Find height changes and compare gravitational energy accounts
Rearrange the gravitational equation and keep reference levels and input limits consistent.
Cross-referenced from the related Cambridge IGCSE Chemistry 0620 or Physics 0625 teaching to the 0652 syllabus. Exact 0652 Core/Supplement objective coverage and this final-series course’s examiner fit still need specialist review. Written practice is self-checked and does not predict a Cambridge grade or demonstrate supervised laboratory competence.
Before you begin
- Calculate mgΔh.
- Rearrange products and differences.
Work on paper. Try each question before revealing help, and explain why your method works.
Quick readiness check
Which rearrangement gives height change?
What you will learn
- Find height change or mass from a stated gravitational energy change.
- Explain why shifting the reference level does not alter a change.
Rearrange with the supplied local field
From ΔEp = mgΔh, height change is Δh = ΔEp/(mg). An increase of 150 J for 3 kg at g = 10 N/kg gives rise 5 m. If the starting height is 2 m above the chosen reference, the final height is 7 m; the calculated 5 m is a change, not the final height.
Mass is m = ΔEp/(gΔh) when the supplied change and height change describe the same event. For 60 J gained over 2 m at g = 10 N/kg, mass is 3 kg. A zero height change supplies zero gravitational change in this model for many masses; do not determine mass using an undefined 0/0 quotient.
Keep references and energy destinations explicit
The same 2 kg lift from 3 m to 7 m can instead be labelled 1 m to 5 m if the reference is moved upward by 2 m. Both differences are 4 m and both give 80 J at g = 10 N/kg. Reference labels change the assigned level values, not the actual displacement or gravitational store change.
If 100 J is supplied to a 2 kg lift at g = 10 N/kg and all of it increases gravitational potential energy, the rise is 5 m. If 20 J instead increases internal stores and the initial and final kinetic stores are equal, only 80 J remains for the rise, giving 4 m. State where input goes before treating it all as a potential gain.
Sources for this lesson
Pause and explain
May a motor's entire input be assigned to potential gain without checking other transfers?
Worked example
A 3 kg load gains 150 J gravitational potential energy with g = 10 N/kg. Find its rise and final height if it starts at 2 m.
Show the worked solution
- Rise = 150/(3 × 10) = 5 m.
- The 5 m value is a height change.
- Final height = 2 + 5 = 7 m.
Answer Rise 5 m; final height 7 m.
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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.
Find rise and final height for 3 kg gaining 150 J at g = 10 N/kg from initial height 2 m.
Give me a hint
Find the rise from energy, then add it to the initial height.
Compare my reasoning
- mg = 30 N.
- Rise = 150/30 = 5 m.
- Final height = 2 + 5 = 7 m.
5 m rise; 7 m final height.
Look for these in your work
- Rearrange correctly.
- Separate change and endpoint.
Find mass when 60 J is gained over a 2 m rise at g = 10 N/kg; compare 3→7 m and 1→5 m labels for a 2 kg lift in the same field.
Give me a hint
Use the rearranged mass formula and two height differences.
Compare my reasoning
- Mass = 60/(10 × 2) = 3 kg.
- Both label pairs give a 4 m rise.
- Each potential-energy increase is 2 × 10 × 4 = 80 J.
3 kg; both reference choices give +80 J.
Look for these in your work
- Use the same event's quantities.
- Shift both endpoint labels consistently.
A 2 kg lift at g = 10 N/kg receives 100 J, transfers 20 J into internal stores, and starts and ends with the same kinetic energy. Find its rise.
Give me a hint
Subtract the other accounted destination before applying mgΔh.
Compare my reasoning
- Potential gain = 100 − 20 = 80 J.
- The mass-field product is 20 N.
- Rise = 80/20 = 4 m.
4 m.
Look for these in your work
- Use the complete input budget.
- Do not assign the internal-energy share twice.
Common mistakes
- Change is not the final height.
- Shift both endpoint labels consistently.
- Not every input joule must reach the gravitational store.
What can you explain now?
Does changing heights 3→7 m into 1→5 m change ΔEp?
Compare with the explanation
No.
Both describe the same 4 m rise for a common shifted reference.
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Make it stick
Rearrange for height, distinguish final height and show how a consistent reference shift leaves potential-energy change unchanged.
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.