GCSE — physicsMedium6 min read

Kinetic and Gravitational Potential Energy

KE = ½mv²; GPE = mgh; energy conservation calculations

Tutor AI Team2026-02-25T05:01:07.746Z

# Kinetic and Gravitational Potential Energy — GCSE Physics

Kinetic energy and gravitational potential energy are two of the most important energy stores in physics. When objects move or are raised to a height, they store energy that can be calculated precisely. The principle of conservation of energy connects these stores, allowing us to solve problems about falling objects, roller coasters, pendulums, and more.

1. Kinetic Energy (KE)

Kinetic energy is the energy an object has due to its motion.

$\boxed{KE = \frac{1}{2}mv^2}$

Where:

$KE$ = kinetic energy (J)
$m$ = mass (kg)
$v$ = velocity/speed (m/s)

Key Points

KE depends on mass (linearly) and speed squared
Doubling the mass → doubles the KE
Doubling the speed → quadruples the KE (because $v$ is squared)
KE is always positive (or zero for stationary objects)
KE is a scalar quantity

Why Does Speed Matter More?

This has huge implications for road safety:

A car at 40 mph has four times the KE of a car at 20 mph
This means it needs four times the braking distance
This is why speeding is so dangerous

2. Gravitational Potential Energy (GPE)

GPE is the energy stored in an object due to its height above a reference point (usually the ground).

$\boxed{GPE = mgh}$

Where:

$GPE$ = gravitational potential energy (J)
$m$ = mass (kg)
$g$ = gravitational field strength (N/kg or m/s²)
$h$ = height above reference point (m)

Key Points

GPE depends on mass, gravitational field strength, and height
Doubling any one of these quantities doubles the GPE
GPE is relative — it depends on where you measure height from
When an object is lifted, work is done against gravity, increasing GPE

3. Conservation of Energy: KE ↔ GPE

When no energy is wasted (no friction or air resistance):

$\text{Decrease in GPE} = \text{Increase in KE}$

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

Notice that mass cancels: $gh = \frac{1}{2}v^2$ $v = \sqrt{2gh}$

This means that (without air resistance) the speed at the bottom does not depend on the mass — a heavy ball and a light ball dropped from the same height reach the same speed!

4. Rearranging the Equations

From KE

$m = \frac{2 \times KE}{v^2} \qquad v = \sqrt{\frac{2 \times KE}{m}}$

From GPE

$m = \frac{GPE}{gh} \qquad h = \frac{GPE}{mg} \qquad g = \frac{GPE}{mh}$

Worked Example: Calculating KE

Problem

Question: Calculate the kinetic energy of a 1200 kg car travelling at 30 m/s.

$KE = \frac{1}{2}mv^2 = \frac{1}{2} \times 1200 \times 30^2 = \frac{1}{2} \times 1200 \times 900 = 540{,}000 \text{ J} = 540 \text{ kJ}$

Solution

Worked Example: Finding Speed from KE

Problem

Question: An object of mass 5 kg has 1000 J of kinetic energy. Calculate its speed.

$v = \sqrt{\frac{2 \times KE}{m}} = \sqrt{\frac{2 \times 1000}{5}} = \sqrt{400} = 20 \text{ m/s}$

Solution

Worked Example: Calculating GPE

Problem

Question: A 3 kg ball is held 12 m above the ground. Calculate its GPE. ( $g = 9.8$ N/kg)

$GPE = mgh = 3 \times 9.8 \times 12 = 352.8 \text{ J}$

Solution

Worked Example: Falling Object (Conservation

Problem

Question: A 2 kg ball is dropped from 20 m. Calculate its speed just before hitting the ground. Ignore air resistance. ( $g = 9.8$ N/kg)

Solution

$GPE_{\text{top}} = KE_{\text{bottom}}$ $mgh = \frac{1}{2}mv^2$ $2 \times 9.8 \times 20 = \frac{1}{2} \times 2 \times v^2$ $392 = v^2$ $v = \sqrt{392} = 19.8 \text{ m/s}$

Worked Example: Roller Coaster

Problem

Question: A 500 kg roller coaster car starts from rest at the top of a 30 m hill. Ignoring friction, calculate its speed at the bottom.

$mgh = \frac{1}{2}mv^2$ $v = \sqrt{2gh} = \sqrt{2 \times 9.8 \times 30} = \sqrt{588} = 24.2 \text{ m/s}$

Note: the mass cancelled, so the speed doesn't depend on the mass of the car!

Solution

Worked Example: Pendulum

Problem

Question: A pendulum bob of mass 0.1 kg is pulled to a height of 0.05 m above its lowest point. Calculate its maximum speed at the lowest point.

$v = \sqrt{2gh} = \sqrt{2 \times 9.8 \times 0.05} = \sqrt{0.98} = 0.99 \text{ m/s}$

Solution

Worked Example: With Energy Loss

Problem

Question: A 0.5 kg ball is dropped from 8 m. Due to air resistance, it reaches the ground at 11 m/s instead of the expected speed. Calculate the energy wasted.

$GPE = mgh = 0.5 \times 9.8 \times 8 = 39.2 \text{ J}$ $KE = \frac{1}{2}mv^2 = \frac{1}{2} \times 0.5 \times 11^2 = 30.25 \text{ J}$ $\text{Energy wasted} = 39.2 - 30.25 = 8.95 \text{ J}$

This energy has been transferred to the thermal energy store of the ball and surrounding air.

Solution

6. KE and Braking Distance

When a car brakes, its kinetic energy is converted to thermal energy in the brakes:

$KE = \text{Work done by brakes} = F \times d$

$\frac{1}{2}mv^2 = F \times d$

$d = \frac{mv^2}{2F}$

Since KE ∝ $v^2$ , the braking distance is proportional to $v^2$ :

Double the speed → four times the braking distance
Triple the speed → nine times the braking distance

7. Practice Questions

1. Calculate the kinetic energy of a 70 kg sprinter running at 10 m/s. (2 marks)
1. A ball of mass 0.4 kg is thrown vertically upwards at 15 m/s. (a) Calculate its KE at launch. (2 marks) (b) Calculate the maximum height reached (ignore air resistance, $g = 9.8$ N/kg). (3 marks)
1. Explain why doubling the speed of a car more than doubles the braking distance. (3 marks)
1. A 200 g apple falls from a tree from a height of 3 m. Calculate its speed just before hitting the ground. (3 marks)
1. A roller coaster starts at rest at a height of 40 m and reaches the bottom of the hill at 25 m/s. The mass of the car is 600 kg. Calculate the energy wasted due to friction. (4 marks)
Answers

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Frequently Asked Questions

Does KE depend on direction?

No — KE is a scalar. It depends on speed (not velocity), so direction doesn't matter.

Why does mass cancel in falling object problems?

In $mgh = \frac{1}{2}mv^2$ , mass appears on both sides and cancels. This means all objects fall at the same rate in a vacuum, regardless of mass (Galileo's insight).

What if there's friction?

Then some GPE is converted to thermal energy instead of KE. The actual KE (and speed) will be less than calculated. Energy wasted = GPE – KE.

Summary

KE = $\frac{1}{2}mv^2$ — depends on mass and speed squared
GPE = $mgh$ — depends on mass, $g$ , and height
Conservation: loss in GPE = gain in KE (if no friction)
$v = \sqrt{2gh}$ for falling objects (mass cancels)
Doubling speed quadruples KE and braking distance
Energy wasted = total initial energy – useful final energy

Tags:

#gcse#physics#energy#kinetic-energy#gravitational-potential-energy#conservation

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# Kinetic and Gravitational Potential Energy — GCSE Physics

1. Kinetic Energy (KE)

Key Points

Why Does Speed Matter More?

2. Gravitational Potential Energy (GPE)

Key Points

3. Conservation of Energy: KE ↔ GPE

4. Rearranging the Equations

From KE

From GPE

Worked Example: Calculating KE

Worked Example: Finding Speed from KE

Worked Example: Calculating GPE

Worked Example: Falling Object (Conservation

Worked Example: Roller Coaster

Worked Example: Pendulum

Worked Example: With Energy Loss

6. KE and Braking Distance

7. Practice Questions

Answers

Frequently Asked Questions

Summary

Related Topics

Evolution and Natural Selection

Biodiversity and Conservation

Sampling and Field Investigations

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