Derive and use gravitational potential energy
Derive the change in gravitational potential energy from the work needed to lift a mass in a uniform field.
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
Which displacement enters ΔEₚ = mgΔh?
What you will learn
- Derive ΔEₚ = mgΔh using work and weight
- Calculate gravitational potential energy change in a uniform field
- Interpret height change and the sign of energy transfer
Lifting work becomes gravitational potential energy
In a uniform gravitational field, the weight of a mass m is mg. If a constant upward force balances the weight while the object rises through height Δh, the work done by the lifting force is W = Fs = mgΔh.
That work increases the object's gravitational potential energy, so ΔEₚ = mgΔh. The result depends on vertical height change, not on the path length, when the lifting is against gravity without other energy changes.
Use vertical height and consistent field strength
A rise gives a positive gravitational potential energy change; a fall gives a decrease. Use the mass, the local value of g and the vertical height difference between the initial and final positions.
If other forces do work or energy is transferred to internal stores, the work done by a particular force need not equal the change in gravitational potential energy alone. The formula describes the gravitational store change in the stated uniform-field model.
Pause and explain
What happens to gravitational potential energy when an object rises in a uniform field?
Worked example
A 2.0 kg bag is lifted vertically by 3.0 m where g = 10 N kg⁻¹. Find its gravitational potential energy increase.
Show the worked solution
- Use the vertical height change Δh = 3.0 m.
- ΔEₚ = mgΔh = 2.0 × 10 × 3.0.
- A rise means the gravitational potential energy increases.
Answer 60 J increase.
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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.
A 5.0 kg box is raised 1.2 m where g = 9.8 N kg⁻¹. Find its gravitational potential energy increase.
Give me a hint
Use mass, local g and vertical rise.
Compare my reasoning
- ΔEₚ = mgΔh.
- ΔEₚ = 5.0 × 9.8 × 1.2.
- Report the energy in joules.
58.8 J, approximately 59 J.
Look for these in your work
- Correct formula
- Vertical height used
- Energy unit stated
A 0.80 kg ball falls 2.5 m in a uniform field where g = 10 N kg⁻¹. Find the change in its gravitational potential energy.
Give me a hint
A downward change in height is negative when upward is chosen as positive.
Compare my reasoning
- Δh = −2.5 m.
- ΔEₚ = 0.80 × 10 × (−2.5).
- Interpret the sign as a decrease.
−20 J; its gravitational potential energy decreases by 20 J.
Look for these in your work
- Height sign chosen
- Product calculated
- Energy decrease interpreted
A 1200 kg car climbs a road and gains 150 m in vertical height. Take g = 10 N kg⁻¹. Find the increase in gravitational potential energy, regardless of the road's length.
Give me a hint
The formula uses vertical height change, not road distance.
Compare my reasoning
- ΔEₚ = mgΔh.
- ΔEₚ = 1200 × 10 × 150.
- State the result in joules or megajoules.
1.8 × 10⁶ J = 1.8 MJ increase.
Look for these in your work
- Vertical rise used
- Mass and g included
- Large energy expressed with scale and unit
Common mistakes
- Using the length of a sloping path instead of vertical height change
- Using mass in grams without conversion
- Giving a negative increase for an upward lift
What can you explain now?
Which displacement enters ΔEₚ = mgΔh?
Compare with the explanation
Vertical height change
Gravity acts vertically, so the relevant displacement is vertical.
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Make it stick
Why does gravitational potential energy depend on vertical height rather than the length of the path?
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