Work & the Work–Energy Theorem

You’ll be able to

Compute work as W = Fd cosθ and identify its sign
Recognize when a force does zero work
Apply the work–energy theorem, W_net = ΔKE

Work is force acting through a distance

Work is the transfer of energy that happens when a force pushes an object through a displacement. Only the part of the force along the displacement counts, so W = Fd cosθ, where θ is the angle between the force and the motion. Work is a scalar measured in joules (J), but it carries a sign: a force that helps the motion does positive work, and a force that opposes it (like friction) does negative work.

When work is zero

Two situations give zero work even when a force is present. First, if there is no displacement (you push a wall that does not move, d = 0). Second, if the force is perpendicular to the motion, because cos90° = 0. That is why gravity does no work on a satellite in a circular orbit, and why the upward force carrying a tray does no work as you walk forward — the force and the motion are at right angles.

Work

W = F·d·cosθ

θ is the angle between the force and the displacement. Force along the motion → cosθ = 1; opposite → cosθ = −1; perpendicular → cosθ = 0.

Work–energy theorem

W_net = ΔKE = ½mv_f² − ½mv_i²

The total work done by all forces equals the change in kinetic energy. Speeding up means positive net work; slowing down means negative.

Worked example

A 2 kg cart starts at rest. A constant 20 N force pushes it 5 m along a frictionless floor in the direction of motion. Find the work done and the cart’s final speed.

1.The force is along the motion, so θ = 0 and cosθ = 1: W = Fd = 20 × 5 = 100 J.
2.With no friction, this is the net work, so W_net = ΔKE = 100 J.
3.Starting from rest, ΔKE = ½mv_f² = 100 J → ½(2)v_f² = 100.
4.Solve: v_f² = 100 → v_f = 10 m/s.

Answer: W = 100 J and the cart’s final speed is 10 m/s

Tip

The work–energy theorem lets you find a speed without ever touching the kinematic equations. If you know the net work, you know the change in kinetic energy — and from there, the speed.

Checkpoint

A 30 N force pushes a crate 4 m in the direction of the force. How much work is done on the crate?

Checkpoint

A waiter carries a tray at a constant height while walking forward at constant speed. How much work does the upward support force do on the tray?

On the exam

Before computing work, check the angle between the force and the displacement. Perpendicular forces (normal force on a flat surface, gravity in horizontal motion, tension in circular motion) do exactly zero work — a favorite AP trap.

Answer the 2 checkpoints as you read.