Potential energy calculator
Potential energy, height or mass from E = mgh.
Fill in the fields and the result will appear here automatically.
Estimate the energy change when lifting a load above a chosen zero level, or solve back for height or mass. E = mgh here assumes constant Earth gravity; it is not an orbital, spring or electric-field energy model. Compare two positions using their vertical height difference. Stored energy is not guaranteed useful output: losses and efficiency are not included.
How it works
Formula and logic
E = m · 9.80665 · h; h = E/(m · 9.80665); m = E/(9.80665 · h). g is standard gravity, not a measured local value. This constant-g model is for height changes small compared with Earth’s radius; direct energy calculation requires positive mass.
Example
5 kg × 9.80665 m/s² × 10 m = 490.3325 J, displayed as 490.33 J. Conversely, 490.3325 J and 5 kg give 10 m. An energy of 98.0665 J at 2 m gives 5 kg. Use unrounded energy for inverse checks.
Fields and units
- What to find — list option
- Mass — kg
- Height — m
- Energy — J
- Mass — kg
- Energy — J
- Height — m
How to use
- — Choose energy, height or mass; each mode needs the other two quantities.
- — Enter kg, m and J. For a lift, use vertical height difference, not stair or ramp length.
- — This interface accepts nonnegative heights and energies relative to your zero level. Solving for mass requires positive height.
Method and limitations
- Calculation method
- Formula and logic
- Data or methodology source
- OpenStax: potential energy with constant g and a chosen zero level JCGM/BIPM: conventional standard acceleration 9.80665 m/s²
- Limitation
- E = mgh with constant standard gravity. Negative reference heights, local gravity and efficiency are not inputs.
FAQ
Must height be above sea level?
Only if sea level is your chosen zero. For a load moved from a floor to a shelf, use the difference between their heights.
What does zero energy mean?
With positive mass at zero height, E = 0 relative to that reference. It does not mean the body has no other energy.
Can I estimate hoist power?
E is the ideal lifting work. Divide by elapsed time for average useful power; input power also depends on losses, which are not modelled here.
Can standard gravity be replaced by a local value?
This interface fixes g at 9.80665 m/s². Local gravity varies with location and altitude, so the calculation is not a geodetic measurement. If a problem specifies a different g, use E = mgh with that value separately.