A.3.10—Gravitational potential energy

Near-Earth gravitational potential energy

For a height change $\Delta h$ in a uniform gravitational field,

$$\Delta E_{p,g}=mg\Delta h$$

Use the height change

Raising an object gives positive change in gravitational potential energy; lowering it gives negative change relative to the chosen reference.

Link to power

If height changes at constant speed, the rate of gravitational potential-energy gain is $mgv$, before accounting for efficiency or other transfers.

Worked example from local Question Bank row 37039

An object's weight is $6.10\times10^2\,\mathrm{N}$ and it rises vertically by $8.0\,\mathrm{m}$. Since $mg$ is its weight,

$$\Delta E_{p,g}=(6.10\times10^2)(8.0)=4.88\times10^3\,\mathrm{J}\approx4.9\,\mathrm{kJ}$$

The positive result means the gravitational potential-energy store increases.

Common trap

Use the local value of $g$ and the vertical height change, not the distance along a slope.