
Find
(a) Find the displacement vector of the particle.
(b) Find the work done by the force.
(c) Find the angle between the force and the displacement.
Hint 1/4
A work calculation needs a displacement, and you have been given two positions instead. Fix that before anything else.
Hint 2/4
$\vec d = \vec r_2 - \vec r_1$, then $W = F_xd_x + F_yd_y$, and finally $\cos\theta = W/(Fd)$.
Hint 3/4
Here $\vec F = (18, -7.0)\ \mathrm{N}$, $\vec r_1 = (2.0, 1.0)\ \mathrm{m}$ and $\vec r_2 = (6.0, 4.0)\ \mathrm{m}$.
Hint 4/4
The displacement is $(4.0, 3.0)$ m, the work is 51 J and the angle is $58.1^{\circ}$.
acute, which agrees with the positive work found above
Answer $$\boxed{\;\vec d = (4.0\,\hat\imath + 3.0\,\hat\jmath)\ \mathrm{m},\quad W = 51\ \mathrm{J},\quad \theta = 58.1^{\circ}\;}$$
Check
Check the work by the other formula: $Fd\cos\theta = (19.31)(5.0)(0.528) = 51$ J, from the magnitudes rather than the components.
Using the positions themselves instead of their difference is the standard trap here, and it gives $(18)(6.0)+(-7.0)(4.0) = 80$ J, which is wrong and looks fine.
3§08.3 — a full ledger for a crate on a rough floor●●●○○
The four-line ledger, which is the single most examinable skill in this section. A 15.0 kg crate is dragged 5.00 m across a level floor by a horizontal rope with a tension of 48.0 N, against a coefficient of kinetic friction of 0.250, starting from rest.
Given
$m = 15.0\ \mathrm{kg}$
$T = 48.0\ \mathrm{N}$, horizontal
$\mu_k = 0.250$
$d = 5.00\ \mathrm{m}$
starts from rest
$g = 9.80\ \mathrm{m/s^{2}}$
Find
(a) Find the work done by each of the four forces on the crate.
(b) Find the net work.
(c) Find the crate's speed after the 5.00 m.
Hint 1/4
Four forces means four lines, and two of them can be written down without any arithmetic. Identify those two first.
Hint 2/4
$N = mg$ here because the rope is horizontal; then $F_{fr} = \mu_k N$, each work is $Fd\cos\theta$, and $W_{\rm net} = \tfrac12 mv^{2}$ from rest.
Hint 3/4
Here $m = 15.0$ kg, $T = 48.0$ N horizontal, $\mu_k = 0.250$ and $d = 5.00$ m, so $mg = 147\ \mathrm{N}$.
Hint 4/4
The rope does $+240$ J, friction $-184$ J, the vertical pair nothing, the net work is 56.3 J and the speed is 2.74 m/s.
Show solutionNormal force and friction
$$N = mg = (15.0)(9.80) = 147\ \mathrm{N}$$
the rope is horizontal, so it has no vertical component to change this
$$F_{fr} = (0.250)(147) = 36.8\ \mathrm{N}$$
sliding, so the kinetic coefficient applies as an equality
Force route: net force $48.0 - 36.75 = 11.25$ N, so $a = 0.750\ \mathrm{m/s^{2}}$ and $v = \sqrt{2(0.750)(5.00)} = 2.74\ \mathrm{m/s}$. Two routes, one answer.
The pattern to carry away is the shape of the ledger, not the numbers: a positive entry from whatever drives, a negative entry from whatever resists, and zeros from everything perpendicular.
4§08.4 — the energy in a served tennis ball●●●○○
A kinetic energy calculation followed by the question every impact problem really asks. A 0.0570 kg tennis ball leaves a racket at 58.0 m/s, having been essentially at rest a moment earlier, and the racket stayed in contact with it over a distance of 0.350 m.
Given
$m = 0.0570\ \mathrm{kg}$
$v = 58.0\ \mathrm{m/s}$ after the hit
the ball was at rest before the hit
contact distance $d = 0.350\ \mathrm{m}$
Find
(a) Find the kinetic energy of the ball as it leaves the racket.
(b) Find the average force the racket exerted on the ball.
Hint 1/4
Part (b) asks for a force from an energy and a distance, so decide which statement in this section connects those three.
Hint 2/4
$KE = \tfrac12 mv^{2}$, and if the racket's force is the only one worth counting, $Fd = \Delta KE$.
Hint 3/4
Here $m = 0.0570$ kg, $v = 58.0$ m/s and $d = 0.350$ m, so $v^{2} = 3364\ \mathrm{m^{2}/s^{2}}$.
Hint 4/4
The ball carries 95.9 J and the average force was 274 N.
square the speed first; the ball is light but 58 m/s is fast and the square is what dominates
The force that put it there
$$F d = \Delta KE = 95.87 - 0 \;\Rightarrow\; F = \frac{95.87}{0.350} = 274\ \mathrm{N}$$
the racket's push is along the ball's motion, and gravity over 0.35 m contributes a fraction of a joule, which is negligible here
Answer $$\boxed{\;KE = 95.9\ \mathrm{J},\qquad F = 274\ \mathrm{N}\;}$$
Check
Plausibility: 274 N on a 57 gram ball is an acceleration of about $4.8\times10^{3}\ \mathrm{m/s^{2}}$, some 490 times $g$, which is the right order for a tennis serve and would be absurd for anything gentler.
Note the phrase average force. The racket's push rises and falls during contact; the work-energy route delivers the average without needing to know the shape of that rise, which is precisely why it is used for impacts.
5§08.5 — braking distance backwards, from a measurement●●●○○
The braking problem run in reverse, which is how accident investigators actually use it. A 1250 kg car travelling at 22.0 m/s brakes to a standstill in 38.0 m on a level road.
Given
$m = 1250\ \mathrm{kg}$
$v_1 = 22.0\ \mathrm{m/s}$, $v_2 = 0$
$d = 38.0\ \mathrm{m}$ on a level road
$g = 9.80\ \mathrm{m/s^{2}}$
Find
(a) Find the net work done on the car while it stops.
(b) Find the average braking force.
(c) Find the coefficient of friction this implies, assuming friction was the only horizontal force.
Hint 1/4
Start from what the car had and what it ended with; the force and the coefficient both follow from that one number.
Hint 2/4
$W_{\rm net} = \tfrac12 mv_2^{2} - \tfrac12 mv_1^{2}$, then $W_{\rm net} = -Fd$, then $F = \mu_k mg$.
Hint 3/4
Here $m = 1250$ kg, $v_1 = 22.0$ m/s, $v_2 = 0$ and $d = 38.0$ m, so $v_1^{2} = 484\ \mathrm{m^{2}/s^{2}}$.
Hint 4/4
The net work is $-3.03\times10^{5}$ J, the force is $7.96\times10^{3}$ N and the coefficient is 0.650.
Cross-check the coefficient through the distance formula instead: $d = v^{2}/(2\mu_k g) = 484/(2)(0.6498)(9.80) = 38.0$ m, which returns the measured distance.
The mass cancelled out of part (c) even though it was used in parts (a) and (b), which is why an investigator can read a coefficient off a skid mark without weighing the car.
6§08.6 — a work read off a three piece graph●●●○○
Varying-force questions on exam papers are usually graphs made of straight pieces, and the skill is cutting them at the corners. A force acts on a 5.00 kg block along the direction of motion on a smooth horizontal track. It rises in a straight line from 0 at $x = 0$ to 40.0 N at $x = 3.00\ \mathrm{m}$, stays at 40.0 N until $x = 7.00\ \mathrm{m}$, then falls in a straight line to 0 at $x = 9.00\ \mathrm{m}$.
Given
$m = 5.00\ \mathrm{kg}$, starting from rest at $x = 0$
$F$ rises linearly from $0$ to $40.0\ \mathrm{N}$ over $0 \le x \le 3.00\ \mathrm{m}$
$F = 40.0\ \mathrm{N}$ for $3.00 \le x \le 7.00\ \mathrm{m}$
$F$ falls linearly from $40.0\ \mathrm{N}$ to $0$ over $7.00 \le x \le 9.00\ \mathrm{m}$
the track is smooth and the force is along the motion throughout
Find
(a) Find the total work done by this force from $x = 0$ to $x = 9.00\ \mathrm{m}$.
(b) Find the block's speed at $x = 9.00\ \mathrm{m}$.
Hint 1/4
Sketch the graph and cut it at the two corners; each piece is a shape whose area you can write down without integrating.
Hint 2/4
The work is the area under the force against position graph, and $W_{\rm net} = \tfrac12 mv^{2}$ from rest.
Hint 3/4
Here the three pieces are a triangle of base 3.00 m and height 40.0 N, a rectangle 4.00 m by 40.0 N, and a triangle of base 2.00 m and height 40.0 N, with $m = 5.00$ kg.
Hint 4/4
The work is 260 J and the speed is 10.2 m/s.
Show solutionOne area per piece
$$W_1 = \tfrac12(3.00)(40.0) = 60.0\ \mathrm{J}$$
a triangle: half the base times the height
$$W_2 = (4.00)(40.0) = 160\ \mathrm{J}$$
a rectangle, and over this stretch the old constant-force rule is exact
$$W_3 = \tfrac12(2.00)(40.0) = 40.0\ \mathrm{J}$$
another triangle, narrower than the first, so worth less
smooth track and force along the motion, so this is the net work
Answer $$\boxed{\;W = 260\ \mathrm{J},\qquad v = 10.2\ \mathrm{m/s}\;}$$
Check
Bound the area: the force never exceeded 40.0 N over 9.00 m, so the work cannot exceed 360 J, and the flat middle alone gives 160 J. The answer sits between those, closer to the upper bound, as the shape suggests.
If a graph question ever tempts you to average the first and last values, look at this one: both ends are zero, and their average would say the force did no work at all.
7§08.7 — a spring measured, then stretched twice●●●○○
Spring questions almost always begin with a measurement that fixes the stiffness, because the stiffness is never handed over directly. A spring is found to stretch 0.080 m when a steady 32 N pull is applied to it.
Given
a pull of $32\ \mathrm{N}$ produces a stretch of $0.080\ \mathrm{m}$
the spring is ideal over the whole range used
all stretches are measured from the natural length
Find
(a) Find the spring's stiffness.
(b) Find the work needed to stretch it from its natural length to 0.25 m.
(c) Find the extra work needed to go from 0.25 m to 0.30 m.
Hint 1/4
The third part is not another application of the same formula; it is a difference between two of them. Notice that before you start.
Hint 2/4
$k = F/x$, then $W = \tfrac12 kx^{2}$ from the natural length, and for a stretch between two points, $W = \tfrac12kx_2^{2} - \tfrac12kx_1^{2}$.
Hint 3/4
Here 32 N gives 0.080 m of stretch, and the two later stretches asked about are 0.25 m and 0.30 m from the natural length.
Hint 4/4
The stiffness is 400 N/m, the first work is 12.5 J and the extra work is 5.50 J.
the strip between the two stretches is the difference of two triangles, not a triangle itself
$$= 18.0 - 12.5 = 5.50\ \mathrm{J}$$
five centimetres of extra stretch cost nearly half of what the first twenty five cost
Answer $$\boxed{\;k = 400\ \mathrm{N/m},\qquad W = 12.5\ \mathrm{J},\qquad \Delta W = 5.50\ \mathrm{J}\;}$$
Check
Check part (c) as a trapezium instead: the pull runs from $kx_1 = 100$ N to $kx_2 = 120$ N over 0.050 m, so the area is $\tfrac12(100+120)(0.050) = 5.50$ J.
A common wrong answer to (c) is $\tfrac12(400)(0.050)^{2} = 0.50$ J, from treating the extra stretch as if it started at the natural length. It is eleven times too small, and the check above catches it instantly.
8§08.7 — a spring launch and the rough patch that stops it●●●●○
The full arc of this section in one question: an area, a principle and a ledger. A 1.20 kg block is held against a spring of stiffness $480\ \mathrm{N/m}$ compressed 0.150 m, on a smooth horizontal surface, and released.
Given
$m = 1.20\ \mathrm{kg}$
$k = 480\ \mathrm{N/m}$
compression $x = 0.150\ \mathrm{m}$
smooth surface for parts (a) and (b); block starts from rest
$g = 9.80\ \mathrm{m/s^{2}}$
Find
(a) Find the work the spring does on the block.
(b) Find the speed at which the block leaves the spring.
(c) If beyond the spring the surface is rough with $\mu_k = 0.200$, find how far the block travels before stopping.
Hint 1/4
Three parts, three different tools, and the third one is a ledger with a single entry in it.
Hint 2/4
$W = \tfrac12kx^{2}$ for the spring; $W_{\rm net} = \tfrac12mv^{2}$ from rest; and on the rough patch $-\mu_k mg\,d = 0 - \tfrac12mv^{2}$.
Hint 3/4
Here $m = 1.20$ kg, $k = 480\ \mathrm{N/m}$, $x = 0.150$ m and later $\mu_k = 0.200$, so $mg = 11.8\ \mathrm{N}$.
Hint 4/4
The spring does 5.40 J, the launch speed is 3.00 m/s and the block travels 2.30 m on the rough patch.
all the energy the spring gave has to be removed, and the patch removes 2.35 J per metre
Answer $$\boxed{\;W = 5.40\ \mathrm{J},\qquad v = 3.00\ \mathrm{m/s},\qquad d = 2.30\ \mathrm{m}\;}$$
Check
Check part (c) through the speed instead of the energy: $a = \mu_k g = 1.96\ \mathrm{m/s^{2}}$, and $d = v^{2}/(2a) = 9.00/3.92 = 2.30$ m, from forces rather than from work.
Part (c) never used the launch speed, and did not need to: what crosses the boundary is an amount of energy, and the rough patch charges a fixed toll per metre until it is gone.
C · exam level 5 questions
1§08.5 — a crate pushed up a ramp, in three parts●●●●○
Exam format: three parts, and the marks are concentrated in the last one. A 22.0 kg crate is pushed 3.50 m up a ramp inclined at $28.0^{\circ}$ to the horizontal by a force of 180 N directed along the ramp. The coefficient of kinetic friction between crate and ramp is 0.180 and the crate starts from rest.
Given
$m = 22.0\ \mathrm{kg}$
ramp angle $28.0^{\circ}$
$F = 180\ \mathrm{N}$ directed up along the ramp
$\mu_k = 0.180$
$d = 3.50\ \mathrm{m}$ up the ramp
starts from rest
$g = 9.80\ \mathrm{m/s^{2}}$
Find
(a) Find the normal force on the crate and the friction force acting on it.
(b) Find the work done by each of the four forces.
(c) Find the crate's speed after 3.50 m, and state what it would have been on a frictionless ramp.
Hint 1/4
Everything depends on part (a), and part (a) depends on one equation you must write rather than recall. Decide which axis that equation belongs to.
Hint 2/4
Across the ramp $N = mg\cos\theta$; then $F_{fr} = \mu_k N$; each work is $Fd\cos(\text{angle to the motion})$; and $W_{\rm net} = \tfrac12 mv^{2}$ from rest.
Hint 3/4
Here $m = 22.0$ kg, $\theta = 28.0^{\circ}$, $F = 180$ N along the ramp, $\mu_k = 0.180$ and $d = 3.50$ m, with $\sin 28.0^{\circ} = 0.4695$ and $\cos 28.0^{\circ} = 0.8829$.
Hint 4/4
The normal force is 190 N, the friction 34.3 N, the works are $+630$, $-354$, $-120$ and 0 joules, and the speed is 3.76 m/s, against 4.66 m/s on a smooth ramp.
Limit test: set the ramp angle to zero and the formulas give $N = 216$ N, $W_G = 0$ and $v = 6.70$ m/s, the fastest of the three cases, exactly as a level floor should be. Set both the angle and the coefficient to zero and $v$ rises again to 7.57 m/s.
On any ramp, compare the two negative entries before you finish: if friction is coming out larger than gravity on a slope steeper than about $\tan^{-1}\mu_k$, one of the two is wrong.
2§08.6 — a force that dies away along the track●●●●○
The integral version, with a force that starts helpful and ends useless. A 3.00 kg body slides along a smooth straight track and is acted on by a force directed along its motion of size $F(x) = 12.0 - 2.00x$ newtons, where $x$ is its position in metres. It starts from rest at $x = 0$.
Given
$m = 3.00\ \mathrm{kg}$
$F(x) = 12.0 - 2.00x$ newtons, along the motion
the body starts from rest at $x = 0$
the track is smooth and straight
Find
(a) Find the work done by this force between $x = 0$ and $x = 6.00\ \mathrm{m}$.
(b) Find the body's speed at $x = 6.00\ \mathrm{m}$.
(c) State where between $x = 0$ and $x = 6.00\ \mathrm{m}$ the body is moving fastest, and why.
Hint 1/4
The force is not constant, so decide which of the two work rules applies before writing anything. For part (c), ask what the force is doing at each stage rather than what the body is doing.
Hint 2/4
$W = \int_{x_1}^{x_2}F(x)dx$, which for a straight-line force is also the area of a trapezium; then $W_{\rm net} = \tfrac12mv^{2}$ from rest.
Hint 3/4
Here $F(0) = 12.0\ \mathrm{N}$, $F(6.00) = 0$ and $m = 3.00$ kg, and the graph of $F$ against $x$ is a straight line falling to zero at $x = 6.00\ \mathrm{m}$.
Hint 4/4
The work is 36.0 J, the speed is 4.90 m/s, and the fastest point is $x = 6.00\ \mathrm{m}$ itself.
the force varies, so the constant-force rule is unavailable and the area is the only route
$$W = 72.0 - 36.0 = 36.0\ \mathrm{J}$$
and the graph is a straight line from 12.0 N to zero over 6.00 m, a triangle of area $\tfrac12(6.00)(12.0) = 36.0$ J
Speed at the end
$$36.0 = \tfrac12(3.00)v^{2} \;\Rightarrow\; v = \sqrt{24.0} = 4.90\ \mathrm{m/s}$$
the track is smooth and the force is along the motion, so this work is the net work
Where the body is fastest
$$F(x) > 0 \ \text{for all } x < 6.00\ \mathrm{m}$$
a positive force along the motion means the body is still gaining speed
$$F(6.00) = 0$$
the speed stops rising exactly where the force runs out, so the maximum is at the end of the run
Answer $$\boxed{\;W = 36.0\ \mathrm{J},\qquad v = 4.90\ \mathrm{m/s},\qquad \text{fastest at } x = 6.00\ \mathrm{m}\;}$$
Check
Bracket the work without integrating: the force lay between 0 and 12.0 N over 6.00 m, so the work is between 0 and 72.0 J, and since the fall is linear the answer must be exactly halfway. It is.
Part (c) is the part worth keeping. The body is slowest to gain speed at the end, but it is still gaining; speed peaks where the force changes sign, not where the force starts to fall.
3§08.5 — two vehicles carrying equal kinetic energy●●●●○
A comparison at exam level, where the tempting answer is the one that reasons about weight rather than about energy. A 1500 kg car and a 3000 kg van happen to carry exactly the same kinetic energy, and both brake to rest on the same road with the same coefficient of friction.
Numerical spot check: take $KE = 3.00\times10^{5}$ J and $\mu_k = 0.700$. The car needs $3.00\times10^{5}/(0.700)(1500)(9.80) = 29.2$ m and the van needs 14.6 m, which is half.
Compare this with the more familiar case where the two vehicles have the same speed rather than the same energy. Then the mass cancels and both stop in the same distance. The two questions look almost identical on the page and have different answers.
4§08.3 — the cable of a lift that is slowing down●●●●○
A sign-heavy problem, which is where careful ledgers earn their keep. A 1150 kg lift is descending and slows uniformly from 3.00 m/s to rest over the last 9.00 m of its travel.
Given
$m = 1150\ \mathrm{kg}$
$v_1 = 3.00\ \mathrm{m/s}$ downward, $v_2 = 0$
$d = 9.00\ \mathrm{m}$ downward
$g = 9.80\ \mathrm{m/s^{2}}$
Find
(a) Find the change in the lift's kinetic energy.
(b) Find the work done by gravity over those 9.00 m.
(c) Find the work done by the cable, and hence the cable tension.
Hint 1/4
Two of the three answers are negative and one is positive. Decide the sign of each from the geometry before you compute its size.
Hint 2/4
$\Delta KE = \tfrac12mv_2^{2} - \tfrac12mv_1^{2}$, $W_G = +mgd$ for a downward move, and the works must add to $\Delta KE$.
Hint 3/4
Here $m = 1150$ kg, $v_1 = 3.00$ m/s, $v_2 = 0$ and $d = 9.00$ m downward, so $mg = 1.127\times10^{4}\ \mathrm{N}$.
Hint 4/4
The kinetic energy falls by $5.18\times10^{3}$ J, gravity does $+1.01\times10^{5}$ J, the cable does $-1.07\times10^{5}$ J and the tension is $1.18\times10^{4}$ N.
Show solutionWhat the motion demands
$$\Delta KE = 0 - \tfrac12(1150)(3.00)^{2} = -5.18\times10^{3}\ \mathrm{J}$$
final minus initial, and the lift ends at rest, so the change is negative
the cable pulls up while the lift moves down, so $W_T = -Td$ and the sign is already accounted for
Answer $$\boxed{\;\Delta KE = -5.18\times10^{3}\ \mathrm{J},\ W_G = +1.01\times10^{5}\ \mathrm{J},\ W_T = -1.07\times10^{5}\ \mathrm{J},\ T = 1.18\times10^{4}\ \mathrm{N}\;}$$
Check
Force route as an independent check: the lift decelerates while descending, so its acceleration points upward with size $a = (3.00)^{2}/(2\times9.00) = 0.500\ \mathrm{m/s^{2}}$, and $T - mg = ma$ gives $T = 1150(9.80+0.500) = 1.18\times10^{4}$ N. The two routes agree.
The tension came out larger than the weight, which is the correct signature of a descending lift being slowed: the cable has to do more than hold the lift up, it has to take motion out of it as well.
5§08.7 — find the two faults in a spring and friction solution●●●●○
A worked solution written by a student, with the answer at the bottom. A 2.00 kg block is launched by a spring of stiffness $500\ \mathrm{N/m}$ compressed 0.120 m, and slides 0.400 m across a rough floor with $\mu_k = 0.250$. The student's four steps are printed below and reach a final speed of 3.03 m/s. Exactly two of the four are faulty.
step 4's method was never wrong; it was fed a wrong number
Answer $$\boxed{\;\text{steps 1 and 3 are faulty};\qquad v = 1.28\ \mathrm{m/s}\;}$$
Check
Sanity check on the size: the friction removes 1.96 J and the spring only supplied 3.60 J, so the block should end up slow, and 1.28 m/s is a slow walk. The student's 3.03 m/s would have needed the spring to supply more than twice what it can.
Both faults inflate the answer, which is the usual pattern: dropped halves and dropped minus signs almost always make a body come out faster than it is. If a mechanical answer looks energetic, check those two places first.
D · interleaved 4 questions
1§08.5 — a car gathering speed on a straight road●●●○○
This set is deliberately mixed, so decide for yourself which tool the question wants before reaching for one. A 1400 kg car accelerates uniformly along a level straight road from 12.0 m/s to 26.0 m/s over a distance of 150 m.
the kinematic relation applies because the acceleration is stated to be uniform
$$F = ma = (1400)(1.7733) = 2.48\times10^{3}\ \mathrm{N}$$
the second law, with no energy anywhere in this route
Answer $$\boxed{\;F = 2.48\times10^{3}\ \mathrm{N}\ \text{by both routes}\;}$$
Check
Order of magnitude: 2.5 kN on a 1400 kg car is $1.8\ \mathrm{m/s^{2}}$, roughly 0 to 100 km/h in 15 s, which is an ordinary family car pressing on rather than a sports car.
When the force is constant and the acceleration uniform, the two routes are the same statement written twice, and either is fine. The energy route earns its place when the force is not constant, where the other route has nothing to say.
2§08.1 — a ball whirled in a horizontal circle●●●○○
Still mixed, and the tool you need may not be the newest one you learned. A 0.250 kg ball on the end of a 0.900 m string is whirled in a horizontal circle at a constant speed of 4.00 m/s, and you are asked about one complete revolution.
Given
$m = 0.250\ \mathrm{kg}$
string length $0.900\ \mathrm{m}$
the ball moves in a horizontal circle at a constant $4.00\ \mathrm{m/s}$
one complete revolution is considered
Find
(a) How much work does the string tension do on the ball during one complete revolution?
Hint 1/4
Do not start with the size of the tension. Start with the direction of the tension compared with the direction the ball is travelling at each instant.
Hint 2/4
$W = Fd\cos\theta$, and a force that stays perpendicular to the motion does no work no matter how long it acts.
Hint 3/4
Here the ball's velocity is always along the tangent of the circle, while the string pulls toward the axis, so the angle between them stays at $90^{\circ}$ throughout.
Hint 4/4
The tension does zero work.
Show solutionCompare the two directions at a general instant
$$\theta = 90^{\circ} \ \text{at every point of the path}$$
the string pulls inward while the ball moves along the tangent, and that stays true all the way round
$$dW = F\,ds\cos 90^{\circ} = 0 \;\Rightarrow\; W = 0$$
every slice of the path contributes zero, so the total over any number of revolutions is zero
Confirm from the other end
$$\Delta KE = \tfrac12(0.250)(4.00)^{2} - \tfrac12(0.250)(4.00)^{2} = 0$$
the speed is the same after a full revolution, so the net work must vanish, which it does
Answer $$\boxed{\;W = 0\;}$$
Check
The two arguments are independent: the first looks only at the geometry of the force, the second only at the speeds at the two ends. Both give zero.
This is the general reason a centripetal force never appears in an energy ledger: it changes the direction of the motion and nothing else. The same argument applies to the normal force on a flat floor and to the gravitational pull on a circular orbit.
3§08.5 — a thrown ball on its way to the top●●●●○
Mixed again, and this one needs a result from an earlier section before the new machinery can be used. A 0.145 kg baseball is thrown at 32.0 m/s at an angle of $40.0^{\circ}$ above the horizontal, and air resistance is ignored.
Given
$m = 0.145\ \mathrm{kg}$
launch speed $32.0\ \mathrm{m/s}$ at $40.0^{\circ}$ above the horizontal
air resistance is ignored
$g = 9.80\ \mathrm{m/s^{2}}$
Find
(a) Find the ball's speed at the highest point of its flight.
(b) Find the work done by gravity between the launch and the highest point.
Hint 1/4
At the top of a projectile's path one component of the velocity is zero and the other is unchanged. Say which is which before starting.
Hint 2/4
The horizontal component $v\cos\theta$ is unchanged throughout the flight; then $W_G = \Delta KE$ between the two points.
Hint 3/4
Here $v = 32.0$ m/s, $\theta = 40.0^{\circ}$ and $m = 0.145$ kg, with $\cos 40.0^{\circ} = 0.7660$.
Hint 4/4
The speed at the top is 24.5 m/s and gravity does $-30.7$ J.
Check by the geometry instead of the speeds: the maximum height is $(32.0\sin 40.0^{\circ})^{2}/(2g) = 21.6$ m, and $W_G = -mgh = -(0.145)(9.80)(21.586) = -30.7$ J, from a route that never mentions kinetic energy.
The ball keeps 59 percent of its kinetic energy at the top, because it keeps all of its horizontal motion. A ball thrown at $80.0^{\circ}$ would keep only 3 percent, which is why steep throws feel so much more expensive for the distance they cover.
4§08.1 — gravity acting on a satellite in a circular orbit●●●○○
The last of the mixed set, and it reaches two sections back. A 720 kg satellite travels in a circular orbit around the Earth, held there by the gravitational pull of the Earth and by nothing else.
Given
$m = 720\ \mathrm{kg}$
the orbit is circular
the only force acting is the Earth's gravitational pull
one complete orbit is considered
Find
(a) True or false: over one complete orbit, the Earth's gravitational pull does positive work on the satellite. Give your reason in one sentence.
Hint 1/4
Ask which way the gravitational pull points and which way the satellite is travelling at the same instant.
Hint 2/4
A force perpendicular to the velocity does no work, and the work over a whole path is the sum of the works over its pieces.
Hint 3/4
Here the pull is directed at the centre of the circular orbit and the satellite's velocity is along the tangent, so the two are at $90^{\circ}$ at every instant.
Hint 4/4
False: the work is zero, not positive.
Show solutionGeometry of a circular path
$$\vec F \perp \vec v \ \text{at every instant}$$
the pull is radial and the velocity is tangential, which is what makes the path a circle rather than a spiral
$$W = 0$$
each slice contributes $F\,ds\cos 90^{\circ} = 0$, so no amount of orbiting accumulates any work
Check against the speed
$$\Delta KE = 0 \;\Rightarrow\; W_{\rm net} = 0$$
a circular orbit is traversed at constant speed, so the kinetic energy never changes
Answer $$\boxed{\;\text{False: } W = 0\;}$$
Check
The two lines are independent: one uses only the direction of the force, the other only the speeds at the two ends of the orbit. Both give zero.
Change the orbit to an ellipse and the answer changes: the pull is then not perpendicular to the motion, the satellite speeds up as it falls inward and slows as it climbs away, and the work over a full lap returns to zero only because it comes back to where it began.
Mistake ledger (14 entries)
⚠ Dropping the cosine when the force is at an angle
the numbers for the force and the distance are both printed in the question and the angle is in a picture, so the two numbers get multiplied and the picture never gets used
every force on the body counted exactly once; the same displacement for all of them
Kinetic energy
$$KE = \tfrac12 m v^{2}$$
$v$ is the speed, so no direction enters; never negative
Work-energy principle
$$W_{\rm net} = \tfrac12 m v_2^{2} - \tfrac12 m v_1^{2}$$
net work, and the change taken as final minus initial; the mass unchanged
Work done by a varying force
$$W = \int_{x_1}^{x_2} F(x)\,dx$$
straight-line motion; $F(x)$ is the component along the motion; areas below the axis are negative
Spring force and the work to stretch it
$$F_{\rm spring} = -kx, \qquad W = \tfrac12 k x^{2}$$
$x$ measured from the natural length; ideal spring; work counted from the natural length
Work between two stretches of the same spring
$$W = \tfrac12 k x_2^{2} - \tfrac12 k x_1^{2}$$
both stretches measured from the natural length; the difference of two triangles
Stopping distance under friction alone
$$d = \frac{v^{2}}{2\mu_k g}$$
level surface; friction the only horizontal force; derived, not memorised
Work done by gravity on a straight move
$$W_G = mg\,d\cos\theta$$
$\theta$ between the downward weight and the displacement; positive going down, negative going up, zero horizontally
Check yourself
Close the page and write out, from memory: the formula for the work done by a constant force and what its angle is measured between; the two ways of computing a scalar product; the one word in the work-energy principle that decides whether you use it correctly; the definition of kinetic energy and what happens to it when the speed doubles; the rule for a force that changes as the body moves; and the work needed to stretch a spring, with the factor that is easiest to drop. Then open the formula card and mark only the ones you missed.
Take a force at an angle and a straight displacement and produce the work, including getting a zero when the two are perpendicular and a negative number when they oppose?
c-work-constant-force
Compute a work from two vectors given in components without converting either to a magnitude and an angle first, and then recover the angle between them if asked?
c-scalar-product
Build a four line ledger for a crate on a rough floor, get the same net work by adding the forces first, and say which entries are zero before computing anything?
c-total-work
Compute a change in kinetic energy as a difference of squares rather than a square of a difference, and say why doubling a speed quadruples the energy?
c-kinetic-energy
Solve a stopping distance or a final speed question from a ledger, and check the answer by the force and kinematics route without borrowing any number from the first?
c-work-energy-principle
Read a work off a graph made of triangles and rectangles, integrate a force that is given as a formula, and say when averaging the two end values is legitimate?
c-varying-force
Find a spring's stiffness from one measurement, compute the work for a stretch from the natural length, and compute it for a stretch between two other points?
c-spring-work
Glossary (18 terms)
workiş
The amount a force delivers to a body over a move, equal to the size of the force times the distance times the cosine of the angle between them. It is a scalar, measured in joules, and it carries a sign.
joulejoule
The unit of work and of energy, equal to one newton metre. Lifting an apple through one metre takes about one joule.
positive workpozitif iş
Work done by a force that leans along the motion, so that the body gains kinetic energy from it. The angle between force and displacement is less than a right angle.
negative worknegatif iş
Work done by a force that leans against the motion, so that the body loses kinetic energy to it. Friction on a sliding body is the standard example.
scalar productskaler çarpım
An operation on two vectors that returns a single number, equal to the product of their magnitudes times the cosine of the angle between them, or equivalently the sum of the products of matching components.
kinetic energykinetik enerji
The quantity one half the mass times the speed squared, carried by any moving body. It is a scalar in joules and it is never negative.
net worknet iş
The sum of the works done by every force acting on one body over the same move. It is the only combination the work-energy principle uses.
work-energy principleiş enerji ilkesi
The statement that the net work done on a body over a move equals the change in its kinetic energy over that move, final value minus initial value.
work ledger
A table with one row per force acting on a body, listing the size, the angle to the motion and the work, so that the entries can be added with their signs.
varying forcedeğişken kuvvet
A force whose size or direction depends on where the body is. Its work is the area under the graph of force against position rather than a simple product.
area under the curve
The signed region between a graph and its horizontal axis, counted positive above the axis and negative below it. For a force against position graph this area is the work.
yay sabiti
The stiffness of a spring, in newtons per metre, giving the force needed per metre of stretch. It is a property of the spring alone and appears as the slope of its force against stretch line.
natural lengthdoğal uzunluk
The length of a spring when nothing is stretching or compressing it. Every stretch used in a work formula is measured from here.
ideal springideal yay
A spring with no mass of its own whose force is proportional to its stretch over the whole range used in the problem.
compressionsıkışma
A squashing of a spring below its natural length, measured as a positive distance. The work needed is the same as for a stretch of the same size.
average forceortalama kuvvet
The constant force that would do the same work over the same distance as the real, varying one. It is found by dividing the work by the distance, not by averaging the two end values.
stopping distancedurma mesafesi
The distance a body travels while a resisting force removes all of its kinetic energy. Under friction alone on the flat it grows as the square of the initial speed.
energyenerji
A scalar quantity in joules that a system can carry and exchange. In this section the only kind considered is the kinetic energy of a body moving as a whole.
What comes next
§09 · Conservation of Energy
Everything here was computed one force at a time: gravity's contribution worked out on the way up a ramp, the spring's worked out as an area, friction's subtracted line by line. The next section notices that two of those three give back exactly what they took whenever the body returns to where it started, and builds an accounting system around that fact which removes the need to compute their work at all.
Sources
D. Giancoli, Physics for Scientists and Engineers with Modern Physics, 5th edition The set textbook for the course. Its chapter on work and energy covers the same ground as this section; 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 8 line and assessment table The scope of this section comes from the week line, which names work and energy and gives no chapter numbers; no chapter number is therefore quoted anywhere here. The weightings on the summary card come from the assessment table and nothing beyond them is claimed.
SI units: the joule as the newton metre, and the newton per metre for stiffness Work, kinetic energy and every quantity in this section that is measured in joules share one unit, which is what makes the work-energy principle an equation rather than a comparison. A spring constant is in newtons per metre and a coefficient of friction is a bare number.