Aim for FivePhysics 1

Unit 3 · Topic 3.2

3.2 Work

Work is energy transferred into or out of a system by a force acting through a distance: W = F∥d = Fd cos θ. Only the part of a force along the displacement does work, so work can be positive, negative or zero. The work–energy theorem says the total work on an object equals its change in kinetic energy.

Key terms

  • work
  • work–energy theorem
  • net work
  • conservative force
  • nonconservative force
  • force–position graph

Calculating work

For a constant force, W = F∥d = Fd cos θ. Here F is the force's magnitude, d is the displacement of the point where the force acts, and θ is the angle between the force and the displacement. F∥ = F cos θ is the force component parallel to the displacement. Work is a scalar, measured in joules (1 J = 1 N·m).

Only the parallel component changes the system's energy. A component perpendicular to the motion can change the object's direction but not its speed.

Positive, negative or zero

If the force has a component in the direction of motion (θ < 90°), the work is positive and the force adds energy. If it opposes the motion (θ > 90°), the work is negative and the force removes energy. If the force is perpendicular to the motion (θ = 90°), or nothing moves, the work is zero.

SituationWork done by the force
You lift a box upwardpositive (your force is along the motion)
Gravity on a rising ballnegative
Kinetic friction on a sliding blocknegative
Normal force on a block sliding across a level floorzero (perpendicular)
Tension on a ball moving in a horizontal circlezero (perpendicular)
You hold a heavy box stillzero (no displacement)

The work–energy theorem

The net work, the sum of the work done by every force on an object, equals its change in kinetic energy: ΔK = ΣW = ΣF∥d. If the net work is positive the object speeds up; if negative, it slows down; if zero, its speed stays the same.

This connects to the object model. When the center of mass and the point where a force acts move the same distance, the object behaves as a single object and only its kinetic energy can change. When they don't, the system's internal energy or shape can change too. When you jump, the floor's force on your feet doesn't move its point of contact, so the floor does no work on you; your kinetic energy comes from energy stored in your own body.

Conservative and nonconservative forces

Work done by a conservative force, like gravity or a spring force, doesn't depend on the path, only on the starting and ending positions. If the object returns to where it started, the conservative force's total work is zero. These are the forces that have potential energies (3.3).

Work done by a nonconservative force, like friction or air resistance, does depend on the path. The energy that kinetic friction removes equals the friction force times the length of the path, F_f × d. A longer, winding route loses more energy than a straight one.

Work from a graph

On a graph of force (parallel to the motion) against position, the work is the area between the curve and the position axis. Area below the axis counts as negative work. This works even when the force changes, which the formula Fd cos θ can't handle. For a spring, the force rises in a straight line from 0 to kx, so the work to stretch it is a triangle's area: ½(kx)(x) = ½kx².

Worked examples

Try each one yourself first, then open the solution.

  1. Example 1Calculator allowed

    Pulling a sled with friction

    You pull an 8.0 kg sled, starting from rest, 12 m across level snow with a 50 N force at 30° above horizontal. Kinetic friction on the sled is 30 N. Find the work done by each force and the sled's final speed.

    Show the solution
    1. Step 1: Your pull: W = Fd cos θ = (50)(12)(cos 30°) ≈ 520 J.
    2. Step 2: Friction points opposite the motion (θ = 180°): W = (30)(12)(−1) = −360 J.
    3. Step 3: Gravity and the normal force are perpendicular to the horizontal motion, so each does zero work.
    4. Step 4: Net work = 520 − 360 ≈ 160 J. By the work–energy theorem, ½mv² − 0 = 160 J, so v = √(2 × 160 ÷ 8.0) ≈ 6.3 m/s.

    Answer: Pull ≈ +520 J, friction = −360 J, gravity and normal = 0; final speed ≈ 6.3 m/s

  2. Example 2Calculator allowed

    Work from a force–position graph

    A force acting along a cart's motion rises steadily from 0 at x = 0 to 20 N at x = 4.0 m, then stays at 20 N until x = 6.0 m. How much work does it do from x = 0 to x = 6.0 m?

    Show the solution
    1. Step 1: The work is the area under the graph.
    2. Step 2: Triangle from 0 to 4.0 m: ½(4.0 m)(20 N) = 40 J.
    3. Step 3: Rectangle from 4.0 to 6.0 m: (2.0 m)(20 N) = 40 J.
    4. Step 4: Total = 40 + 40 = 80 J.

    Answer: 80 J

  3. Example 3Calculator allowed

    Lifting and carrying (classic trap)

    You lift a 2.0 kg box 1.5 m straight up at constant speed, then carry it 10 m across a room at constant speed and height. Use g = 9.8 m/s². How much work does your hand's force do in each part, and what is the net work on the box?

    Show the solution
    1. Step 1: Lifting at constant speed: your upward force equals mg = 19.6 N and points along the motion. W = (19.6)(1.5) ≈ 29 J.
    2. Step 2: Carrying: your force is still upward, but the motion is horizontal, so θ = 90° and W = 0. It feels like effort, but no energy goes into the box.
    3. Step 3: Net work: the speed never changes, so ΔK = 0 and the net work is zero in both parts. While lifting, gravity does −29 J, cancelling your +29 J.

    Answer: About 29 J while lifting, 0 J while carrying; net work 0 in both parts

Common mistakes

  • Using the full force when only part of it is along the motion. Use F cos θ, where θ is the angle between the force and the displacement.
  • Thinking holding or carrying something horizontally does work on it. With no displacement, or a force perpendicular to the displacement, the work is zero.
  • Forgetting that work can be negative. Friction and gravity on a rising object remove energy.
  • Setting net work equal to the work of one force. The work–energy theorem uses the sum of all forces' work.

On the exam

  • Expect force–position graphs: find the work from the area, then use the work–energy theorem to get a speed.
  • Questions often ask whether a particular force does positive, negative or zero work. Compare the force's direction with the displacement's.
  • In lab questions, you might measure work from a force sensor and a motion sensor and compare it with the change in kinetic energy.

Connected topics

Videos

Check yourself

4 questions on 3.2 Work. Pick an answer to see if you got it, and why.

Question 1 of 4Calculator allowed

A child pulls a wagon 10 m along level ground with a 50 N force directed 60° above the horizontal. How much work does the child's force do on the wagon?

Question 2 of 4Calculator allowed

A box slides 5.0 m across a level floor. How much work does gravity do on the box during the slide?

Question 3 of 4Calculator allowed

A 2.0 kg cart speeds up from 3.0 m/s to 5.0 m/s. How much net work is done on it?

Question 4 of 4Calculator allowed

A force acting on a cart along its direction of motion increases steadily from 0 N to 8.0 N as the cart moves from x = 0 to x = 0.50 m, then stays at 8.0 N from x = 0.50 m to x = 1.0 m. How much work does the force do from x = 0 to x = 1.0 m?

0 of 4 answered