Use vertical height in gravitational potential energy
Calculate mgΔh, distinguish height from distance along a slope and keep energy differences unchanged when the reference level changes.
Exam scope and assessment notes
Mechanics (Paper 4, AS & A Level) · Mathematics 9709, 2026–2027, section 4.5. Use g = 10 m s⁻² throughout these problems. Masses are positive and constant. Work by a constant force is Fd cos θ, with θ between force and displacement; scalar-product notation is not required. Kinetic energy is ½mv² and gravitational potential-energy change is mgΔh. State the system and count gravitational work or potential-energy change once, not both. Mechanical energy is conserved only under the stated loss-free model; friction and supplied resistance transfer mechanical energy to other forms. Smooth curved-slide examples use overall energy changes and assume contact throughout, rather than requiring a normal-reaction calculation. Average power is work divided by elapsed time; P = Fv applies to a force along the motion. Constant driving power generally gives variable force and acceleration; instantaneous calculations do not justify constant-acceleration formulae over a whole journey. Check string, contact and motion-state changes before continuing a model. Written practice is self-checked; human subject-teacher review remains pending.
Before you begin
- Find vertical height from slope distance and sine.
- Calculate signed changes between two levels.
Work on paper. Try each question before revealing help, and explain why your method works.
Quick readiness check
A particle travels distance s up a plane inclined at θ to horizontal. What height change enters mgΔh?
What you will learn
- Use vertical height change in gravitational potential-energy calculations.
- Relate gravitational work to potential-energy change without double counting.
Potential-energy differences depend on vertical height
In a uniform gravitational field, potential energy relative to a chosen reference level is U = mgh. Its change is mg(h₂ − h₁). For a 4 kg particle moved 5 m up a 30° slope, the vertical rise is 5 sin 30° = 2.5 m, so ΔU = 4 × 10 × 2.5 = 100 J. Using 5 m directly would confuse distance along the plane with vertical height.
Rising gives positive ΔU; falling gives negative ΔU. A 2 kg particle falling 4 m has ΔU = −80 J. Gravity then does +80 J of work. For any change in this field, gravitational work is −ΔU. The path shape does not alter this height-based gravitational contribution, though friction or other forces can add path-dependent work.
Choose a datum but do not count gravity twice
A 5 kg particle rises from height 2 m to 5 m above ground. With ground as datum, U changes from 100 to 250 J, giving +150 J. Choosing a datum 3 m above ground instead gives heights −1 and +2 m, so U changes from −50 to +100 J. The same +150 J difference is physically relevant; negative reference potential energy is allowed.
In a kinetic-energy work balance, include gravitational work with its sign. In a mechanical-energy balance using K + U, put gravity into ΔU and include only the other work terms on the work side. Including both −mgΔh as work and +mgΔh as potential change in the same balance counts gravity twice. State the chosen form and system before inserting numbers.
Pause and explain
A particle gains potential energy 100 J. What work does gravity do?
Worked example
A 4 kg particle moves 5 m up a 30° plane. Find potential-energy change and work by gravity using g = 10.
Show the worked solution
- Vertical rise is Δh = 5 sin 30° = 2.5 m.
- Potential-energy change is ΔU = 4 × 10 × 2.5 = +100 J.
- Gravity does the opposite signed work, −100 J.
Answer ΔU = +100 J; gravitational work −100 J.
Try a topic test
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.
A 3 kg particle rises vertically by 2 m. Find its potential-energy change and work by gravity.
Give me a hint
Use the signed vertical change and opposite signs for potential change and gravity work.
Compare my reasoning
- Δh = +2 m, so ΔU = 3 × 10 × 2 = +60 J.
- Gravitational work is −ΔU = −60 J.
- The rise increases stored gravitational energy while gravity opposes the displacement.
ΔU = +60 J; gravitational work −60 J.
Look for these in your work
- Use g = 10 as supplied.
- Keep the two quantities equal in magnitude and opposite in sign.
A 2 kg particle falls from one level to another 4 m lower. Find potential-energy change and work by gravity.
Give me a hint
A fall gives a negative final-minus-initial height change.
Compare my reasoning
- Δh = −4 m.
- ΔU = 2 × 10 × (−4) = −80 J.
- Gravity does +80 J of work as the particle moves down.
ΔU = −80 J; gravitational work +80 J.
Look for these in your work
- Use signed height change.
- Do not report a positive potential gain for a fall.
A 5 kg particle rises from 2 m to 5 m above ground. Calculate both endpoint potential energies and their difference using first ground, then a level 3 m above ground, as the datum.
Give me a hint
Subtract the same reference height from both endpoints.
Compare my reasoning
- Ground datum gives energies 5 × 10 × 2 = 100 J and 5 × 10 × 5 = 250 J, difference 150 J.
- The higher datum gives heights −1 and +2, energies −50 J and +100 J.
- Their difference is again +150 J; the reference choice changes endpoint values, not this change.
Ground: 100 to 250 J; higher datum: −50 to +100 J; ΔU = +150 J in both.
Look for these in your work
- Use the same datum for both endpoints in each calculation.
- Retain negative relative height where appropriate.
Common mistakes
- Use vertical height rather than slope distance.
- Negative reference potential energy is possible.
- Count gravity as work or potential-energy change once.
What can you explain now?
Does changing the potential-energy datum change the energy difference between the same two positions?
Compare with the explanation
No; the additive reference shift cancels from the difference.
Both endpoint values change by the same constant for the same mass.
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Make it stick
Find vertical height change, calculate mgΔh and include the gravitational contribution exactly once in the balance.
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.