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Learn4PH1Paper 1P · Paper 1 (core content)Energy resources and energy transfersEnergy, work and power
Energy resources and energy transfers · Subtopic 2 of 3

Energy, Work and Power

Energy is stored in kinetic, gravitational, elastic, chemical, thermal, nuclear and other stores, and transferred by forces, heating, electrical currents and waves. It's never created or destroyed. Work done is energy transferred by a force, W=FdW=Fd. Power is the rate of transfer, P=WtP=\dfrac Wt. Efficiency is the fraction of input energy that ends up useful.

Cambridge 5054 Paper 1Cambridge 5054 Paper 2Cambridge 0625 Papers 2, 4Cambridge 0625 Papers 1, 3Edexcel 4PH1 Paper 1PEdexcel 4PH1 Paper 2P≈ 35 minRevision 5 · updated 2026-10-27
Section 1 of 14

Learning objectives

  • Name energy stores and describe energy transfers.
  • Calculate kinetic energy and changes in gravitational potential energy.
  • Use conservation of energy to find speeds and heights.
  • Calculate work done, power and efficiency.
Pause and think

A ball is dropped from a height. Just before it hits the ground, where did its kinetic energy come from? And after a few bounces it stops: where did the energy go?

Section 2 of 14

Core idea

Start hereA roller coaster has no engine on the way down

A coaster is hauled to the top once. After that, every hill and loop is paid for by the energy stored at the top, swapped back and forth between height and speed. Energy is never created or destroyed, only moved between stores.

The big idea

Ek=12mv2E_k=\frac12mv^2; Ep=mghE_p=mgh; work W=FdW=Fd; power P=EtP=\frac Et; efficiency =useful outtotal in=\frac{\text{useful out}}{\text{total in}}.

1. Kinetic and potential energy

A 2 kg ball falling 20 m: energy changes store
0.511.522.533.541002003004000energy (J)gravitational PEkinetictotal
mgh=2×10×20=400mgh=2\times10\times20=400 J at the top. Just before landing all 400 J is kinetic: 12(2)v2=400\tfrac12(2)v^2=400, v=20v=20 m/s.
Where 12mv2\frac12mv^2 comes from
  1. W=Fs=ma⋅sW=Fs=ma\cdot s
    Work to accelerate from rest.
  2. s=v22as=\frac{v^2}{2a}
    Constant acceleration from rest.
  3. W=ma⋅v22a=12mv2W=ma\cdot\frac{v^2}{2a}=\tfrac12mv^2
    aa cancels.
Worked example2 kg ball dropped from 20 m

GPE at the top; speed at the bottom (no air resistance).

Goal
EpE_p, vv.
Strategy
All GPE becomes KE.
  1. Ep=2×10×20=400 JE_p=2\times10\times20=400\text{ J}
    mghmgh.
  2. 12(2)v2=400⇒v=20 m/s\tfrac12(2)v^2=400\Rightarrow v=20\text{ m/s}
    Energy conserved.
Answer400400 J; 2020 m/s.
✓ Checkv=2gh=400v=\sqrt{2gh}=\sqrt{400}; independent of mass ✓.

2. Work and power

Worked exampleLifting 50 kg

Up 12 m in 20 s.

Goal
Work and power.
Strategy
Force = weight.
  1. W=500×12=6000 JW=500\times12=6000\text{ J}
    mg=500mg=500 N.
  2. P=6000÷20=300 WP=6000\div20=300\text{ W}
    Energy per second.
Answer60006000 J; 300300 W.
✓ CheckEquals the GPE gained, mghmgh ✓.

3. Efficiency

Energy flow in a filament lamp: only 10% useful
electrical 100 Jlight 10 J (useful)heat 90 J (wasted)
Efficiency =10100=10%=\dfrac{10}{100}=10\%. An LED gives much more light for the same input.
Worked exampleA 60 W lamp at 10%

Energy per second.

Goal
Useful and wasted.
Strategy
Efficiency × input.
  1. 0.10×60=6 J0.10\times60=6\text{ J}
    Useful (light).
  2. 60−6=54 J60-6=54\text{ J}
    Wasted (heat).
Answer66 J light; 5454 J heat.
✓ CheckTotal conserved: 6060 J ✓.
Common misconceptionEfficiency over 100% Impossible: check you divided useful by total.
Pause and think

A motor uses 500 J to give 200 J useful. Efficiency?

Summary

  • Energy conserved; stores change.
  • W=FdW=Fd; P=E/tP=E/t; efficiency.
Key words
Joule
Unit of energy.
Watt
Joule per second.
HistoryJoule's paddle wheel. In 18451845 James Joule let falling weights turn a paddle wheel in a tank of water, and measured the tiny temperature rise. He showed that a fixed amount of mechanical work always produced the same amount of heating, proving heat is a form of energy transfer. The unit of energy is named after him.
Fun factA chocolate bar could lift you up a skyscraper. A 5050 g chocolate bar stores about 1 000 0001\,000\,000 J. Lifting a 6060 kg person 11 m takes about 600600 J, so in theory the bar could lift you 17001700 m. Muscles are only about 25%25\% efficient, so you'd manage around 400400 m.
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