Conserve energy through multiple stages
Balance intermediate transfers and final stores without double-counting the same energy.
Mapped to the named 0654 section using teaching from the corresponding single-science course. A 0654-specific review of every Core and Supplement objective is still in progress. Written practice is self-checked; it does not demonstrate supervised laboratory competence or award an official Cambridge grade.
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.
Sources for this lesson
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.