Potential energy in a uniform gravitational field
Key idea: In a uniform gravitational field, gravitational potential-energy change depends only on vertical height change, not on the path taken.
Continue where you stopped
The core idea
H1 Physics 8867 · Lesson 2 of 3
Check your understandingBy the end of this lesson, you should be able to
- Derive ΔEp = mgΔh from work done.
- Use signed height changes consistently.
- Explain why path length does not determine gravitational potential-energy change.
Learn the idea
Big question: Why does gravitational potential-energy change depend on height but not on the path taken?
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
- Define the two endpoints before choosing Δh.
- Use vertical height, not distance along a slope or trajectory.
- The zero of potential energy is arbitrary; changes are physically useful.
Relationships to know
ΔEp = mgΔhwork by gravity = −ΔEp
Follow 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.
Now try it with support
Practise with support
Derive the potential-energy change for a mass raised slowly through height h by an angled ramp.
Hints
- The component of weight down a frictionless ramp is mg sin θ.
- Ramp length L satisfies L sin θ = h.
View the guided answer
Work against gravity = (mg sin θ)L = mg(L sin θ) = mgh, so ΔEp = mgh independent of ramp angle.
Your turn
Practise independently
A 0.40 kg ball moves from a platform 1.2 m high to a point 0.30 m below the launch level. Find ΔEp for the ball–Earth system.
Check your answer
The endpoint is 0.30 m below the launch level, so Δh = −0.30 m. ΔEp = 0.40(9.81)(−0.30) = −1.18 J. The platform’s absolute height is not needed.
Common mistakes and exam guidance
Watch out for
- Using the full path length as Δh.
- Dropping the sign when the final point is lower than the initial point.
In an exam
- Write Δh = hfinal − hinitial before substituting.
- If only an energy change is asked for, do not invent an absolute zero level.
Put the ideas together
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.
View the marking points and model answer
Marking points
- Finds ΔEp = 441 J.
- States route A work = 441 J for a slow ideal lift.
- Finds resistive work = 108 J.
- Adds it for route B.
- Obtains route B work = 549 J.
- Explains that gravity's contribution is path-independent but resistance adds a path-dependent thermal transfer.
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.
Finish from memory
Three-question recap
Write ΔEp in a uniform gravitational field.
Check
ΔEp = mgΔh.
What does Δh mean?
Check
Final vertical height minus initial vertical height.
Why is the zero of potential energy arbitrary?
Check
Only changes between states affect measurable energy transfers.
Continue with the next resource in this course.
Course and syllabus information
- Course
- GCE A-Level H1 Physics
- Edition
- GCE A-Level H1 Physics 2027