Potential Energy (Gravitational)

Multiply the mass in kilograms by the local factor g ≈ 9.81 m/s² and by the height h in metres: E_pot = m·g·h. The result is the positional energy in joules. A body of 50 kg at 4 m height has E_pot = 50·9.81·4 ≈ 1962 J. The height h is always understood relative to a freely chosen zero level, usually the ground or a table top. Only height differences are physically meaningful, since the absolute height depends on the chosen reference point. Insert g as 9.81 or, rounded, 10 m/s² depending on the task.

Because only energy differences act physically, not the absolute energy value. When a body falls from one height to another, only the height difference matters for the energy converted. Whether you place the zero level at the ground or at the table edge changes the numerical value of E_pot, but the difference between start and finish stays the same. Therefore you deliberately fix the zero level before each calculation and keep it. A common mistake is to interpret h as absolute height above sea level, although the problem means the height above the ground.

Without friction the total mechanical energy stays constant, so E_pot + E_kin = constant. During the fall the body loses height and thus potential energy, but gains exactly the same amount of kinetic energy. Setting m·g·h = ½·m·v² cancels the mass and gives the impact speed v = √(2·g·h). From 5 m height, for example, v = √(2·9.81·5) ≈ 9.9 m/s, independent of the mass. This energy approach is often faster than the equations of motion, because you do not even need to know the fall time.

The formula assumes a nearly uniform gravitational field, that is a constant local factor g. This holds well near the ground as long as the height is small compared with the Earth radius. If you rise to several hundred kilometres, for example into a satellite orbit, g decreases noticeably and you need the general gravitational potential energy, E_pot = −G·m·M/r. On other celestial bodies you must also insert their own local factor, on the Moon about 1.62 m/s². For everyday heights from metres to a few kilometres, however, E_pot = m·g·h gives very accurate values.

Divide the potential energy by the product of mass and local factor: h = E_pot/(m·g). The height is therefore the ratio of positional energy to weight, since m·g is exactly the weight F_G. Insert the energy in joules, the mass in kilograms and g in m/s², then h comes out in metres. Example: 1962 J at 50 kg give h = 1962/(50·9.81) = 4 m. If you want the mass instead, rearrange to m = E_pot/(g·h). Always check the result for plausibility; negative heights usually point to a sign or reference error.