Potential Energy Calculator
Gravitational energy from mass and height, with drop speed, both systems.
What it does
Lifted objects bank energy that gravity repays on the way down. A 5 kg toolbox on a 10 m scaffold holds 490.3 J before anyone touches it, enough to explain why small drops still hit hard. The tool above prices that storage from mass and height in either unit system.
The metric default multiplies straight through: 5 x 9.80665 x 10 = 490.3 J, with 361.7 ft-lb beside it for shop units. Freed in vacuum, the same drop lands at 14 m/s, which reads 50.4 km/h. Imperial mode mirrors the setup at 11 lb and 33 ft and lands within a few joules.
Two reference points anchor the page. OpenStax college physics defines the change as m g h and works 0.500 kg at 1.00 m to 4.90 J, which this tool reproduces digit for digit. The drop speed comes from energy conservation: v equals the square root of 2 g h, with mass canceling out.
Vacuum wording matters because air brakes real falls. A toolbox and a beach ball share the same vacuum speed from one height yet land worlds apart outdoors. Read the joule figure as stored energy and the speed as an upper bound that drag only trims.
How to use
- Pick metric (kg, m) or imperial (lb, ft).
- Type the mass of the object.
- Type height above the landing or reference level.
- Read joules first for energy math.
- Read drop speed as the vacuum upper bound.
Formula
- m
- Mass in kg (pounds convert behind the scenes).
- h
- Vertical height in m (feet convert behind the scenes).
- g
- Standard gravity 9.80665 m/s2.
Reference level is arbitrary; only height differences carry energy. Paths do not matter, so ramps and stairs with the same rise store the same joules.
Worked example
Example: 5 kg at 10 m
Metric defaults: mass 5 kg, height 10 m.
- Multiply: 5 x 9.80665 x 10 = 490.3 J.
- Convert: 490.3 x 0.73756 = 361.7 ft-lb.
- Speed: sqrt(2 x 9.80665 x 10) = 14 m/s.
- Pace: 14 x 3.6 = 50.4 km/h.
- Tool shows 490.3 J, 361.7 ft-lb, 14 m/s (50.4 km/h).
Reference table
| Height | Joules | Foot pounds |
|---|---|---|
| 1 m | 9.8 J | 7.2 ft-lb |
| 5 m | 49.0 J | 36.2 ft-lb |
| 10 m | 98.1 J | 72.3 ft-lb |
| 20 m | 196.1 J | 144.7 ft-lb |
Limits
- Near surface only; g drifts on tall mountains and orbits.
- Vacuum speeds; air drag lowers real impacts.
- Static storage only, no springs or pressure.
- Reference level choice shifts totals but never differences.
Common mistakes
- Using weight in pounds as mass in kilos.
- Measuring slope length instead of vertical height.
- Expecting real falls to match vacuum speed.
- Forgetting g in the product and quoting m x h.
- Mixing joules with foot pounds mid problem.
Try it yourself
Quick checks. Answers sit tucked under each question.
2 kg at 3 m?
2 x 9.80665 x 3 = 58.8 J.
0.5 kg at 1 m?
0.5 x 9.80665 x 1 = 4.9 J, the textbook check.
150 lb person climbs 10 ft?
68.04 kg x 9.80665 x 3.048 m = 2,033.7 J stored.
Questions
What is 5 kg at 10 m in joules?
490.3 J, or 361.7 ft-lb.
Why does mass cancel in drop speed?
Energy and inertia both scale with m, so it divides out.
Joules or foot pounds?
Same energy: multiply joules by 0.73756 for ft-lb.
Does the path up matter?
No. Only vertical rise sets the stored joules.
Why is real impact slower?
Air drag bleeds energy, badly for light wide objects.
What g does this use?
Standard 9.80665 m/s2 in both systems.