IGCSE Physics 0625 · Topic 1.7

IGCSE Physics: Energy, Work and Power Practice Questions

Energy is transferred between stores but is never created or destroyed. Work done equals force multiplied by distance moved in the direction of the force, and power is the rate at which work is done or energy is transferred.

Cambridge IGCSE Physics (0625) · Topic 1.7: Energy, Work and Power

Sub-topic 1.7 of Cambridge IGCSE Physics 0625 carries four equations that appear on nearly every paper. The energy conversion question, where gravitational potential energy becomes kinetic energy, is the one that separates grades because it requires equating two expressions. The questions below build up to it.

What you need to know for Energy, Work and Power

IGCSE Physics Energy, Work and Power questions and answers

4 exam-style questions written to the 0625 syllabus. Try each one on paper first, then open the worked answer to check your method against the marks.

Question 1[3 marks]

A car of mass 1400 kg is travelling at 20 m/s. Calculate its kinetic energy.

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Answer: 280000 J, or 280 kJ.
  1. Use KE = half x mass x velocity squared.
  2. Square the velocity first: 20 squared = 400.
  3. KE = 0.5 x 1400 x 400.
  4. KE = 280000 J, which is 280 kJ.
How the marks are awarded. 1 mark for using KE = half x m x v squared. 1 mark for squaring the velocity correctly. 1 mark for 280000 J with the unit.
Where students lose the mark. Squaring the whole expression or forgetting to square v at all. Only the velocity is squared, and the half applies to the entire product.
Question 2[4 marks]

A ball of mass 0.20 kg is dropped from a height of 1.8 m. Calculate its speed just before it hits the ground, assuming air resistance is negligible. Take g as 9.8 N/kg.

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Answer: 5.9 m/s
  1. As the ball falls, gravitational potential energy is transferred to kinetic energy. With no air resistance, all of it is transferred.
  2. Change in GPE = mg delta h = 0.20 x 9.8 x 1.8 = 3.528 J.
  3. Set this equal to the kinetic energy: 3.528 = 0.5 x 0.20 x v squared, so v squared = 3.528 divided by 0.10 = 35.28.
  4. v = square root of 35.28 = 5.94, which is 5.9 m/s to 2 significant figures. Note that the mass cancels, so any mass dropped from this height reaches the same speed.
How the marks are awarded. 1 mark for stating that GPE is transferred to KE. 1 mark for a GPE of 3.528 J. 1 mark for equating GPE to KE and rearranging for v squared. 1 mark for 5.9 m/s with the unit.
Where students lose the mark. Forgetting to take the square root at the end. Check that the final answer is a speed, not a speed squared.
Question 3[4 marks]

A crane lifts a 250 kg load through a vertical height of 12 m in 40 s. Calculate the useful power output of the crane. Take g as 9.8 N/kg.

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Answer: 735 W
  1. The useful work done is the gain in gravitational potential energy of the load.
  2. Work done = mg delta h = 250 x 9.8 x 12 = 29400 J.
  3. Power = work done divided by time taken = 29400 divided by 40.
  4. Power = 735 W.
How the marks are awarded. 1 mark for identifying the work done as the gain in GPE. 1 mark for 29400 J. 1 mark for using power = work divided by time. 1 mark for 735 W with the unit.
Where students lose the mark. Using the horizontal distance or the cable length instead of the vertical height. Only the vertical height counts when calculating a change in gravitational potential energy.
Question 4[3 marks]

An electric motor is supplied with 4800 J of electrical energy and does 1200 J of useful work. Calculate its efficiency as a percentage and state what happens to the remaining energy.

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Answer: 25 per cent. The rest is dissipated to the surroundings, mainly as thermal energy.
  1. Efficiency = useful energy output divided by total energy input.
  2. Efficiency = 1200 divided by 4800 = 0.25.
  3. As a percentage: 0.25 x 100 = 25 per cent.
  4. The remaining 3600 J is transferred to the surroundings, mostly as thermal energy through friction in the bearings and heating in the wires, and some as sound. The total energy is unchanged, which is why efficiency can never exceed 100 per cent.
How the marks are awarded. 1 mark for using efficiency = useful output divided by total input. 1 mark for 25 per cent. 1 mark for stating that the rest is dissipated to the surroundings as thermal energy and sound.
Where students lose the mark. Dividing the input by the output, giving 400 per cent. Efficiency is always output over input and can never be above 100 per cent.

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Energy, Work and Power FAQs

What is the equation for kinetic energy?

Kinetic energy equals half multiplied by mass multiplied by velocity squared. Mass is in kilograms, velocity in metres per second and the energy comes out in joules. Because velocity is squared, doubling the speed of an object multiplies its kinetic energy by four.

How do I calculate the speed of a falling object using energy?

Calculate the change in gravitational potential energy using mass multiplied by g multiplied by the drop in height. If air resistance is negligible, all of it becomes kinetic energy, so set it equal to half multiplied by mass multiplied by velocity squared, rearrange for velocity squared, and take the square root.

What is the difference between work and power?

Work done is the energy transferred when a force moves an object, calculated as force multiplied by distance in the direction of the force, and measured in joules. Power is the rate at which that work is done, calculated as work divided by time, and measured in watts.

How do I calculate efficiency?

Divide the useful energy output by the total energy input, then multiply by 100 for a percentage. Efficiency can never exceed 100 per cent, because the energy that is not usefully transferred is dissipated to the surroundings, usually as thermal energy and sound.

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Written to the published Cambridge IGCSE Physics (0625) syllabus. Check your school entry code and syllabus year, because Core and Extended candidates are assessed on different content. Last reviewed 2026-08-12.