3.2 Work

Four little picture stories

Press Next (or Play) to walk through each story one small step at a time. The numbers come last.

1. Pushing a box: it moves, so you do work

2. Pushing a wall: no motion, no work

Which of these does work on the object?

3. Pulling a sled with a tilted rope

Read the story as text

You pull a friend on a sled across flat snow. The rope is tilted up at 30°. You pull hard, but the sled speeds up only a little. Some of your pull is "wasted" lifting on the rope, and the snow rubs on the runners the whole way.

Your arms get tired, so you are clearly giving energy to something. How much energy does a force actually put into the sled, and how much does friction take back out?

A rope pulls a sled forward and up while the sled slides along flat snow. Which part of the pull does work on the sled?

4. Check yourself

Think of your answer first, then tap to see it.

a) You hold a heavy bag perfectly still for a whole minute. How much work do you do on the bag?

Show answer

Zero. The bag does not move, so W = F × 0 = 0. Your tired arms use energy inside your body, but none goes into the bag.

b) You push a box with a 20 N force along the floor, and it moves 3 m the same way. How much work does your push do?

Show answer

W = 20 N × 3 m = 60 J. The push points the same way the box moves, so all of it counts.

c) A box slides across the floor and slows to a stop. Is the work done by friction positive, negative or zero?

Show answer

Negative. Friction points backward, against the motion, so it takes energy out of the box.

Already know this?

Three quick questions. Get all three right on the first try and you can skip ahead.

A 10 N push moves a box 2 m in the same direction as the push. How much work does the push do?

You hold a bag still for 30 s. How much work do you do on the bag?

A box slides across a rough floor. What kind of work does friction do on it?

How a force moves energy into or out of a system: W = F d cos θ, work by friction, gravity and springs, the area under a force-position graph, and the work-energy theorem.

The idea

In pictures

  • A force does work when the thing it pushes moves. Push a box and it slides: work. Push a wall and it stays put: no work.
  • Only the part of the force along the motion counts. A tilted rope pulls forward and up; only the forward part does work on a sled sliding along flat snow.
  • A force with the motion puts energy in (positive work). A force against the motion, like friction, takes energy out (negative work).
  • A force sideways to the motion (straight up while you walk along) does no work at all.

Below is the same idea in words, and then with numbers.

Work is energy moved into or out of a system by a force acting while the object moves. Only the part of the force along the motion counts.

W = F∥ d = F d cos θ

F is the size of the force, d is the size of the displacement, and θ is the angle between the force and the displacement. Work is a scalar, measured in joules (1 J = 1 N·m).

F F cos θ (does work) F sin θ (no work) θ displacement d

Four forces you meet all the time

Work is the area under a force-position graph

When the force changes, split the motion into tiny steps. Each step adds F × (tiny distance), which is a thin strip of area under the F∥-x graph. Add the strips: work = area under the F∥-x graph. Area below the axis is negative work. For a spring pulled from 0 to x, the graph is a straight line from 0 to kx, so the area is a triangle: W = ½ (x)(kx) = ½ k x².

The work-energy theorem

Add up the work done by every force on an object (or, the same thing, the work done by the net force). That total is the change in its kinetic energy:

Wnet = ΔK = ½ m vf² − ½ m vi²

Positive net work: the object speeds up. Zero net work: its speed does not change (even if forces act). Negative net work: it slows down.

With numbers: the sled

m = 20 kg, pull F = 100 N at θ = 30° above horizontal, pulled d = 10 m across flat snow, μk = 0.20, starts from rest. Take right (the direction of motion) as positive.

Show all steps as text
  1. Work by your pull: WF = F d cos θ = 100 × 10 × cos 30° = 100 × 10 × 0.866 = 866 J. θ is the angle between the rope and the displacement.
  2. Normal force: the rope lifts a little, so N = m g − F sin θ = 20 × 9.8 − 100 × 0.5 = 196 − 50 = 146 N. Vertical forces balance (no vertical acceleration): N + F sin θ = m g.
  3. Friction: fk = μk N = 0.20 × 146 = 29.2 N, so Wf = −29.2 × 10 = −292 J. Friction points backwards, θ = 180°, cos 180° = −1.
  4. Gravity and normal force: 0 J each. Both are vertical and the motion is horizontal.
  5. Net work: Wnet = 866 − 292 + 0 + 0 = 574 J. Work is a scalar, so just add the numbers with their signs.
  6. Speed at the end: ½ (20) v² = 574, so v² = 57.4 and v = 7.6 m/s. Work-energy theorem with Ki = 0.
  7. Check with forces: a = (86.6 − 29.2) / 20 = 2.87 m/s², v² = 2 a d = 2 × 2.87 × 10 = 57.4. Same answer. Two methods agreeing is a strong check.

Play: the sled and the spring

Pull the sled. The bars on the right are the work done by each force so far, and the change in kinetic energy. Watch the "net work" bar and the "ΔK" bar: they always match. Try θ = 0°, then 60°. Try friction 0.

Now stretch a spring slowly from its natural length. You must pull harder the further it stretches. The shaded area on the force-position graph is the work you have done.

Try: double the stretch. The force at the end doubles, but the area (the work) becomes four times bigger. That is the ½ k x² pattern.

Worked examples

Pushing a box basic

You push a box with a horizontal 50 N force for 4.0 m across a floor. How much work does your push do?

Show solution
Show all steps as text
  1. The push is along the motion, θ = 0°, cos 0° = 1. Same direction.
  2. W = 50 N × 4.0 m × 1 = 200 J. N·m = J.

Carrying a book across a room basic

You carry a 2 kg book 5 m across a room at constant height and constant speed. How much work does your hand do on the book?

Show solution
Show all steps as text
  1. Your hand pushes up (to balance gravity); the book moves sideways. θ = 90°. Force and displacement are perpendicular.
  2. W = F d cos 90° = 0 J. Your muscles get tired for biology reasons, but no energy goes into the book.
  3. Check: the book's speed and height do not change, so its energy does not change. Consistent with zero work.

Lifting a box medium

You lift a 2.0 kg box 1.5 m straight up at constant speed. Find the work done by you, by gravity, and the net work.

Show solution
Show all steps as text
  1. Constant speed means net force zero, so your force = m g = 2.0 × 9.8 = 19.6 N, upward. Newton's first law.
  2. Wyou = 19.6 × 1.5 × cos 0° = +29.4 J. Force up, motion up.
  3. Wgravity = −m g Δy = −19.6 × 1.5 = −29.4 J. Gravity points down, motion up.
  4. Wnet = 0, so ΔK = 0. Speed is constant, as the work-energy theorem says.

Area under a force-position graph medium

A 5.0 kg cart starts at rest on a frictionless track. The force on it (along the track) grows steadily from 0 N at x = 0 to 10 N at x = 2.0 m, then stays at 10 N until x = 5.0 m. Find the work done and the final speed.

Show solution
Show all steps as text
  1. From 0 to 2 m the graph is a triangle: area = ½ × 2.0 m × 10 N = 10 J. Area of a triangle = ½ base × height.
  2. From 2 to 5 m it is a rectangle: area = 3.0 m × 10 N = 30 J. Constant force.
  3. Total work W = 40 J. This is the only force doing work, so Wnet = 40 J. Gravity and normal force are perpendicular to the track.
  4. ½ (5.0) v² = 40, v² = 16, v = 4.0 m/s. Work-energy theorem from rest.

Sliding down a rough ramp AP

A 2.0 kg block starts from rest and slides 3.0 m down a 30° ramp. μk = 0.25. Use work to find its speed at the bottom.

Show solution
Show all steps as text
  1. Gravity: the block drops Δy = −3.0 × sin 30° = −1.5 m, so Wg = −m g Δy = 2.0 × 9.8 × 1.5 = +29.4 J. Only the height change matters for gravity.
  2. Normal force: N = m g cos 30° = 2.0 × 9.8 × 0.866 = 16.97 N. Friction f = 0.25 × 16.97 = 4.24 N. On a ramp, N balances the part of gravity into the ramp.
  3. Wf = −4.24 × 3.0 = −12.7 J. WN = 0. Friction is opposite the motion; N is perpendicular.
  4. Wnet = 29.4 − 12.7 = 16.7 J = ½ (2.0) v², so v² = 16.7 and v = 4.1 m/s. Work-energy theorem.
  5. Check: a = g (sin 30° − 0.25 cos 30°) = 9.8 × (0.5 − 0.2165) = 2.78 m/s², v² = 2 × 2.78 × 3.0 = 16.7. Same as Unit 2 forces and kinematics.

Practice

  1. A 40 N force pulls a box 5.0 m across the floor. The force is 60° above the horizontal. How much work does it do?

    Worked answer

    W = F d cos θ = 40 × 5.0 × cos 60° = 40 × 5.0 × 0.5 = 100 J.

  2. A satellite moves in a circular orbit at constant speed. How much work does Earth's gravity do on it during one half orbit?

    Worked answer

    In a circle, gravity points to the centre and the velocity is along the circle, so θ = 90° at every moment. Zero work, which matches the constant speed (ΔK = 0).

  3. A box slides to a stop on a rough floor. Which force does negative work on the box?

    Worked answer

    Friction points opposite the displacement (θ = 180°), so its work is negative. Gravity and the normal force are perpendicular to the motion and do zero work. The box's K drops to zero, so the net work is negative, which needs a force doing negative work.

  4. A spring (k = 200 N/m) is already stretched 0.10 m. How much work is needed to stretch it from 0.10 m to 0.20 m?

    Worked answer

    Area under F = kx from 0.10 to 0.20 m: ½ k (0.20² − 0.10²) = ½ × 200 × (0.040 − 0.010) = 3.0 J. The first 0.10 m only took 1.0 J: the second one is harder because the force is bigger.

  5. A crate moves across a floor at constant velocity while several forces act on it. What is the net work done on the crate?

    Worked answer

    Constant velocity means constant speed, so ΔK = 0, and the work-energy theorem gives Wnet = 0. (Individual forces can still do positive and negative work that cancel.)

  6. The force along the motion on an object is 6.0 N from x = 0 to x = 3.0 m, then falls steadily to 0 at x = 5.0 m. How much work is done from 0 to 5.0 m?

    Worked answer

    Rectangle: 6.0 × 3.0 = 18 J. Triangle: ½ × 2.0 × 6.0 = 6 J. Total area = 24 J.

  7. Short answer. A 3.0 kg box starts at rest. The net work done on it is 54 J. (a) Find its final speed. (b) The box slides on a level floor. Explain why the normal force does no work on it.

    Worked answer

    (a) Wnet = ΔK: 54 = ½ (3.0) v², v² = 36, v = 6.0 m/s.

    (b) The normal force is vertical and the displacement is horizontal, so θ = 90° and cos 90° = 0. A force perpendicular to the motion cannot speed the box up or slow it down.

  8. Short answer (translate a graph). The net force on a cart (positive = to the right) is +4 N from x = 0 to 2 m, then −2 N from x = 2 m to 6 m. The cart starts at 0 m moving right with 2 J of kinetic energy. Find its kinetic energy at x = 2 m and at x = 6 m and describe its motion in words.

    Worked answer

    0 to 2 m: area = +4 × 2 = +8 J, so K = 2 + 8 = 10 J (it speeds up).

    2 to 6 m: area = −2 × 4 = −8 J, so K = 10 − 8 = 2 J. It slows down but is still moving right at 6 m with the same speed it started with.

    Scoring idea: 1 point for each area with the correct sign, 1 point for adding to the starting K, 1 point for the description.

AP question types for this topic

These four free response questions match the four question types on the AP Physics 1 exam, each sized for work. Work each part on paper first, then open the worked answer to check it and see how points are given.

1. Mathematical Routines

Derive with symbols first, then put in numbers. Hana pulls a sled of mass m from rest across level snow with a rope at angle θ above the horizontal. The rope tension is T and a friction force of size f acts. She pulls it a distance d.

  1. Derive expressions for the work done by the rope and the net work on the sled.
  2. Use the work-energy theorem to derive the sled's final speed.
  3. Take m = 12 kg, T = 40 N, θ = 30°, f = 20 N, d = 8.0 m. Calculate both works and the speed.
Worked answer and scoring

(a) Only the part of T along the motion does work: Wrope = Td cosθ. Friction points backward: Wf = −fd. Gravity and the normal force are perpendicular, so they do no work. Wnet = (T cosθ − f)d.

(b) Wnet = ΔK = ½mv² − 0, so v = √(2(T cosθ − f)d/m).

(c) T cos30° = 34.6 N. Wrope = 34.6 × 8.0 = 277 J; Wf = −20 × 8.0 = −160 J; Wnet = 117 J. v = √(2 × 117 / 12) = 4.4 m/s.

Scoring (6 points)

  • 1 point for Td cosθ
  • 1 point for −fd
  • 1 point for Wnet = (T cosθ − f)d
  • 1 point for the speed expression from Wnet = ΔK
  • 1 point for 277 J and 117 J
  • 1 point for 4.4 m/s

2. Translation Between Representations

Move between graphs, pictures, equations and words, and justify. Owen pushes a 4.0 kg box from rest along a frictionless floor. His force along the motion changes with position as in the graph below.

The force graph: F rises in a straight line from 0 N at x = 0 to 30 N at x = 2 m, then stays at 30 N until x = 5 m.30 NF02 m5 mx

Four possible graphs of the box's kinetic energy K (vertical) against position x (horizontal):

KxA
Graph A: curves up from the origin, then becomes a straight rising line.
KxB
Graph B: a straight rising line, then flat.
KxC
Graph C: one straight line through the origin.
KxD
Graph D: rises steeply, then levels off.
  1. Which graph, A to D, shows K against x? Justify using what the slope of a K-x graph means.
  2. Calculate the work Owen does from x = 0 to 5.0 m.
  3. Calculate the box's speed at x = 5.0 m.
Worked answer and scoring

(a) Graph A. ΔK = FΔx, so the slope of K against x equals the force. From 0 to 2 m the force grows, so the slope grows: K curves up. From 2 to 5 m the force is constant, so the slope is constant: a straight rising line. It never goes flat because the force never drops to zero.

(b) Work = area under F-x: triangle ½ × 2.0 m × 30 N = 30 J, plus rectangle 3.0 m × 30 N = 90 J. Total 120 J.

(c) No friction, so ΔK = 120 J: v = √(2 × 120 / 4.0) = √60 = 7.7 m/s.

Scoring (5 points)

  • 1 point for choosing A
  • 1 point for slope of K-x = force
  • 1 point for work = area under F-x
  • 1 point for 120 J
  • 1 point for 7.7 m/s

3. Experimental Design and Analysis

Plan a measurement, make a straight-line graph and read its slope. Elena wants to find the coefficient of kinetic friction μ between a wooden block and the floor using work and energy. She has the block, a photogate with a flag on the block, a metre stick and a balance.

Her data:

v0 (m/s)1.01.52.02.5
stopping distance d (m)0.170.390.671.07
  1. Describe her procedure.
  2. Derive what to plot for a straight line and what the slope means.
  3. Use the data to find μ.
Worked answer and scoring

(a) Slide the block by hand so it is moving freely as it passes the photogate; speed v0 = flag width / gate time. Measure with the metre stick how far it slides past the gate before stopping. Repeat for different push speeds, three trials each.

(b) Friction does all the work: −μmgd = 0 − ½mv0², so d = v0² / (2μg). Plot d against v0²; slope = 1/(2μg). Mass cancels, so the balance is not even needed.

(c) v0² = 1.0, 2.25, 4.0, 6.25 m²/s². Best-fit slope ≈ 0.17 s²/m. μ = 1/(2 × 9.8 × 0.17) ≈ 0.30.

Scoring (6 points)

  • 1 point for v0 from flag width / gate time
  • 1 point for measuring d, with repeats
  • 1 point for −μmgd = −½mv0²
  • 1 point for plotting d against v0²
  • 1 point for slope = 1/(2μg)
  • 1 point for μ ≈ 0.30

4. Qualitative/Quantitative Translation

Explain a claim in words, back it with a derivation, and connect the two. Diego lowers a 15 kg box 1.2 m to the floor with a rope, at constant speed. He says: Gravity does positive work on the box the whole way down, yet the net work on the box is zero.

  1. In words, explain whether both parts of Diego's claim can be true.
  2. Derive the work done by gravity and by the rope, and calculate both.
  3. Connect your result in (b) to your reasoning in (a).
Worked answer and scoring

(a) Yes. Gravity points down along the motion, so it does positive work. The rope pulls up against the motion, so it does negative work. The speed does not change, so K does not change, and the net work must be zero.

(b) Wg = mgh = 15 × 9.8 × 1.2 = +176 J. Constant velocity means T = mg, so Wrope = −Th = −176 J. Sum: 0.

(c) The two works are equal and opposite, which is exactly the balance described in (a), and their sum of zero matches ΔK = 0 from the work-energy theorem.

Scoring (5 points)

  • 1 point for the signs (gravity +, rope −)
  • 1 point for net work = 0 because K is constant
  • 1 point for T = mg
  • 1 point for ±176 J
  • 1 point for connecting the numbers to the reasoning

Common mistakes

The mistake: using the angle of the ramp (or the angle to the vertical) as θ in W = F d cos θ.

Why it is wrong: θ must be the angle between the force and the displacement.

How to spot it: sketch the force arrow and the displacement arrow tail to tail and measure the angle between them.

The mistake: giving friction positive work, or forgetting its sign when adding works.

Why it is wrong: kinetic friction on a sliding object points against the motion, so its work is negative.

How to spot it: your object speeds up more with friction than without. That cannot be right.

The mistake: "Holding a heavy box still is hard work, so I am doing work on it."

Why it is wrong: no displacement means W = F × 0 = 0 on the box. Your muscles use chemical energy inside your body, but none goes into the box.

How to spot it: ask "did the object move along the force?" If not, the work on it is zero.

The mistake: W = k x × x = k x² for a spring.

Why it is wrong: the spring force grows from 0 to kx, it is not kx the whole way. The area is a triangle, ½ k x².

How to spot it: whenever a force changes with position, find the area under the F-x graph.

The mistake: only counting one force in the work-energy theorem.

Why it is wrong: ΔK equals the work done by all forces together (the net work).

How to spot it: list every force on your free-body diagram and give each a work value, even if it is zero.