Work when the force is at an angle — and when it is zero

5 minAS Level5.1.1

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

By the end of this lesson

By the end you can find the work done by a force that is not along the motion, by using only the component of the force in the direction of the displacement, and you can explain why a perpendicular force does no work.

Only the part along the motion counts

A suitcase is dragged along the floor by a strap that goes up at an angle. The pull does two different jobs:

the horizontal component moves the case along the floor — this part transfers energy;
the vertical component just lifts a little — it does not move the case horizontally.

So we split the force up (resolve it) and use only the component along the displacement.

Definition

Work done by a force at an angle

If a force $F$ acts at an angle $\theta$ to the displacement $s$ , the work done is

$W = F s \cos\theta$

because $F\cos\theta$ is the component of the force along the displacement. When $\theta = 0^\circ$ this is $W = Fs$ , the case you already know; when $\theta = 90^\circ$ it gives zero.

Why cos θ, and not sin θ

Put the displacement along the x-axis. The component of the force along that axis is next to the angle, so it is the cosine:

$F_\text{along} = F\cos\theta$

Multiplying that component by the distance gives the work:

$W = (F\cos\theta)\,s = Fs\cos\theta$

Only that component can transfer energy along the motion.

Worked example: Worked example: pulling a suitcase

A suitcase is pulled $4.0\ \text{m}$ along the floor by a strap with a force of $50\ \text{N}$ at $30^\circ$ above the horizontal. How much work does the pull do on the case?

Have a go first?

Quick check

Fill in the blanksQuestion 1 · Recall

CalculateQuestion 2 · Apply

Answer the 3 questions above to continue.