Conserve energy through multiple stages
Balance intermediate transfers and final stores without double-counting the same energy.
Section 1.7.1 · Energy
Exam scope and assessment notes
Extended only (Supplement) · Physics 0972, 2026–2028, section 1.7.1. Energy stores, transfers and simple conservation are Core; kinetic and gravitational potential energy equations, multistage conservation and Sankey diagrams are Supplement. State the system, interval and model assumptions. Written practice is self-checked. Supervised practical experience, cumulative assessment and human subject-teacher review remain pending. Cambridge states that IGCSE 0972 has the same subject content as IGCSE 0625, with 9–1 grading. This StudyForA pathway maps the 2026–2028 syllabus only. This 9–1 graded qualification is available only in a limited number of administrative zones. Confirm that your centre can enter you for this code and exam series with Cambridge International. Cambridge has published a separate syllabus for 2029. StudyForA has not reconciled that version yet; this page supports exams in 2026, 2027 and 2028 only.
Before you begin
- Calculate kinetic and potential changes.
- Balance distinct energy destinations.
Work on paper. Try each question before revealing help, and explain why your method works.
Quick readiness check
How much is 120 J plus 180 J?
What you will learn
- Apply conservation to a multistage chain.
- Combine kinetic and gravitational calculations with stated other transfers.
An intermediate transfer is not an extra final output
In an original two-stage lifting account, a source releases 500 J. The generator transfers 300 J electrically onward and 200 J to internal stores. The motor uses that 300 J to increase a gravitational store by 120 J and other internal stores by 180 J. The final destinations are 120 + 180 + 200 = 500 J; adding the intermediate 300 again would count energy twice.
At g = 10 N/kg, a 3 kg load gains 120 J by rising 4 m if its kinetic store has no net change. Each stage needs its own balance, and the complete chain needs the same input amount and interval. An unexplained output discrepancy should prompt a check of missing transfers and stored changes rather than accepting energy creation.
Several stores can change in one stage
A 2 kg object falls 5 m from rest at g = 10 N/kg. Its gravitational potential decreases by 100 J. If air resistance is negligible and the entire decrease becomes kinetic energy, ½ × 2 × v² = 100 gives speed 10 m/s. If instead 36 J is transferred to internal stores and waves, its final kinetic energy is 64 J and speed is 8 m/s.
For an object already moving, include its initial kinetic energy as well as the potential decrease. A 2 kg object starting at 3 m/s has 9 J kinetic energy; a subsequent 1 m descent at g = 10 N/kg releases another 20 J, so its final kinetic energy is 29 J if other transfers are negligible. Do not silently assume rest or infer a direction from the energy magnitude.
Pause and explain
Should an intermediate 300 J transfer be added to all its final destination gains as extra output?
Worked example
A 500 J release becomes 300 J electrical transfer and 200 J internal gain; the 300 J then becomes 120 J gravitational gain and 180 J internal gain. Check the chain.
Show the worked solution
- First stage: 300 + 200 = 500 J.
- Second stage: 120 + 180 = 300 J.
- Final destinations: 120 + 180 + 200 = 500 J; omit the intermediate 300 from that final sum.
Answer 500 J accounted for: 120 J gravitational and 380 J internal gains.
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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.
Check the final budget for 500 J split into 300 J onward and 200 J internal, then the 300 J split into 120 J gravitational and 180 J internal.
Give me a hint
Use final destinations once each.
Compare my reasoning
- The first stage balances 500 = 300 + 200.
- The second balances 300 = 120 + 180.
- Final gains total 120 + 180 + 200 = 500 J.
120 J gravitational plus 380 J internal = 500 J.
Look for these in your work
- Balance each stage.
- Do not add the intermediate 300 twice.
A 2 kg object falls 5 m from rest at g = 10 N/kg. Find final speed if other transfers are negligible, then if 36 J is transferred elsewhere.
Give me a hint
Compute potential decrease, allocate the kinetic share and take a square root.
Compare my reasoning
- Potential decrease magnitude is 100 J.
- With no other transfers, final Ek = 100 J and speed = 10 m/s.
- With 36 J elsewhere, Ek = 64 J and speed = 8 m/s.
10 m/s; 8 m/s.
Look for these in your work
- Use the stated initially resting condition.
- Subtract other destinations before solving speed.
A 2 kg object starts at 3 m/s and descends 1 m at g = 10 N/kg with negligible other transfers. Find final kinetic energy and explain why using only mgΔh is incomplete.
Give me a hint
Add the initial kinetic store to the released gravitational amount.
Compare my reasoning
- Initial kinetic energy is 9 J.
- The descent releases 20 J.
- Final kinetic energy is 29 J; starting motion cannot be omitted.
29 J; the initial 9 J must be included.
Look for these in your work
- Include the initial store.
- Use conservation with all stated changes.
Common mistakes
- Do not add intermediate energy twice.
- Include initial kinetic energy when non-zero.
- Equality of two store changes needs other transfers to be negligible.
What can you explain now?
What speed follows from 64 J kinetic energy for 2 kg?
Compare with the explanation
8 m/s.
v = √(2 × 64/2) = 8 m/s.
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Make it stick
Balance a two-stage chain, avoid intermediate double-counting and include initial kinetic energy in a falling-object account.
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.