Kinetic, Elastic and Gravitational Potential Energy

Three stores can be calculated directly at GCSE: the kinetic store of a moving object, the gravitational potential store of a raised object, and the elastic potential store of a stretched spring.

Store	Equation	Recall or given?
Kinetic	$E_k=\frac{1}{2}m{v}^2$	Recall and apply
Gravitational potential	$E_p=mgh$	Recall and apply
Elastic potential	$E_e=\frac{1}{2}k{e}^2$	Given on the Physics equation sheet

You must recall and apply $E_k=\frac{1}{2}m{v}^2$ and $E_p=mgh$. The elastic equation $E_e=\frac{1}{2}k{e}^2$ is given on the equation sheet, but you still have to use it correctly.

Every quantity must be in SI units before you substitute: mass in kilograms, speed in metres per second, height and extension in metres. The energy then comes out in joules.

The kinetic store of any moving object depends on its mass and, crucially, on the square of its speed.

$E_k=\frac{1}{2}m{v}^2$

where $E_k$ is kinetic energy in joules (J), $m$ is mass in kilograms (kg) and $v$ is speed in metres per second (m/s).

Because the speed is squared, kinetic energy grows very quickly with speed. Doubling the speed multiplies the kinetic energy by four, since $2^2=4$. This is why a small increase in a car's speed causes a large increase in the energy that must be removed to stop it.

Square the speed before multiplying by the mass and the half. Multiplying first and squaring later is the most common slip in these questions.

Worked example 1 — a car of mass 1500 kg travels at 20 m/s. Find its kinetic energy.

$E_k=\frac{1}{2}m{v}^2=\frac{1}{2}\times 1500\times 20^2$

Square the speed first: $20^2=400$.

$E_k=\frac{1}{2}\times 1500\times 400=300000\text{J}=300\text{kJ}$

Worked example 2 — a golf ball of mass 45 g is hit so that it leaves the tee at 40 m/s. Find its kinetic energy.

First convert the mass to kilograms: $45\text{g}=0.045\text{kg}$.

$E_k=\frac{1}{2}\times 0.045\times 40^2=\frac{1}{2}\times 0.045\times 1600=36\text{J}$

Convert grams to kilograms (divide by 1000) before substituting. Leaving the mass in grams inflates the answer by a factor of 1000.

When an object is lifted, energy is transferred to the gravitational potential store of the object–Earth system. The higher and heavier the object, the more energy is stored.

$E_p=mgh$

where $E_p$ is gravitational potential energy in joules (J), $m$ is mass in kilograms (kg), $g$ is the gravitational field strength in newtons per kilogram (N/kg), and $h$ is the vertical height gained in metres (m).

On the Earth's surface $g$ is about 9.8 N/kg. The value of $g$ is given in a question when you need it, so read it carefully; some questions use 9.8 and older papers use 10.

$h$ is the vertical height gained, not the distance travelled along a slope or path. Only the change in height above the starting point matters.

Worked example 1 — a 2 kg book is lifted onto a shelf 1.5 m above the floor. Take $g=9.8$ N/kg. Find the gain in gravitational potential energy.

$E_p=mgh=2\times 9.8\times 1.5=29.4\text{J}$

Worked example 2 — a climber of mass 58 kg climbs a vertical cliff of height 3 m. Take $g=9.8$ N/kg. Find the gravitational potential energy gained.

$E_p=mgh=58\times 9.8\times 3=1705.2\text{J}\approx 1710\text{J}$

The answer rounds sensibly to 3 significant figures. Keep the full value if you are carrying it into a later step, and only round at the end.

Stretching or compressing a spring stores energy in its elastic potential store. As long as the spring is not stretched past its limit of proportionality, the energy stored is:

$E_e=\frac{1}{2}k{e}^2$

where $E_e$ is elastic potential energy in joules (J), $k$ is the spring constant in newtons per metre (N/m), and $e$ is the extension in metres (m).

This equation is given on the Physics equation sheet. It only applies while the spring obeys Hooke's law, that is, up to the limit of proportionality.

Worked example — a spring with a spring constant of 200 N/m is stretched by 15 cm. Find the elastic potential energy stored.

Convert the extension to metres: $15\text{cm}=0.15\text{m}$.

$E_e=\frac{1}{2}\times 200\times 0.15^2=\frac{1}{2}\times 200\times 0.0225=2.25\text{J}$

As with kinetic energy, square the extension before multiplying. Here $0.15^2=0.0225$, not 0.15.

1. Forgetting to square the speed or extension

Both $E_k=\frac{1}{2}m{v}^2$ and $E_e=\frac{1}{2}k{e}^2$ square a quantity. Work out $v^2$ or $e^2$ first, then multiply. Squaring at the wrong point is the single biggest source of lost marks.

2. Leaving mass in grams

Mass must be in kilograms. A 45 g ball is 0.045 kg. Substituting 45 gives an answer a thousand times too big.

3. Using the slope distance instead of the vertical height

In $E_p=mgh$, $h$ is the vertical height gained. If an object moves 5 m up a ramp but only rises 2 m vertically, use 2 m.

4. Confusing which equations must be recalled

$E_k=\frac{1}{2}m{v}^2$ and $E_p=mgh$ must be memorised. $E_e=\frac{1}{2}k{e}^2$ is given on the equation sheet. Do not waste time hunting for the ones you should already know.

5. Rounding too early

Carry the full value through multi-step calculations and round only the final answer to a sensible number of significant figures, usually two or three.