IB Physics HLTopic 1 — Motion, Forces & EnergyPaper 1 & 2Work, Energy & Power~9 min read
Gravitational Potential Energy
Lift something up and you’ve given it energy — energy it will hand straight back the moment you let go. That stored “height energy” is gravitational potential energy. It’s what a raised hammer holds before it swings down, what water at the top of a dam carries, and what turns into speed every time something falls. The bigger the mass and the higher you raise it, the more it stores — and near the Earth’s surface, one tidy equation captures the lot.
📘 What you need to know
Gravitational potential energy (GPE) is the energy stored in a mass because of its position in a gravitational field
Lift a mass up and it gains GPE; let it fall and it loses GPE
Near the Earth’s surface: ΔEp = mgΔh
g is the gravitational field strength, about 9.8 N kg−1 at the surface
Only the change in height matters — you can pick any level as your zero
GPE and height have a linear relationship — the graph is a straight line
This equation only works in a uniform field (close to the surface)
The gravitational potential energy equation
Close to the Earth’s surface, where gravity is effectively uniform, the change in gravitational potential energy when a mass moves up or down is:
Change in gravitational potential energy
ΔEp = mgΔh
where ΔEp is the change in GPE in joules (J), m is the mass in kilograms (kg), g is the gravitational field strength (about 9.8 N kg−1 near the surface), and Δh is the change in height in metres (m). It makes intuitive sense: heavier objects and bigger lifts both store more energy, so mass and height both multiply in.
Raising a mass m through a height Δh, against gravity g, stores gravitational potential energy ΔEp = mgΔh.
WE 1
A 12 kg box is lifted 2.5 m from the floor onto a shelf. Calculate the gain in gravitational potential energy. (Take g = 9.8 N kg−1.)
Step 1 — write the equation
ΔEp = mgΔh
Step 2 — substituteΔEp = 12 × 9.8 × 2.5ΔEp = 294 JThat 294 J is stored while the box sits on the shelf — knock it off and it converts to kinetic energy on the way down.
Only the change in height matters
Notice the equation uses Δh — a change in height, not an absolute height. That’s because there’s no universal “zero” for GPE. We usually take ground level as zero for convenience, but you’re free to choose any level, as long as you’re consistent. What always counts is how far up or down the mass moves.
Choosing your zero is like choosing where to start a stopwatch — it doesn’t change the physics, only the numbers you write down. Set your zero at the floor, the tabletop, or the bottom of a cliff; the change in GPE between two points comes out exactly the same either way. Pick whatever makes the arithmetic easiest.
WE 2
A 0.45 kg ball is lifted straight up and gains 66 J of gravitational potential energy. How high was it raised? (Take g = 9.8 N kg−1.)
Step 1 — rearrange ΔEp = mgΔh for Δh
Δh = ΔEp ÷ (mg)
Step 2 — substituteΔh = 66 ÷ (0.45 × 9.8)Δh = 66 ÷ 4.41Δh = 15 m (2 s.f.)Rearrange first, then substitute — it keeps the working clean.
GPE against height: a straight line
Because m and g are both constant for a given object near the surface, ΔEp = mgΔh is just “GPE = constant × height”. Plot GPE against height and you get a straight line through the origin — a linear relationship. The steeper the line, the heavier the object (its gradient is mg).
GPE rises in direct proportion to height — a straight line through the origin. The gradient equals mg, so heavier objects give steeper lines.
WE 3
A 0.20 kg apple falls from a branch 8.0 m above the ground. Ignoring air resistance, use energy conservation to find its speed just before it lands. (Take g = 9.8 N kg−1.)
Step 1 — all the GPE lost becomes kinetic energy
mgΔh = ½mv²
Step 2 — mass cancels, rearrange for v
v = √(2gΔh)
Step 3 — substitutev = √(2 × 9.8 × 8.0) = √156.8v = 12.5 m s⁻¹ (3 s.f.)The GPE lost (0.20 × 9.8 × 8.0 = 15.7 J) all turns into kinetic energy, and the mass cancels out.
🛠️ Solving a GPE problem
Choose your zero height — usually the lowest point in the problem.
Find the change in height Δh between start and finish.
Use ΔEp = mgΔh, rearranging first if you’re solving for mass or height.
For falling problems, set the GPE lost equal to the kinetic energy gained (drag ignored).
Check g — use the value given, about 9.8 N kg−1 near the Earth’s surface.
💡 Top tips
It’s a change. The equation gives the change in GPE, so always use the change in height Δh, not an absolute height.
Pick a sensible zero. Any level works — choose the one that makes Δh easiest to read off.
Height is vertical. On a slope, Δh is the vertical rise, not the distance travelled along the slope (use h = length × sin θ).
Only near the surface. This mgΔh form assumes a uniform field — it doesn’t hold far out in space.
Quick recap: Gravitational potential energy is the energy stored in a mass because of its height in a gravitational field. Near the Earth’s surface, ΔEp = mgΔh, using the change in height. GPE is directly proportional to height (a straight-line graph with gradient mg), and when a mass falls, its lost GPE turns into kinetic energy.
⚠ Common mistakes
Using the distance along a slope instead of the vertical height — Δh is always vertical
Treating GPE as an absolute value rather than a change from a chosen zero
Confusing mass and weight — put mass (kg) into the equation, and g supplies the gravity
Applying mgΔhfar from the surface, where the field is no longer uniform
Forgetting that a falling object’s lost GPE becomes kinetic energy, not “disappears”
You’ve now got both halves of mechanical energy that swap during a fall: kinetic (moving) and gravitational potential (height). The third member of the family is stored in stretched and squashed springs — elastic potential energy — which is what we tackle next. After that, we can put all three together and watch mechanical energy stay perfectly conserved.
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