Calculating Gravitational Potential Energy

Gravitational potential energy (GPE) is the energy stored in an object due to its position in a gravitational field. It increases when the object moves higher.

The equation

$\Delta E_p = mg\Delta h$

Symbol	Quantity	Unit
$\Delta E_p$	Change in gravitational potential energy	J
$m$	Mass	kg
$g$	Gravitational field strength	N/kg (or m/s$^2$)
$\Delta h$	Change in height	m

Use $g = 9.8$ N/kg (or the value given in the question).

Worked example

A 60 kg student climbs stairs through a vertical height of 3.5 m. Calculate the gain in gravitational potential energy. Use $g = 9.8$ N/kg.

$\Delta E_p = 60 \times 9.8 \times 3.5 = 2058\text{ J} \approx 2100\text{ J}$

Linking kinetic and gravitational potential energy

When an object falls freely (no air resistance), its gravitational potential energy converts to kinetic energy. At any point:

$\text{loss in GPE} = \text{gain in KE}$

$mg\Delta h = \frac{1}{2}mv^2$

This allows you to find the speed at any height without knowing the time or acceleration.

Worked example: falling object

A 2.0 kg object falls from rest through a height of 5.0 m. Assuming no air resistance, find its speed at the bottom. Use $g = 9.8$ N/kg.

$mg\Delta h = \frac{1}{2}mv^2$ $9.8 \times 5.0 = \frac{1}{2}v^2$ $v = \sqrt{2 \times 9.8 \times 5.0} = \sqrt{98} = 9.9\text{ m/s}$

Common errors and how to correct them

Using the total height instead of the change in height. $\Delta h$ is the vertical distance the object moves, not its height above the ground (unless it falls to ground level).

Forgetting that height must be vertical. If a ball rolls down a slope of length 10 m at an angle of 30 degrees, the vertical height is $10 \times \sin 30 = 5.0$ m, not 10 m.

How to apply this in an exam

Check whether $g$ is given in the question or whether you should use 9.8 N/kg. Show the equation and substitution. When linking GPE and KE, state the assumption “no air resistance” or “no friction.”