Uniform-field potential energy
Key idea: H2 Physics lessons on weight, projectile components, gravitational potential energy, air resistance and terminal velocity.
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The core idea
Build the idea
Learn the idea
Big question: How does height in a uniform field store and release energy?
Near Earth's surface, gravitational field strength is approximately constant, so a height change Δh gives ΔEp = mgΔh. The sign follows the chosen initial and final levels. If resistance is negligible, the decrease in gravitational potential energy equals the increase in kinetic energy, providing a useful alternative to kinematics.
Derive rather than memorise mgh
In a uniform field, lifting a mass slowly needs an upward force equal to mg. Work against gravity through vertical height Δh is therefore mgΔh, and this increases the gravitational potential-energy store of the mass–Earth system.
The sign comes from Δh = hfinal − hinitial. Rising gives a positive change; falling gives a negative change. The zero level is a convenient choice, but differences do not depend on that choice.
Check your understanding: A ball finishes 2 m below its start. What is the sign of ΔEp?
Negative, because Δh is negative.
See why the route does not matter
On a frictionless ramp, the downslope weight component is mg sinθ and the ramp length L satisfies L sinθ = h. Their product is therefore mgh whatever the angle.
Gravity is conservative in this model: work depends only on endpoints. A longer path can change the force required, but not the gravitational-store change between the same heights.
Check your understanding: Does a gentler frictionless ramp reduce the work done against gravity?
No. It reduces the required force but increases the distance so the work remains mgh.
Key ideas to keep
- Only a change in height matters; the path length does not.
- Choose and state a zero level, although energy differences do not depend on that choice.
- Energy methods give speed but not directly the direction of velocity.
See the reasoning
Worked example
Use height change within a projectile journey
Question: A 0.50 kg ball is launched from a platform 1.2 m above the ground, rises another 3.0 m and then lands. Find ΔEp from launch to the highest point, from highest point to landing, and overall. Use g = 9.81 m s⁻².
Step 1: Choose one reference and list heights
Why: Consistent endpoints prevent mixing path length with vertical change.
Working: Take ground as zero: hlaunch = 1.2 m, htop = 4.2 m, hfinal = 0.
Step 2: Calculate each change
Why: Each interval uses final height minus initial height.
Working: Launch→top: 0.50(9.81)(4.2−1.2) = +14.7 J. Top→ground: 0.50(9.81)(0−4.2) = −20.6 J.
Step 3: Check the complete journey
Why: Intermediate height must cancel when changes are added.
Working: Overall = 0.50(9.81)(0−1.2) = −5.89 J; +14.7 − 20.6 = −5.9 J.
Answer: +14.7 J, −20.6 J and −5.89 J respectively.
Check: The overall change depends only on launch and landing heights, not the maximum height reached between them.
Another worked model
Question
A 2.5 kg load is raised slowly through 6.0 m where g = 9.8 N kg⁻¹. Derive the relationship used, then find its potential-energy increase and work done by gravity.
Check the worked solution
The lifting force is mg, so transferred work W = force × upward displacement = mgΔh = ΔEₚ. Thus ΔEₚ = 2.5(9.8)(6.0) = +147 J. Gravity acts opposite the displacement, so its work is −147 J.
Use a hint if needed
Practise with support
Try this
A 5.0 kg object is lowered by 3.0 m in a uniform field where g = 9.8 N kg⁻¹. Find ΔEₚ and the work done by gravity.
Hint: Take upward height change as positive.
Check your answer
Δh = −3.0 m, so ΔEₚ = 5.0(9.8)(−3.0) = −147 J. Work done by gravity is −ΔEₚ = +147 J.
Now work without the hint
Practise independently
Your turn
Explain why ΔEₚ = mgΔh follows from work done, including the assumptions, sign of Δh and why the chosen zero height does not affect the answer.
Check your answer
In a uniform field, weight mg is constant. Moving slowly through vertical displacement Δh requires opposite applied force mg, so transferred work is mgΔh and equals the potential-energy change. Upward Δh is positive and downward negative. Changing the zero adds the same constant to both endpoint energies, so their difference is unchanged.
Avoid these traps
Common mistakes
Common mistake
The equation mgh gives one absolute potential energy.
What is wrong with this reasoning?
Show better thinking
In a uniform field, ΔEₚ = mgΔh gives a change relative to chosen heights. Only the height difference matters; raising gives positive Δh and lowering gives negative Δh.
Common mistake
Gravitational potential-energy change depends on the path taken between two heights.
What is wrong with this reasoning?
Show better thinking
Gravity is conservative. In the uniform-field model, ΔEₚ depends only on the vertical height change: mgΔh, not the route length.
Write for the examiner
Exam guidance
Keep ΔEp signed, or write a clear store-transfer statement and use positive magnitudes consistently.
Exam-style practice [6 marks]
A 25 kg crate is moved slowly to a shelf 1.8 m higher. Route A is vertical. Route B is a 6.0 m ramp with a constant 18 N resistive force. Compare the work done by the mover on the crate for the two routes. Use g = 9.81 m s⁻².
Plan before you answer
- Find the common gravitational-store increase.
- Add the transfer to internal stores only for the ramp.
- Explain the difference using conservation.
Mark your answer and compare the model
Marking points
Tick each point only if your answer states it clearly.
Model answer
Both routes raise the crate by 1.8 m, so ΔEp = 25(9.81)(1.8) = 441 J. Route A therefore needs 441 J in the ideal slow lift. On route B, 18(6.0) = 108 J is also transferred to internal stores, so the mover does 441 + 108 = 549 J. The gravitational-store change is path-independent; the extra work is due to resistance.
Come back in three days
Check what stayed with you
Recall question 1
Write ΔEp in a uniform gravitational field.
Check the answer
ΔEp = mgΔh.
Recall question 2
What does Δh mean?
Check the answer
Final vertical height minus initial vertical height.
Recall question 3
Why is the zero of potential energy arbitrary?
Check the answer
Only changes between states affect measurable energy transfers.
Syllabus and review details
This lesson covers the listed H2 Physics 9478 outcomes. Treat horizontal and vertical motion separately, then connect them through their shared time. In the ideal model, air resistance is negligible and gravitational acceleration is uniform; questions about drag must state that a resistive force is present.
- GCE A-Level H2 PhysicsTopic 5(c) / Topic 5(d) · 2027Checked against the syllabus · partial topic coverageOfficial 9478 syllabus
Course and syllabus information
- Course
- GCE A-Level H2 Physics
- Edition
- GCE A-Level H2 Physics 2027