Gravitational potential energy
Work, Energy & Power - OCR A-Level Physics
Key Definition
Gravitational potential energy
The energy stored in an object due to its position in a gravitational fieldA region of space in which a mass experiences a gravitational force.. The GPE store increases when an object moves up and decreases when it moves down. Near the Earth's surface, the change depends only on the vertical height change.
The energy stored in an object due to its position in a gravitational fieldA region of space in which a mass experiences a gravitational force.. The GPE store increases when an object moves up and decreases when it moves down. Near the Earth's surface, the change depends only on the vertical height change.
$$\Delta E_p = mg\Delta h$$
Derivation from work done
The GPE gained when lifting an object equals the work done against its weight. Weight is $W = mg$, so lifting through a vertical height $\Delta h$ gives $W_{\text{against gravity}} = mg \Delta h$, which becomes the GPE stored.
- The equation only applies near the Earth's surface, where $g \approx 9.81 \text{ m s}^{-2}$ is approximately constant. For large vertical distances, $g$ varies with altitude.
- Only the vertical change in height $\Delta h$ matters. The path taken and the distance along a slope are irrelevant.
- On a slope of angle $\theta$, if the object slides a distance $x$ along the slope, the vertical height change is $\Delta h = x\sin\theta$.
- The reference point where $h = 0$ can be chosen freely. Only changes in GPE are physically meaningful.
- Lift a $0.50 \text{ kg}$ book by $1.2 \text{ m}$: $\Delta E_p = 0.50 \times 9.81 \times 1.2 \approx 5.9 \text{ J}$.
Common Mistake
MEDIUM
Wrong: Using the distance along a slope as $\Delta h$ on inclined plane problems.
Right: $\Delta h$ is the vertical height change. For a slope of angle $\theta$ and length $x$, use $\Delta h = x\sin\theta$. The path along the slope is longer than the vertical drop.
Right: $\Delta h$ is the vertical height change. For a slope of angle $\theta$ and length $x$, use $\Delta h = x\sin\theta$. The path along the slope is longer than the vertical drop.
Worked Example [3 marks]
A $65 \text{ kg}$ climber walks $120 \text{ m}$ up a slope inclined at $25^\circ$ to the horizontal. Calculate the gain in gravitational potential energy. Use $g = 9.81 \text{ m s}^{-2}$.
Show Solution
1
Find the vertical height change
$\Delta h = x\sin\theta = 120 \times \sin 25^\circ = 120 \times 0.4226 = 50.7 \text{ m}$.
[1]2
$\Delta E_p = mg\Delta h = 65 \times 9.81 \times 50.7$.
[1]3
$\Delta E_p = 32\,300 \text{ J} \approx 3.2 \times 10^{4} \text{ J}$.
[1]Answer
$3.2 \times 10^{4} \text{ J}$
Examiner Tips and Tricks
- If the question gives a slope distance, your first step should be to resolve to find the vertical height with $\sin\theta$. A quick triangle sketch prevents the most common error.
- $\Delta E_p$ is independent of how the object got there. A direct lift and a long zigzag path that end at the same height give the same GPE change.