Work done by a force: W = Fd cos θ

7 minAS Level4.5.1

AI-drafted, independently checked · not yet teacher-reviewed

By the end of this lesson

By the end you can define work done by a constant force, calculate work using $W = Fd\cos\theta$ , and understand when work is positive, negative, or zero.

Key to the symbols

Symbol	Meaning	Unit
$W$	work done (here $W$ is not weight)	joules, J
$F$	the size of a constant force	newtons, N
$d$	the distance moved by the point of application	metres, m
$\theta$	the angle between the force and the direction of motion	degrees

What is work?

In physics, work has a precise meaning: it is the energy transferred by a force when its point of application moves. (The point of application is the point where the force acts on the object. Energy, measured in joules, is defined fully in the next lesson.) The definition: $W = Fd \cos\theta$

where:

$F$ = magnitude of the constant force (N)
$d$ = magnitude of the displacement of the point of application (m)
$\theta$ = angle between the force vector and the displacement vector

Key points:

Work is a scalar (has magnitude, no direction).
Unit: joule (J). $1\text{ J} = 1\text{ N m}$ .
Only the component of force in the direction of displacement does work.
If the force varies, we need calculus (not in this syllabus). Here, forces are constant.

Why cos θ?

Imagine pulling a sled with a rope at an angle.

If you pull horizontally ( $\theta = 0^\circ$ ), all your force goes into moving the sled: $W = Fd$ .
If you pull at right angles to the motion ( $\theta = 90^\circ$ ), none of your pull is along the motion, so it does no work: $\cos 90^\circ = 0$ , so $W = 0$ .
If you pull backwards ( $\theta = 180^\circ$ ), you oppose the motion: $\cos 180^\circ = -1$ , so $W = -Fd$ (negative work).

The force component parallel to the displacement is $F\cos\theta$ . Work = (parallel component) × distance.

Positive, negative, zero work

Angle $\theta$	$\cos\theta$	Work $W$	Meaning
$0^\circ$	$+1$	$+Fd$	Force helps motion
$0^\circ < \theta < 90^\circ$	$>0$	positive	Force partly helps
$90^\circ$	$0$	$0$	Force perpendicular — no energy transfer
$90^\circ < \theta < 180^\circ$	$<0$	negative	Force partly opposes
$180^\circ$	$-1$	$-Fd$	Force directly opposes

Negative work means energy is taken away from the object (e.g., friction, braking). Zero work means the force transfers no energy. For example, the normal reaction on an object sliding along a horizontal floor is at $90°$ to the motion.

Worked example: Work at an angle

A force of $20\text{ N}$ acts at $30^\circ$ to the horizontal. The object moves $5\text{ m}$ horizontally. Find the work done by the force.

Have a go first?

Worked example: Work done by friction

A block slides $4\text{ m}$ along a rough horizontal surface. The frictional force is $8\text{ N}$ . Find the work done by friction.

Have a go first?

Check yourself

CalculateQuestion 1 · Apply

CalculateQuestion 2 · Recall

Answer the 3 questions above to continue.