Potential Energy Calculator

The formula

PE = m × g × h

Mass in kilograms, gravitational acceleration in metres per second squared, height in metres. The result is in joules, the same unit as work and kinetic energy — which is the point: lifting something is doing work, and the work is stored as potential energy.

Worked examples

5 kg, 10 m, 9.8 → 490 J
2.5 kg → 245 J
10 kg → 980 J
5 m high → 245 J
20 m high → 980 J
g = 4.9 → 245 J
g = 19.6 → 980 J

Energy is linear in every input, so doubling any one of them doubles the answer. That makes the formula easy to reason about without a calculator: 1 kg lifted 1 m at 9.8 m/s² stores 9.8 J, and everything else is a multiple.

Putting 490 J in perspective

490 J is small. It would raise 1 kg by 50 m, or heat one kilogram of water by about 0.12 °C. A 60 W light bulb uses it in eight seconds. Climbing a flight of stairs costs a person of 70 kg about 2,000 J — four times this figure — which is why stair-climbing burns more than lifting a 5 kg box.

For pumped hydro, the arithmetic scales: a cubic metre of water is 1,000 kg, so lifting it 100 m stores about 980,000 J, roughly 0.27 kWh before losses. Real installations run at 70–85% round-trip efficiency, so a good fraction of that is lost coming back down.

Where the model stops

It assumes g is constant. Over a 10 m lift that is fine; over a satellite orbit it is not, because g falls with distance. On the Moon, entering 1.62 m/s² gives the correct answer for a 5 kg mass at 10 m — 81 J.

Height must be a vertical difference, not distance travelled. Carrying the mass up a ramp, on an escalator or along a slope stores the same energy as lifting it straight up, provided the vertical rise is the same; the path does not matter in this model because gravity is conservative.

Air resistance, friction and the speed of the lift are all absent. If you want the energy needed from a motor, divide by the mechanism’s efficiency — a winch at 80% needs 612 J of input to deliver 490 J of lift.

Frequently Asked Questions