Calculate gravitational potential energy changes
Use the vertical height change with mass and the stated local gravitational field strength.
Section 1.7.1 · Energy
Exam scope and assessment notes
Extended only (Supplement) · Physics 0625, 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.
Before you begin
- Distinguish mass and weight.
- Convert height changes into metres.
Work on paper. Try each question before revealing help, and explain why your method works.
Quick readiness check
What vertical change is represented by heights 3 m and then 7 m?
What you will learn
- Recall and use ΔEp = mgΔh.
- Distinguish a vertical change from route length and an absolute height label.
Use the change in vertical height
Near Earth's surface in an approximately uniform local gravitational field, ΔEp = mgΔh. Use mass in kg, g in N/kg and vertical height change in m to obtain J. A 2 kg load raised from 3 m to 7 m above a chosen reference rises 4 m, so for g = 10 N/kg its gravitational potential increase is 80 J.
The route can be a longer sloping path or a vertical lift; the same endpoints give the same gravitational potential change in this model. The path length is not the vertical height change. If a 0.50 kg load rises 60 cm, convert to 0.60 m: at g = 9.8 N/kg its increase is 2.94 J.
A descent gives a decrease
Take Δh = hfinal − hinitial. The 2 kg load descending from 7 m to 3 m gives Δh = −4 m and ΔEp = −80 J. The negative change means the chosen store decreases; it is not a negative mass or a requirement that all possible definitions of its absolute potential energy are negative.
Use the supplied g rather than silently replacing it by 10. A consistent reference shift changes both height labels by the same amount and leaves Δh unchanged. An increase in gravitational potential energy does not alone establish the total input from a motor when friction, internal-store changes or kinetic-energy changes also matter.
Pause and explain
A longer sloping route reaches the same final height. Does that alone increase ΔEp?
Worked example
A 2 kg load rises from height 3 m to 7 m with g = 10 N/kg. Find ΔEp.
Show the worked solution
- Vertical change = 7 − 3 = 4 m.
- ΔEp = 2 × 10 × 4.
- The increase is +80 J.
Answer +80 J.
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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 ΔEp for 2 kg raised from 3 m to 7 m with g = 10 N/kg.
Give me a hint
Subtract the heights before multiplying by mg.
Compare my reasoning
- Δh = +4 m.
- mg = 20 N.
- ΔEp = 20 × 4 = +80 J.
+80 J.
Look for these in your work
- Use a height difference.
- Report an energy change in J.
Find the gain for 0.50 kg rising 60 cm with g = 9.8 N/kg, and the change for 2 kg descending 4 m with g = 10 N/kg.
Give me a hint
Convert the rise and use a negative vertical change for descent.
Compare my reasoning
- 60 cm = 0.60 m.
- First change = 0.50 × 9.8 × 0.60 = +2.94 J.
- Second change = 2 × 10 × (−4) = −80 J.
+2.94 J; −80 J.
Look for these in your work
- Use each supplied field strength.
- Distinguish gains and decreases.
A 2 kg load rises 4 m vertically along a 5 m ramp, with g = 10 N/kg. A student uses 2 × 10 × 5. Correct the potential-energy calculation without claiming the ramp has no friction.
Give me a hint
Use the vertical rise while leaving other input requirements separate.
Compare my reasoning
- The formula uses Δh = 4 m.
- The potential-energy increase is 80 J, not 100 J.
- Friction or other changes can affect total input, but not this endpoint potential-energy change.
ΔEp = 80 J; other transfers can make total input greater.
Look for these in your work
- Use vertical height.
- Do not turn a potential-energy calculation into an unsupported friction-free claim.
Common mistakes
- Use vertical change rather than path length.
- Convert cm to m.
- Use the stated g and chronological height difference.
What can you explain now?
What is ΔEp for the same load descending from 7 m to 3 m?
Compare with the explanation
−80 J.
Δh = 3 − 7 = −4 m, so the store decreases.
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Make it stick
Recall ΔEp, explain rise and descent signs, and distinguish vertical change, path length and total motor input.
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.