The big idea: Lift something up and it stores gravitational potential energy (PE) — energy because of its height.
Let it fall and that PE turns into kinetic energy (KE) — energy because of its motion.
No energy is lost: the PE the object loses becomes the KE it gains.
Spot it: Top of a fall = all PE, no KE. Bottom = all KE, no PE. At any height in between, PE + KE adds up to the same total — that total is the mechanical energy, and it stays constant.
The change in gravitational PE when an object moves up or down a height Δh is given in the data booklet:
- change in gravitational PE (J)
- mass (kg)
- gravitational field strength (≈ 9.8 N kg⁻¹ on Earth)
- change in height (m)
What 'conservation of energy' means here: For a falling body with no air resistance, the mechanical energy (PE + KE) is conserved — it stays the same.
So the PE lost = the KE gained: mgΔh = ½mv².
The kinetic energy of a moving object is also given in the booklet:
- kinetic energy (J)
- mass (kg)
- speed (m s⁻¹)
Worked example — PE at the top
A 2.0 kg ball is lifted 5.0 m above the ground. Take g = 9.8 N kg⁻¹. Find the gravitational PE it gains.
Solution
- Start with the given formula:
- Put in the numbers (m = 2.0, g = 9.8, Δh = 5.0):
- Work it out — keep the unit:
Final answer
PE gained = 98 J.
Practice with real exam questions
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How this is tested: Falling-body energy is tested in two predictable forms.
- Paper 1A: a quick MCQ — often a ratio of KE part-way down to the PE at the top, or 'where is PE = KE?'. - Paper 2: a short calculation that sets the PE lost equal to the KE gained (mgΔh = ½mv²) to find a landing speed.
Classic trap: the mass cancels — speed at the bottom doesn't depend on m. And drop half the height ⇒ only half the energy is KE so far, not all of it.
The ratio shortcut: Because KE gained = PE lost, an object that has fallen a fraction f of its total height has turned a fraction f of its starting PE into KE. Fall half the height → KE = ½ of the starting PE.
IB-style question — ratio of KE to starting PE
A stone is released from rest at the top of a cliff. Air resistance is negligible. When it has fallen to a point one-quarter of the way down from the top, what is the ratio (KE there) : (PE at the release point)?
Solution
- Energy is conserved, so the KE gained = the PE lost:
- It has fallen Δh = ¼ of the total height H, so the PE lost is ¼ of the starting PE (mgH):
- Divide by the starting PE (mgH) — the mass and g cancel:
Final answer
The ratio is 1 : 4. (Fallen ¼ of the way ⇒ ¼ of the starting PE is now KE.)