Changes in Energy Stores — Kinetic and Gravitational

In this lesson you will learn how to calculate kinetic energy and gravitational potential energy, and how these two stores are linked when objects rise or fall. These equations are essential for the AQA GCSE Combined Science Trilogy specification (8464), Section 6.1.

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Kinetic Energy

Kinetic energy is the energy stored in a moving object. The faster the object moves and the greater its mass, the more kinetic energy it has.

The Equation

$E_k = \frac{1}{2} m v^2$Ek​=21​mv2

Where:

• $E_k$Ek​ = kinetic energy in joules (J)
• $m$m = mass in kilograms (kg)
• $v$v = speed in metres per second (m/s)

Key Features of the Equation

• Kinetic energy is proportional to mass: double the mass → double the kinetic energy.
• Kinetic energy is proportional to the square of speed: double the speed → four times the kinetic energy.

Exam Tip: The $v^2$v2 relationship is tested frequently. If a car doubles its speed, its kinetic energy quadruples — this is why braking distance increases so sharply at higher speeds.

Worked Example 1

A cyclist and bike have a combined mass of 80 kg and travel at 10 m/s. Calculate the kinetic energy.

$E_k = \frac{1}{2} \times 80 \times 10^2 = \frac{1}{2} \times 80 \times 100 = 4000 \text{ J}$Ek​=21​×80×102=21​×80×100=4000 J

Worked Example 2

A ball has 50 J of kinetic energy and a mass of 0.4 kg. What is its speed?

$v = \sqrt{\frac{2 E_k}{m}} = \sqrt{\frac{2 \times 50}{0.4}} = \sqrt{250} \approx 15.8 \text{ m/s}$v=m2Ek​​​=0.42×50​​=250​≈15.8 m/s

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Gravitational Potential Energy

Gravitational potential energy (GPE) is the energy stored in an object because of its height above a reference point (usually the ground).

The Equation

$E_p = m g h$Ep​=mgh

Where:

• $E_p$Ep​ = gravitational potential energy in joules (J)
• $m$m = mass in kilograms (kg)
• $g$g = gravitational field strength in N/kg (on Earth $g = 9.8$g=9.8 N/kg; in calculations use the value given in the question, often 10 N/kg)
• $h$h = height in metres (m)

Worked Example 3

A 5 kg box is lifted to a shelf 2.5 m above the floor. Calculate the GPE gained. (Use $g = 9.8$g=9.8 N/kg.)

$E_p = 5 \times 9.8 \times 2.5 = 122.5 \text{ J}$Ep​=5×9.8×2.5=122.5 J

Worked Example 4

A 0.2 kg ball is 12 m above the ground. How much GPE does it have? (Use $g = 10$g=10 N/kg.)

$E_p = 0.2 \times 10 \times 12 = 24 \text{ J}$Ep​=0.2×10×12=24 J

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Linking Kinetic and Gravitational Potential Energy

When an object falls freely (ignoring air resistance), gravitational potential energy is transferred to kinetic energy. When an object rises, kinetic energy is transferred to gravitational potential energy.

In a closed system with no dissipation:

$E_p \text{ lost} = E_k \text{ gained}$Ep​ lost=Ek​ gained

$m g h = \frac{1}{2} m v^2$mgh=21​mv2

This equation lets you find the speed of a falling object or the maximum height reached by a projected object.

graph LR
    A["Gravitational potential\nstore"] -->|"Object falls"| B["Kinetic\nstore"]
    B -->|"Object rises"| A
    B -->|"On impact / friction"| C["Internal (thermal)\nstore of surroundings"]
    style C fill:#ffcccc,stroke:#cc0000

Worked Example 5

A 0.5 kg ball is dropped from 20 m. Assuming no air resistance, what is its speed just before hitting the ground? (Use $g = 10$g=10 N/kg.)

$mgh = \frac{1}{2}mv^2$mgh=21​mv2

Mass cancels:

$v = \sqrt{2gh} = \sqrt{2 \times 10 \times 20} = \sqrt{400} = 20 \text{ m/s}$v=2gh​=2×10×20​=400​=20 m/s

Worked Example 6

A stone is thrown vertically upward at 14 m/s. What maximum height does it reach? (Use $g = 10$g=10 N/kg, ignore air resistance.)

$h = \frac{v^2}{2g} = \frac{14^2}{2 \times 10} = \frac{196}{20} = 9.8 \text{ m}$h=2gv2​=2×10142​=20196​=9.8 m

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Effect on Braking Distance

Because $E_k \propto v^2$Ek​∝v2, doubling the speed of a vehicle quadruples the kinetic energy that the brakes must dissipate. This is why braking distance increases with the square of the speed.

Speed (m/s)	Relative Kinetic Energy	Braking Distance Factor
10	1×	1×
20	4×	4×
30	9×	9×

Exam Tip: AQA often combines energy calculations with real-world applications such as braking. Be prepared to explain why higher speeds are more dangerous using the $v^2$v2 relationship.

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Common Mistakes

Mistake	Correction
Forgetting to square the speed in $E_k = \frac{1}{2}mv^2$Ek​=21​mv2	Always square $v$v first, then multiply
Using the wrong units (g, km/h)	Convert mass to kg, speed to m/s, height to m
Using $g = 10$g=10 when the question gives $g = 9.8$g=9.8	Always read the question carefully and use the value provided
Forgetting mass cancels when equating $E_p$Ep​ and $E_k$Ek​	In free-fall problems you can cancel $m$m on both sides

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Summary

• Kinetic energy: $E_k = \frac{1}{2}mv^2$Ek​=21​mv2. Proportional to mass and to the square of speed.
• Gravitational potential energy: $E_p = mgh$Ep​=mgh. Proportional to mass, $g$g, and height.
• In free fall (no air resistance), $E_p$Ep​ lost = $E_k$Ek​ gained.
• The $v^2$v2 relationship explains why braking distance increases sharply with speed.
• These equations are given on the AQA equation sheet for Combined Science (8464).

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Extended Worked Examples

Worked Example 7 — Comparing Two Vehicles

A 1,200 kg car travels at 15 m/s. A 40,000 kg lorry travels at 8 m/s. Which has more kinetic energy?

Car:

$E_k = \tfrac{1}{2} \times 1200 \times 15^2 = 0.5 \times 1200 \times 225 = 135{,}000 \text{ J}$Ek​=21​×1200×152=0.5×1200×225=135,000 J

Lorry:

$E_k = \tfrac{1}{2} \times 40{,}000 \times 8^2 = 0.5 \times 40{,}000 \times 64 = 1{,}280{,}000 \text{ J}$Ek​=21​×40,000×82=0.5×40,000×64=1,280,000 J

The lorry has roughly 9.5 times the kinetic energy of the car, even though it travels slower. This is why heavy vehicles require much longer stopping distances and dissipate far more energy through their brakes.

Worked Example 8 — Roller Coaster Hill

A 500 kg roller-coaster carriage is at the top of a 35 m hill moving at 3 m/s. It descends to a point 5 m above the ground with a speed of 22 m/s. How much energy is dissipated? ($g = 10$g=10 N/kg.)

Initial energy: