Calculate the kinetic energy of a 2.0\ kg ball moving at 6.0\ m/s. [2 marks]

Find the gravitational potential energy gained when a 3.0\ kg mass is raised 2.5\ m, g = 9.8\ N/kg. [2 marks]

CCEA CCEA style-style practice: Calculate the kinetic energy of a 1200 kg car travelling at 15 m/s.

Use the kinetic energy equation: E_k = 12 m v^2 = 12 × 1200 × 15^2 = 12 × 1200 × 225 = 135000\ J. So the kinetic energy is 135000 J (135 kJ). Markers reward the equation, squaring the speed (225), and the value 135000 J.

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How do we calculate the energy of a moving object and the energy of a raised object?

Kinetic energy from E = half m v squared, gravitational potential energy from E = mgh, and using conservation of energy to link them.

A CCEA GCSE Physics answer on calculating kinetic energy with half m v squared and gravitational potential energy with mgh, and using conservation of energy to link the two stores in falling objects.

Generated by Claude Opus 4.811 min answerUpdated 2026-06-14

Reviewed by: AI editorial process; not yet individually human-reviewed

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Examples in context
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What this dot point is asking

CCEA wants you to calculate kinetic energy with $E_k = \tfrac{1}{2}mv^2$ and gravitational potential energy with $E_p = mgh$ , and to use conservation of energy to link the two, for example finding the speed of a falling object.

The answer

Kinetic energy

This is why braking distance rises sharply with speed: a faster car has much more kinetic energy to remove.

Gravitational potential energy

Lifting an object transfers energy from your chemical store (via work done) to its gravitational potential store.

Linking the two with conservation of energy

Worked example: a roller-coaster drop

Speed at the bottom of a drop

A roller-coaster car of mass $500\ \text{kg}$ starts almost at rest at the top of a $20\ \text{m}$ drop. Take $g = 9.8\ \text{N/kg}$ and ignore friction. Find the speed at the bottom.

step Find the gravitational potential energy lost

E_p = mgh = 500 \times 9.8 \times 20 = 98000\ \text{J}.

step Equate it to the kinetic energy gained

\tfrac{1}{2}mv^2 = 98000 \Rightarrow \tfrac{1}{2} \times 500 \times v^2 = 98000.

step Solve for the speed

250 v^2 = 98000 \Rightarrow v^2 = 392 \Rightarrow v = \sqrt{392} = 19.8\ \text{m/s} \approx 20\ \text{m/s}.

So the car reaches about $20\ \text{m/s}$ at the bottom of the drop.

Examples in context

Example 1. A pendulum: At the top of its swing a pendulum bob has maximum gravitational potential energy and zero kinetic energy; at the bottom it has maximum kinetic energy and minimum potential energy. The two stores swap back and forth, conserving total energy.
Example 2. A cyclist freewheeling downhill: Gravitational potential energy is transferred to a kinetic store, so the cyclist speeds up without pedalling, though some energy is dissipated by friction and air resistance.
Example 3. A diver on a high board: Standing on the board, the diver has gravitational potential energy $mgh$ . As they fall, this transfers to kinetic energy, so they hit the water moving fast. Doubling the height of the board doubles the potential energy and so doubles the kinetic energy on entry, but only increases the entry speed by a factor of $\sqrt{2}$ , because kinetic energy depends on speed squared.

These two equations are among the most common in CCEA calculations, and they are often combined. A typical exam chain is: find the gravitational potential energy at the top, set it equal to the kinetic energy at the bottom, then solve for the speed. Always keep mass in kilograms, height in metres and speed in metres per second so the energy comes out in joules.

Try this

Q1. Calculate the kinetic energy of a $2.0\ \text{kg}$ ball moving at $6.0\ \text{m/s}$ . [2 marks]

Cue. $E_k = \tfrac{1}{2} \times 2.0 \times 6.0^2 = \tfrac{1}{2} \times 2.0 \times 36 = 36\ \text{J}$ .

Q2. Find the gravitational potential energy gained when a $3.0\ \text{kg}$ mass is raised $2.5\ \text{m}$ , $g = 9.8\ \text{N/kg}$ . [2 marks]

Cue. $E_p = mgh = 3.0 \times 9.8 \times 2.5 = 73.5\ \text{J}$ .

Q3. Why does doubling a car's speed have such a big effect on its kinetic energy? [2 marks]

Cue. Kinetic energy depends on speed squared, so doubling the speed gives four times the energy.

Exam-style practice questions

Practice questions written in the style of CCEA exam questions on this dot point, with worked answer explainers. The year tag is the paper they imitate, not the source.

CCEA style3 marksCalculate the kinetic energy of a 1200 kg car travelling at 15 m/s.

Show worked answer →

Use the kinetic energy equation:

$E_k = \tfrac{1}{2} m v^2 = \tfrac{1}{2} \times 1200 \times 15^2 = \tfrac{1}{2} \times 1200 \times 225 = 135000\ \text{J}.$

So the kinetic energy is 135000 J (135 kJ).

Markers reward the equation, squaring the speed (225), and the value 135000 J.

CCEA style4 marksA 0.50 kg ball is dropped from a height of 1.8 m. Using conservation of energy, calculate its speed just before it hits the ground. Take g as 9.8 N/kg and ignore air resistance.

Show worked answer →

Gravitational potential energy lost equals kinetic energy gained:

$mgh = \tfrac{1}{2} m v^2.$

The mass cancels, so $gh = \tfrac{1}{2} v^2$ , giving $v = \sqrt{2gh}$ .

$v = \sqrt{2 \times 9.8 \times 1.8} = \sqrt{35.28} = 5.9\ \text{m/s}.$

Markers reward equating $mgh$ with kinetic energy, cancelling mass (or finding the energy then the speed), and the value 5.9 m/s.

Related dot points

Sources & how we know this

CCEA GCSE Physics specification — CCEA (2017)