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Learn7367Paper 3 · Mechanics optionWork, energy and powerThe work–energy principle
Work, energy and power · Subtopic 3 of 4

The Work–Energy Principle

Energy is never lost, only moved around. Work done by a driving force goes into kinetic energy, potential energy, and the work done against resistances. Balance the books and the unknown drops out.

Cambridge 9709 Paper 4Edexcel YMA01 WME02Edexcel 9FM0 OptionAQA 7367 Paper 3≈ 40 minRevision 5 · updated 2026-10-27
Section 1 of 14

Learning objectives

  • State and apply the work–energy principle.
  • Include work against friction and air resistance in an energy equation.
  • Find an unknown speed, distance, resistance or coefficient of friction.
  • Choose between an energy method and a Newton's-laws method.
Pause and think

A ball dropped from 2020 m would hit the ground at 20 m s−120\ \text{m s}^{-1} with no air resistance. It actually lands at 18 m s−118\ \text{m s}^{-1}. Where did the missing energy go?

Section 2 of 14

Core idea

Start hereWhere does the energy go?

A block slides down a rough slope. It loses height (PE) and gains speed (KE), but gains less KE than the PE it lost. The difference went into heating the surfaces: work done against friction. Track every joule and the problem solves itself.

The big idea

Energy in = energy out: KE1+PE1+Wdriving=KE2+PE2+Wresistance\text{KE}_1+\text{PE}_1+W_{\text{driving}}=\text{KE}_2+\text{PE}_2+W_{\text{resistance}}.

1. The energy budget

KE1+PE1+Wdriving=KE2+PE2+Wagainst resistance\text{KE}_1+\text{PE}_1+W_{\text{driving}}=\text{KE}_2+\text{PE}_2+W_{\text{against resistance}}
Energy budget: a 5 kg block slides 10 m down a rough slope, dropping 6 m
1234501001502002503003500energy (J)PE lost 300friction 160KE gained 140
PE lost =mgh=5(10)(6)=300=mgh=5(10)(6)=300 J. Friction 1616 N over 1010 m takes 160160 J. The rest, 140140 J, becomes KE: 12(5)v2=140\tfrac12(5)v^2=140, v=7.48v=7.48 m/s.
Worked example5 kg block, 10 m down a rough slope, dropping 6 m, friction 16 N

From rest; final speed?

Goal
vv.
Strategy
Budget.
  1. PE lost=5(10)(6)=300\text{PE lost}=5(10)(6)=300
    In.
  2. friction work=16×10=160\text{friction work}=16\times10=160
    Out.
  3. 12(5)v2=140⇒v=7.48\tfrac12(5)v^2=140\Rightarrow v=7.48
    Rest becomes KE.
Answer7.487.48 m/s.
✓ CheckWithout friction: v=120=10.95v=\sqrt{120}=10.95; friction reduces it ✓.
Worked exampleEngine's work

800 kg car, 15 → 25 m/s over 200 m on level road, resistance 400 N. Work by engine?

Goal
WW.
Strategy
KE gain + resistance work.
  1. 12(800)(625−225)=160000\tfrac12(800)(625-225)=160000
    KE gain.
  2. 400×200=80000400\times200=80000
    Resistance.
Answer240 000240\,000 J.
✓ CheckAverage driving force 240000200=1200\frac{240000}{200}=1200 N ✓ plausible.
Pause and think

3 kg box slows from 6 m/s to rest in 9 m on level ground. Friction?

Summary

  • Track every energy and work.
Key words
Work–energy principle
Work done = change in energy.
HistoryWhere the half came from. Before 1829, physicists measured "living force" as mv2mv^2. Coriolis, working on machines, saw that the work done by a force from rest equals 12mv2\tfrac12mv^2, not mv2mv^2, and introduced the half. With it, work and kinetic energy could be compared directly, and the work–energy principle was born.
Fun factGlowing brakes. A Formula 1 car braking from about 300300 km/h to 8080 km/h turns roughly 33 million joules of kinetic energy into heat in about two seconds. The carbon brake discs can glow orange at over 1000∘1000^\circC: the work–energy principle made visible.
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