Learn7408Paper 1 · Paper 1Mechanics and materialsWork, power and efficiency

Mechanics and materials · Subtopic 2 of 11

Work, Power and Efficiency

Work is done when a force moves its point of application: $W=Fs\cos\theta$ , the force times the displacement in its direction. It's the area under a force–displacement graph, and $W=p\Delta V$ for a gas expanding at constant pressure. Power is the rate of doing work: $P=\dfrac Wt$ , and for a force moving at steady speed, $P=Fv$ . Efficiency compares useful output with total input.

Cambridge 9702 Paper 1Cambridge 9702 Paper 2Edexcel YPH11 WPH11AQA 7408 Paper 1≈ 35 minRevision 5 · updated 2026-10-27

Section 1 of 14

Learning objectives

Calculate work done using $W=Fs\cos\theta$ and as an area under a graph.
Calculate work done by an expanding gas, $W=p\Delta V$ .
Use $P=\dfrac Wt$ and derive and use $P=Fv$ .
Calculate efficiency, and apply conservation of energy.

Pause and think

You carry a heavy bag along a flat corridor at steady speed. Your arms ache. How much work do you do on the bag?

Section 2 of 14

Core idea

Start hereWhy cars have a top speed

Floor the accelerator and a car speeds up, but less and less, until it can go no faster. The engine's power is fixed; at higher speeds the same power gives a smaller driving force ( $P=Fv$ ), until it only just matches air resistance.

The big idea

$W=Fs\cos\theta$ = area under $F$ – $s$ . $P=\frac Wt=Fv$ . Efficiency = useful ÷ total.

1. Work

Work done by a varying force is the area under the F–x graph

Stretching a spring 0.1 m to a final force of 40 N: work

=\tfrac12\times40\times0.1=2

J (the triangle's area).

W=Fs\cos\theta

$\theta$: angle between force and displacement.
varying $F$: area under the force–displacement graph.

Edge caseNo work A force perpendicular to motion (

\theta=90^\circ

) does no work, e.g. the normal force on a sliding box.

2. Power

Driving force P/v against resistance for a 60 kW car

At low speed there is spare force to accelerate; it runs out at 30 m/s.

P=Fv

$P=\frac{\Delta W}{\Delta t}=\frac{F\Delta s}{\Delta t}$
Constant force.
$=Fv$
$\frac{\Delta s}{\Delta t}=v$ .

Worked exampleA 60 kW car (1200 kg)

Top speed 30 m s $^{-1}$ .

Goal: Resistance; acceleration at 20 m s $^{-1}$ (same resistance).
Strategy: $P=Fv$ .

$F=\frac{60000}{30}=2000\text{ N}$
At top speed, driving force = resistance.
$F=\frac{60000}{20}=3000\text{ N}$
Driving force at 20 m s $^{-1}$ .
$a=\frac{3000-2000}{1200}=0.83\text{ m s}^{-2}$
$F=ma$ .

Answer

2000

0.83

m s

^{-2}

✓ CheckAcceleration falls to 0 as

v\to30

✓.

Worked exampleMotor efficiency

Lifts 80 kg by 15 m in 30 s; input 600 W.

Goal: Efficiency.
Strategy: Useful power = $mgh/t$ .

$\frac{80\times9.81\times15}{30}=392\text{ W}$
Useful.
$\frac{392}{600}=65\%$
Efficiency.

Answer

65\%

✓ CheckBelow 100% ✓.

Summary

$W=Fs\cos\theta$ ; area.
$P=Fv$ ; efficiency.

Power: Rate of energy transfer.

HistoryWatt's horsepower. James Watt needed to sell his steam engines to mine owners who used horses, so in the

1780

s he measured how fast a horse could lift loads. His "horsepower" (

\approx746

W) became a unit of power, and the SI unit, the watt, was later named after him.

Fun factYour brain runs on 20 W. Your brain uses about

20

W, about a fifth of your resting energy use, less than an old filament light bulb. A supercomputer trying to match a brain's activity needs millions of watts.

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