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How is energy transferred when forces do work?

Work done by a force, kinetic and gravitational potential energy, the principle of conservation of energy, power as the rate of doing work, and efficiency.

A focused answer to the Edexcel 9PH0 work and energy content, covering work done by a force, kinetic and gravitational potential energy, conservation of energy, power as the rate of doing work, and efficiency.

Generated by Claude Opus 4.810 min answer

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  1. What this dot point is asking
  2. The answer
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What this dot point is asking

Edexcel wants you to calculate the work done by a force (including at an angle), use kinetic energy and gravitational potential energy, apply the principle of conservation of energy, define power as the rate of doing work, and calculate efficiency.

The answer

Work done

When the force is along the motion (θ=0\theta = 0), W=FsW = Fs. A force at right angles to the motion (θ=90\theta = 90) does no work, which is why the centripetal force in circular motion and the magnetic force on a moving charge do no work. Work done is the area under a force-displacement graph.

Kinetic and potential energy

These two stores convert into each other freely under gravity: a falling object loses mghmgh of potential energy and gains 12mv2\frac{1}{2}mv^2 of kinetic energy, so for a frictionless fall v=2ghv = \sqrt{2gh} regardless of mass.

Conservation of energy

This principle lets you solve problems without tracking forces: equate the total energy at the start and end, accounting for any work done against resistive forces.

Power and efficiency

No real machine is 100%100\% efficient because some input energy is always dissipated, usually as heat through friction or resistance. The wasted energy is not destroyed; it simply ends up in a less useful store.

Examples in context

Pumped-storage hydroelectric stations convert electrical energy to gravitational potential energy by pumping water uphill, then recover most of it as electricity at peak demand, with an overall efficiency around 75%75\%. A car engine is only about 30%30\% efficient, most of the fuel energy leaving as heat in the exhaust and coolant. Regenerative braking in electric vehicles recovers kinetic energy that would otherwise become heat in the brakes. The P=FvP = Fv relation explains why a car needs far more power to maintain high speed against drag.

Try this

Q1. State the principle of conservation of energy. [1 mark]

  • Cue. Energy cannot be created or destroyed, only transferred between stores; the total in a closed system is constant.

Q2. A 2.02.0 kg object moves at 5.05.0 m per second. Find its kinetic energy. [2 marks]

  • Cue. Ek=12mv2=12×2.0×5.02=25E_k = \frac{1}{2}mv^2 = \frac{1}{2} \times 2.0 \times 5.0^2 = 25 J.

Q3. A motor takes in 500500 W and delivers 400400 W of useful power. Find its efficiency. [2 marks]

  • Cue. Efficiency =400500=0.80=80%= \frac{400}{500} = 0.80 = 80\%.

Exam-style practice questions

Practice questions written in the style of Pearson Edexcel exam questions on this dot point, with worked answer explainers. The year tag is the paper they imitate, not the source.

Edexcel 20184 marksA crane lifts a 500500 kg load vertically through 1212 m in 8.08.0 s at constant speed. Calculate the work done against gravity and the useful output power. Take g=9.81g = 9.81 m per second squared.
Show worked answer →

Work against gravity: W=mgh=500×9.81×12=5.9×104W = mgh = 500 \times 9.81 \times 12 = 5.9 \times 10^{4} J.

Useful output power: P=Wt=5.886×1048.0=7.4×103P = \frac{W}{t} = \frac{5.886 \times 10^{4}}{8.0} = 7.4 \times 10^{3} W.

Markers reward W=mghW = mgh, the power as work over time, and the value about 7.47.4 kW.

Edexcel 20215 marksA 0.200.20 kg ball is dropped from rest from a height of 1.81.8 m and rebounds to 1.21.2 m. Calculate its speed just before impact and determine the efficiency of the bounce in terms of energy. Take g=9.81g = 9.81 m per second squared.
Show worked answer →

Speed before impact from energy conservation: 12mv2=mgh\frac{1}{2}mv^2 = mgh, so v=2gh=2×9.81×1.8=35.3=5.9v = \sqrt{2gh} = \sqrt{2 \times 9.81 \times 1.8} = \sqrt{35.3} = 5.9 m per second.

Efficiency uses the ratio of rebound height to drop height (since gravitational PE is proportional to height): efficiency=1.21.8=0.67=67%\text{efficiency} = \frac{1.2}{1.8} = 0.67 = 67\%.

Markers reward v=2ghv = \sqrt{2gh}, the height ratio for efficiency, and the values 5.95.9 m per second and 67%67\%.

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