Kinetic and Potential Energy

Three energy stores turn up again and again in mechanics problems, and this lesson brings them together. Kinetic energy is the energy of anything that is moving; gravitational potential energy is the energy of anything that has been lifted up; and elastic potential energy is the energy of anything that has been stretched or squashed. Each has its own equation, and — crucially — each swaps its energy with the others as objects fall, rise, launch and bounce. A rollercoaster, a pendulum, a bouncing ball and a catapult all rely on energy flowing between these three stores. This lesson, part of Topic P5 (Energy) of OCR Gateway Combined Science A, introduces $E_k = \tfrac{1}{2}mv^2$ , $E_p = mgh$ and $E_e = \tfrac{1}{2}ke^2$ , works through calculations with each, and shows how to solve transfer problems that link them.

By the end of this lesson you should be able to use and rearrange the three energy equations, explain why doubling a speed or an extension quadruples the stored energy, and solve transfer problems using $mgh = \tfrac{1}{2}mv^2$ and $\tfrac{1}{2}ke^2 = \tfrac{1}{2}mv^2$ .

This lesson is calculation-heavy AO2 (rearranging $E_k = \tfrac{1}{2}mv^2$ , $E_p = mgh$ and $E_e = \tfrac{1}{2}ke^2$ and solving transfer problems), with AO2 reasoning about why the squared term makes energy quadruple.

Kinetic Energy

Kinetic energy is the energy an object has because it is moving. Any moving object — whatever its direction — carries energy in its kinetic store. The amount depends on two things: the mass of the object (a heavier object at the same speed has more) and its speed (a faster object has more, and speed matters far more than mass). Kinetic energy, like all energy, is measured in joules (J).

The kinetic energy of a moving object is given by:

$E_k = \tfrac{1}{2}mv^2$

where $E_k$ is the kinetic energy (in joules, $\text{J}$ ), $m$ is the mass (in kilograms, $\text{kg}$ ) and $v$ is the speed (in metres per second, $\text{m/s}$ ). Two features catch people out: the $\tfrac{1}{2}$ at the front (you must halve, or the answer is twice too big) and the $v^2$ (you square the speed only, not the mass). Work out $v^2$ first, then multiply by the mass, then halve.

Worked Example 1 — Kinetic energy of a car

Calculate the kinetic energy of a $1200 \text{ kg}$ car travelling at $20 \text{ m/s}$ .

Step 1 — write the equation: $E_k = \tfrac{1}{2}mv^2$ .

Step 2 — square the speed first: $v^2 = 20^2 = 400$ .

Step 3 — substitute: $E_k = \tfrac{1}{2} \times 1200 \times 400$ .

Step 4 — calculate: $E_k = 0.5 \times 1200 \times 400 = 240\,000 \text{ J}$ .

Answer: the car has $240\,000 \text{ J}$ (= $240 \text{ kJ}$ ) of kinetic energy.

Worked Example 2 — Finding the speed

A ball of mass $0.5 \text{ kg}$ has $25 \text{ J}$ of kinetic energy. Calculate its speed.

To find the speed, rearrange $E_k = \tfrac{1}{2}mv^2$ to $v = \sqrt{\dfrac{2E_k}{m}}$ (multiply $E_k$ by 2, divide by the mass, then take the square root).

Step 1 — substitute: $v = \sqrt{\dfrac{2 \times 25}{0.5}} = \sqrt{\dfrac{50}{0.5}}$ .

Step 2 — simplify inside the root: $\dfrac{50}{0.5} = 100$ .

Step 3 — take the square root: $v = \sqrt{100} = 10 \text{ m/s}$ .

Answer: the ball is moving at $10 \text{ m/s}$ .

Exam Tip: Always square the speed first, then multiply by the mass, then halve. To find a speed from kinetic energy, do not forget the final square root — it is the step most often missed. Keep the mass in kilograms, or the answer will not come out in joules.

The effect of doubling the speed

Because the speed is squared, changing it has a dramatic effect on the kinetic energy — one of the most important ideas in the topic. If you double the speed, the kinetic energy is multiplied by $2^2 = 4$ ; if you treble it, the energy grows $3^2 = 9$ times. In general, multiplying the speed by a factor multiplies the kinetic energy by the square of that factor.

Speed / $\text{m/s}$	$v^2$	$E_k = \tfrac{1}{2}mv^2$ for a $1000 \text{ kg}$ car / $\text{J}$
10	100	50 000
20	400	200 000
30	900	450 000

Going from $10 \text{ m/s}$ to $20 \text{ m/s}$ doubles the speed but makes the kinetic energy four times bigger. This has a vital real-world consequence you meet again with stopping distances: because a faster car has so much more kinetic energy, its brakes must do far more work to stop it, so the braking distance grows much faster than the speed.

Exam Tip: If the speed is multiplied by a number, the kinetic energy is multiplied by the square of that number: ×2 speed → ×4 energy, ×3 speed → ×9 energy. A common misconception is that doubling the speed doubles the kinetic energy — it actually quadruples it.

Gravitational Potential Energy

Gravitational potential energy (GPE) is the energy stored in an object because of its height above a chosen level (usually the ground) in a gravitational field. Lifting an object higher does work against gravity and fills its gravitational potential store; letting it fall empties that store again. The higher you raise something, and the heavier it is, the more gravitational potential energy it gains.

The gravitational potential energy of an object is given by:

$E_p = mgh$

where $E_p$ is the gravitational potential energy (in joules, $\text{J}$ ), $m$ is the mass (in kilograms, $\text{kg}$ ), $g$ is the gravitational field strength (equal to $9.8 \text{ N/kg}$ on Earth) and $h$ is the height (in metres, $\text{m}$ ). It rearranges to $m = \dfrac{E_p}{gh}$ and $h = \dfrac{E_p}{mg}$ .

Worked Example 3 — Lifting a box

A box of mass $20 \text{ kg}$ is lifted onto a shelf $2.5 \text{ m}$ high. Calculate the gain in gravitational potential energy. ( $g = 9.8 \text{ N/kg}$ .)

Step 1 — write the equation: $E_p = mgh$ .

Step 2 — substitute: $E_p = 20 \times 9.8 \times 2.5$ .

Step 3 — calculate: $E_p = 490 \text{ J}$ .

Answer: the box gains $490 \text{ J}$ of gravitational potential energy.

Exam Tip: Keep the mass in kilograms and the height in metres, and use $g = 9.8 \text{ N/kg}$ on Earth. The height $h$ is the vertical rise — for an object pulled up a slope, it is the height gained, not the distance along the slope.

Transfer between gravitational and kinetic stores

The real power of gravitational potential energy is what happens when an object falls. As it drops, its height falls so its gravitational potential store empties, while at the same time it speeds up so its kinetic store fills. Energy is transferred mechanically (gravity does the work) from the gravitational potential store to the kinetic store. This is what happens for a falling object, a swinging pendulum and a rollercoaster racing down a hill.

If we ignore air resistance and friction, no energy is dissipated, so the gravitational potential energy lost exactly equals the kinetic energy gained:

$mgh = \tfrac{1}{2}mv^2$

Because the mass $m$ appears on both sides, it cancels, leaving $gh = \tfrac{1}{2}v^2$ , so $v = \sqrt{2gh}$ . The speed of a freely falling object therefore depends only on $g$ and the height fallen — not on its mass.

Worked Example 4 — Speed of a falling stone

A stone is dropped from a height of $5 \text{ m}$ . Calculate its speed just before it hits the ground, ignoring air resistance. ( $g = 9.8 \text{ N/kg}$ .)

Step 1 — equate the stores: $mgh = \tfrac{1}{2}mv^2$ ; the mass cancels, giving $v = \sqrt{2gh}$ .

Step 2 — substitute: $v = \sqrt{2 \times 9.8 \times 5} = \sqrt{98}$ .

Step 3 — calculate: $v = 9.9 \text{ m/s}$ (to 2 significant figures).

Answer: the stone is travelling at about $9.9 \text{ m/s}$ just before impact.

Kinetic and Potential Energy

Kinetic and Potential Energy

Kinetic Energy

Worked Example 1 — Kinetic energy of a car

Worked Example 2 — Finding the speed

The effect of doubling the speed

Gravitational Potential Energy

Worked Example 3 — Lifting a box

Transfer between gravitational and kinetic stores

Worked Example 4 — Speed of a falling stone

More in Combined Science