Apply energy conservation and efficiency
Track energy transfers through a system and calculate efficiency as useful output divided by total input.
Section 5 · Work, energy and power
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.
Pause and explain
A device is 60% efficient. What does that tell you?
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
- Efficiency = useful output / total input.
- Efficiency = 600/800 = 0.75.
- Convert the ratio to a percentage.
Answer Efficiency = 75%. The remaining 200 J is transferred to other stores or the surroundings.
Try a topic test
Choose Easy, Normal or Difficult. Save your answers and open worked review after finishing.
Test this topic →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.
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
- Efficiency = 350/500.
- Efficiency = 0.70.
- Convert to a percentage.
70%.
Look for these in your work
- Useful output identified
- Correct ratio formed
- Percentage calculated
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
- Efficiency = 30/120 = 0.25 = 25%.
- Other energy transfer = 120 − 30 = 90 J.
- 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
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
- Convert useful energy to 2400 J.
- Input = useful output / efficiency = 2400/0.80.
- 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
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.