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A Level · Cambridge (CIE) · 9702

Apply energy conservation and efficiency

Track energy transfers through a system and calculate efficiency as useful output divided by total input.

Work, energy and power · Cambridge International AS & A Level Physics 9702 · 5.1

AS Level; Cambridge International Physics 9702, 2025–2027 syllabus, Topic 5. These original explanations and calculations support theory study; full timed assessment and human subject-teacher review remain outstanding.

Before you begin

  • Use force, displacement, speed, mass and gravitational field strength with consistent SI units.

Work on paper. Try each question before revealing help, and explain why your method works.

Quick readiness check

What belongs in the denominator of efficiency?

What you will learn

  • Apply conservation of energy to system changes
  • Define efficiency using useful output and total input
  • Calculate useful energy, input energy or percentage efficiency

Energy is transferred between stores

The principle of conservation of energy says that energy is not created or destroyed. It is transferred between stores or between a system and its surroundings. A calculation should account for all relevant energy inputs and outputs.

An object's mechanical energy may decrease when resistive forces do work; the energy has been transferred, often to internal energy of the object and surroundings. A model that counts only useful output can appear to lose energy if the other transfers are omitted.

Efficiency compares useful output with total input

Efficiency is the ratio of useful energy output to total energy input. Multiply the ratio by 100% to express it as a percentage. The useful output cannot exceed the total input for a passive system under this accounting.

Rearrange the relationship to find a missing input or useful output. Keep the numerator and denominator tied to the same process and interval. A low efficiency means a smaller fraction reaches the chosen useful output, not that the remaining energy has ceased to exist.

efficiency = useful energy output / total energy input

Pause and explain

A device is 60% efficient. What does that tell you?

Put the idea to work

Worked example

A motor receives 800 J of electrical energy and transfers 600 J usefully to a load. Find its efficiency.

Show the worked solution
  1. Efficiency = useful output / total input.
  2. Efficiency = 600/800 = 0.75.
  3. Convert the ratio to a percentage.

Answer Efficiency = 75%. The remaining 200 J is transferred to other stores or the surroundings.

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From guided practice to a new situation

Write your answer and reasoning first. Use a hint only if you are stuck. The model solution is for self-checking; your written work is not automatically marked.

1 · Guided

A lifting system uses 500 J of input energy and transfers 350 J to the raised load. Calculate its percentage efficiency.

Give me a hint

Divide useful output by total input, then multiply by 100%.

Compare my reasoning
  1. Efficiency = 350/500.
  2. Efficiency = 0.70.
  3. Convert to a percentage.

70%.

Look for these in your work

  • Useful output identified
  • Correct ratio formed
  • Percentage calculated
2 · Independent

A lamp transfers 120 J of electrical energy; 30 J becomes visible light usefully. Find the useful efficiency and the energy transferred to other forms.

Give me a hint

Calculate the ratio and use conservation to account for the remainder.

Compare my reasoning
  1. Efficiency = 30/120 = 0.25 = 25%.
  2. Other energy transfer = 120 − 30 = 90 J.
  3. The 90 J is transferred to other forms or the surroundings, not destroyed.

Useful efficiency 25%; 90 J is transferred to other forms or the surroundings.

Look for these in your work

  • Useful fraction calculated
  • Remainder found
  • Energy conservation stated
3 · Transfer

A winch is 80% efficient and delivers 2.4 kJ of useful energy to a load. Find the total energy input.

Give me a hint

Rearrange efficiency = useful output / input.

Compare my reasoning
  1. Convert useful energy to 2400 J.
  2. Input = useful output / efficiency = 2400/0.80.
  3. Check that input is greater than useful output.

3000 J, or 3.0 kJ, input energy.

Look for these in your work

  • Efficiency used as 0.80
  • Equation rearranged
  • Input greater than output

Common mistakes

  • Using total output instead of useful output in the numerator
  • Reversing input and output in the ratio
  • Treating energy that is not useful as destroyed
Recall without your notes

What can you explain now?

What belongs in the denominator of efficiency?

Compare with the explanation

Total energy input

Efficiency compares useful output with all supplied input.

After trying it yourself, choose your next review. This is your self-assessment.

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Make it stick

Where does the energy go when it is not transferred to a system's selected useful output?

If you needed the solution, close it and solve the problem again from a blank page. A correct answer is stronger when you can explain the reason for each step.

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Course scope and assessment

One hundred and sixty-two original lessons cover all 12 published subject-content outcomes in Topic 1 (sections 1.1–1.4), all nine outcomes in Topic 2 (section 2.1), all 13 outcomes in Topic 3 (sections 3.1–3.3), all 13 outcomes in Topic 4 (sections 4.1–4.3), all 11 outcomes in Topic 5 (sections 5.1–5.2), all 10 outcomes in Topic 6 (sections 6.1–6.2), all 16 outcomes in Topic 7 (sections 7.1–7.5), all 12 outcomes in Topic 8 (sections 8.1–8.4), all 15 outcomes in Topic 9 (sections 9.1–9.3), all 16 outcomes in Topic 10 (sections 10.1–10.3), all 18 outcomes in Topic 11 (sections 11.1–11.2), all seven outcomes in Topic 12 (sections 12.1–12.2), all 12 outcomes in Topic 13 (sections 13.1–13.4), all eight outcomes in Topic 14 (sections 14.1–14.3), and all nine outcomes in Topic 15 (sections 15.1–15.3), plus all four outcomes in Topic 16 (sections 16.1–16.2), and all ten outcomes in Topic 17 (sections 17.1–17.3), plus all 12 outcomes in Topic 18 (sections 18.1–18.5), all nine outcomes in Topic 19 (sections 19.1–19.3), all 19 outcomes in Topic 20 (sections 20.1–20.5), and all eight outcomes in Topic 21 (sections 21.1–21.2), and all 17 outcomes in Topic 22 (sections 22.1–22.4), and all 13 outcomes in Topic 23 (sections 23.1–23.2), and all 16 outcomes in Topic 24 (sections 24.1–24.3), and all 11 outcomes in Topic 25 (sections 25.1–25.3), of the 2025–2027 syllabus. Each lesson has guided, independent and application practice with explained answers. All 25 syllabus topics now have mapped teaching and practice; the 2028–2030 changes outside Topics 23–25 still need reconciliation. Online study cannot demonstrate supervised Paper 3 laboratory competence. Paper 5 planning and analysis, full timed cumulative assessment and human subject-teacher review remain outstanding.

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