How this instrument works
Raise a mass and you bank energy in the Earth–object system: PE = m·g·h. Each symbol earns its keep — more mass stores more, more height stores more, and g, the strength of gravity, sets the exchange rate at 9.80665 joules per kilogram per metre of lift. The relationship is strictly linear in every term, so doubling any one of the three doubles the result, no curves, no surprises.
The formula measures a difference, not an absolute — you choose the zero. Floor, ground, sea level: potential energy only ever appears in physics as a change between two heights, so the reference level is yours to pick and only the vertical rise h matters. That is also why the route is irrelevant. Haul a crate up a ramp, a ladder, or a spiral staircase, and the stored figure at the top is identical; the path changes your effort and your friction losses, never the m·g·h.
The instrument uses the conventional standard value g = 9.80665 m/s², fixed by international agreement in 1901 and published by NIST, rather than your local reading — real gravity varies by roughly ±0.3% across the globe. Treating g as constant also limits the formula to near-surface work: ten kilometres up it overstates the answer by only about 0.16%, but for orbits and escape trajectories the full Newtonian expression takes over.
- Enter the mass — 75 kg is preloaded, and the unit menu also takes pounds, grams, or tonnes.
- Enter the lift height in metres or feet — the vertical rise only, not the distance travelled.
- Read the potential energy in joules, or flip the output unit to kilojoules or kilocalories.
- Check the working block to see m·g·h substituted with your own numbers.
Worked example — a climber on a 10-metre wall
A 75 kg climber tops out on a 10-metre wall. PE = 75 × 9.80665 × 10 = 7,354.99 J — call it 7.35 kJ. In food units that is about 1.76 kcal, less than a bite of apple, which says more about how energy-dense food is than about the climb.
The body pays more than the physics: muscles turn food into mechanical work at roughly 20–25% efficiency, so those 1.76 kcal of stored height cost the climber about 7–9 kcal of metabolism. And the deposit is fully refundable — step off the top and all 7,354.99 J convert to motion, arriving at v = √(2gh) ≈ 14 m/s if nothing intervenes.
Questions
Why does the calculator use g = 9.80665 m/s²?
Because that is standard gravity — the exact conventional value fixed by the 3rd CGPM in 1901 and published by NIST, the one engineering tables assume. Your local g differs by at most about ±0.3% (Earth's shape and rotation do the varying), which is usually smaller than the uncertainty in your mass and height figures anyway.
Where should I measure the height from?
From whatever level you define as zero — potential energy is always relative to a reference you choose. Physics only ever uses differences, so pick the floor, the ground, or the tabletop, measure the vertical rise to the object's centre of mass, and the result is the stored energy relative to that level. A different zero shifts every value by the same constant and changes nothing physical.
Does the path I take to lift the object matter?
No — gravity is a conservative force, so only the net vertical rise counts. A ramp, a pulley, or a winding staircase all store exactly the same m·g·h at the top. The path changes the force needed along the way and the friction you fight, which cost extra effort but never enter the stored total.
How do joules relate to the calories on food labels?
One food calorie (kcal) is 4,184 joules, and the kcal output on this instrument makes the comparison direct. Lifting stores remarkably little in food terms: raising 75 kg by 10 m banks about 1.76 kcal, and a single 100 kcal snack could in principle hoist that same mass nearly 570 m straight up.
What happens to the potential energy when the object falls?
It converts to kinetic energy, joule for joule, when drag is negligible: m·g·h = ½·m·v², which rearranges to v = √(2gh) — independent of mass. From 10 m that means 14 m/s at the bottom. Machinery runs the exchange both ways: pumped-storage power stations bank surplus electricity as water hauled uphill, then reclaim it through turbines on demand.
Is PE = mgh exact at any height?
It is an approximation that treats g as constant, and near the surface it is excellent — at airliner cruising altitude the error is only about 0.16%, because g weakens slowly with distance from Earth's centre. For satellites, the Moon, or escape-velocity problems, the full form U = −GMm/r replaces it.