How to Calculate Potential Energy

PE = m g h

Mass in kilograms, g in m/s², height in meters, result in joules. This is the work you would do (slowly, no leftover KE) lifting the mass through height h in a uniform gravitational field, or the work gravity would do if the mass dropped through that same h.

The potential energy calculator multiplies m, g, and h in the browser. No account. It will not choose your zero of height, and it will not switch to ½kx² if the object is a spring.

h is a difference, not an altitude above sea level

You pick PE = 0 at some level. Then h is how far above that level the object is. A 15 kg crate on a 1.8 m bench, with the floor as zero, has

PE = (15 kg)(9.80 m/s²)(1.8 m) = 264.6 J

which you should report as 2.6 × 10² J if 1.8 m has two significant figures.

Move the zero to the bench: the crate now has PE = 0, and the floor is at h = −1.8 m, PE = −265 J. The change when the crate falls to the floor is still −265 J of PE (and, with no losses, +265 J of KE). Graders who see you obsess over sea level on a classroom bench will dock for missing the point.

A ramp does not change the formula. h is the vertical rise, not the distance along the slope. A 5.0 m ramp at 20° has h = 5.0 sin 20° ≈ 1.7 m, not 5.0 m.

g = 9.80 m/s² unless they say otherwise

Intro courses use 9.8 m/s² or 9.80 m/s² or even 10 m/s² “to make the arithmetic nice.” Use what the sheet specifies. On the Moon, g ≈ 1.6 m/s²; PE = mgh still holds locally with that g.

Do not put g = 9.80 into a problem that already gave weight in newtons and asked for PE. If weight W = mg is 147 N and h = 1.8 m, PE = W h = 265 J. Using 147 as mass would be the mass/weight swap from the force chapter.

g is not the object’s acceleration if the object is not in free fall. A crate being lifted at constant speed still changes PE by mgh; the crane does positive work, gravity does negative work, KE stays put.

A 15 kg crate lifted 1.8 m

You raise the crate from the floor to the bench at constant speed.

ΔPE = (15)(9.80)(1.8) = 264.6 J ≈ 265 J

The crane (or your arms) puts in about 265 J of work against gravity. If you then drop the crate, 265 J is the KE just before it hits, if air and the bench edge do not steal energy:

½ m v² = 265 J
v² = 2 × 265 / 15 ≈ 35.3
v ≈ 5.9 m/s

That is the energy-method check: mgh at the top equals ½mv² at the bottom when PE is zero on the floor. The kinetic energy calculator is the second half of that check. If a lab measures 5.1 m/s, about 20% of the energy went somewhere else (bounce, sound, you did not drop from a dead stop at 1.8 m).

When mgh is the wrong expression

Springs. Elastic PE = ½kx². A mass on a spring has both if it also changes height; you add them with a consistent zero.

Large heights. g drops with altitude. From the surface to low Earth orbit, mgh with surface g wildly overestimates the energy. Use ΔU = GMm(1/r₁ − 1/r₂) or the −GMm/r form.

Buoyancy and “effective g.” A submerged object’s gravitational PE in a simple model still uses mgh for the object’s mass; the fluid is a separate energy and force problem. Do not invent g = 9.80 − something unless the course defined an effective weight that way.

Chemical / nuclear / thermal. Those are not mgh. “Potential energy” in biology slides is not this calculator.

Mistakes that keep showing up

Using path length instead of vertical h. Using cm for h and still expecting joules (1.8 m is 180 cm; 15 × 9.80 × 180 is 26 000 - off by 100). Using 9.80 as a mass. Forgetting that PE can be negative below the zero. Adding mgh to ½mv² when both are already measured from inconsistent zeros and calling it “total energy” without stating the zeros.

Another: thinking a horizontal move at constant height changes gravitational PE. It does not. Work by gravity is zero on a horizontal displacement.

Multiply locally, then compare to KE

The potential energy calculator is mgh with the numbers you typed. Pair it with kinetic energy when the problem is a drop, a pendulum, or a roller-coaster hill. If the next question is the force of gravity, that is mg, which is F = ma with a = g, not a new energy formula.