Kinetic Energy

This lesson covers the kinetic energy store and how to calculate kinetic energy using the equation $KE = \frac{1}{2}mv^2$KE=21​mv2, as required by the Edexcel GCSE Combined Science specification (1SC0). You will also explore how changes in mass and speed affect kinetic energy.

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What Is Kinetic Energy?

Kinetic energy (KE) is the energy stored in a moving object. Every object that is in motion has energy in its kinetic store.

• A moving car, a rolling ball and a flying bird all have kinetic energy.
• The faster an object moves, the more kinetic energy it has.
• The heavier an object is, the more kinetic energy it has at the same speed.

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The Kinetic Energy Equation

The kinetic energy of a moving object is calculated using:

$KE = \frac{1}{2}mv^2$KE=21​mv2

where:

• KE = kinetic energy in joules (J)
• m = mass in kilograms (kg)
• v = speed (or velocity) in metres per second (m/s)

Quantity	Unit	Symbol
Kinetic energy	joule	J
Mass	kilogram	kg
Speed	metres per second	m/s

Exam Tip: This equation is on the GCSE equation sheet, but you should still memorise it. Make sure you square the speed before multiplying by mass and dividing by 2. A very common error is to forget to square the speed.

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Worked Examples

Worked Example 1

Calculate the kinetic energy of a 1200 kg car travelling at 15 m/s.

1. Write the equation: $KE = \frac{1}{2}mv^2$KE=21​mv2
2. Substitute values: $KE = \frac{1}{2} \times 1200 \times 15^2$KE=21​×1200×152
3. Calculate $15^2 = 225$152=225
4. $KE = \frac{1}{2} \times 1200 \times 225 = 135\,000 \text{ J}$KE=21​×1200×225=135000 J
5. Answer: KE = 135 000 J (or 135 kJ)

Worked Example 2

A 0.5 kg ball has 10 J of kinetic energy. What is its speed?

1. Write the equation: $KE = \frac{1}{2}mv^2$KE=21​mv2
2. Rearrange for v: $v^2 = \frac{2 \times KE}{m}$v2=m2×KE​
3. Substitute: $v^2 = \frac{2 \times 10}{0.5} = 40$v2=0.52×10​=40
4. $v = \sqrt{40} = 6.32 \text{ m/s}$v=40​=6.32 m/s
5. Answer: v ≈ 6.3 m/s (2 s.f.)

Worked Example 3

A 70 kg runner has 2800 J of kinetic energy. Calculate the runner's speed.

1. $v^2 = \frac{2 \times KE}{m} = \frac{2 \times 2800}{70} = 80$v2=m2×KE​=702×2800​=80
2. $v = \sqrt{80} = 8.94 \text{ m/s}$v=80​=8.94 m/s
3. Answer: v ≈ 8.9 m/s (2 s.f.)

Exam Tip: When rearranging the equation to find speed, remember to take the square root at the end. Show each step clearly for full marks.

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The Effect of Doubling Speed

Because speed is squared in the kinetic energy equation, doubling the speed has a dramatic effect.

Speed (m/s)	Speed²	KE (for m = 1000 kg)
10	100	50 000 J
20	400	200 000 J
30	900	450 000 J
40	1600	800 000 J

• When the speed doubles, the kinetic energy increases by a factor of 4 (2² = 4).
• When the speed triples, the kinetic energy increases by a factor of 9 (3² = 9).

This is why high-speed collisions are so much more dangerous — the energy involved rises with the square of the speed.

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The Effect of Doubling Mass

Mass has a linear (directly proportional) relationship with kinetic energy.

• If the mass doubles (at the same speed), the kinetic energy doubles.
• If the mass triples, the kinetic energy triples.

flowchart LR
    subgraph "Effect on KE"
        A["Double mass"] -->|"×2"| B["KE doubles"]
        C["Double speed"] -->|"×4"| D["KE quadruples"]
    end

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Kinetic Energy and Braking

The kinetic energy equation explains why braking distance increases with the square of speed:

• To stop a moving vehicle, all its kinetic energy must be transferred to the thermal store of the brakes and tyres (via friction).
• Since $KE = \frac{1}{2}mv^2$KE=21​mv2, doubling the speed means the brakes must transfer four times as much energy.
• This means the braking distance is roughly proportional to speed squared.

Worked Example 4

A car of mass 1500 kg is travelling at 20 m/s. The braking force is 6000 N. Estimate the braking distance.

1. Calculate KE: $KE = \frac{1}{2} \times 1500 \times 20^2 = \frac{1}{2} \times 1500 \times 400 = 300\,000 \text{ J}$KE=21​×1500×202=21​×1500×400=300000 J
2. Work done by brakes = KE: $W = F \times d$W=F×d
3. Rearrange: $d = \frac{W}{F} = \frac{300\,000}{6000} = 50 \text{ m}$d=FW​=6000300000​=50 m
4. Answer: Braking distance ≈ 50 m

Exam Tip: Questions often combine kinetic energy with work done (W = Fd). Be ready to link these two equations together — this is a favourite examiner technique.

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Rearranging the Equation

You may need to rearrange the kinetic energy equation to find mass or speed.

To find	Rearranged equation
KE	$KE = \frac{1}{2}mv^2$KE=21​mv2
m	$m = \frac{2 \times KE}{v^2}$m=v22×KE​
v	$v = \sqrt{\frac{2 \times KE}{m}}$v=m2×KE​​

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Summary

• Kinetic energy is the energy in the store of a moving object.
• The equation is $KE = \frac{1}{2}mv^2$KE=21​mv2.
• KE depends on mass (linear) and speed squared (quadratic).
• Doubling the speed increases KE by a factor of four.
• Doubling the mass doubles the KE.
• KE is linked to braking distance through the work done equation.
• Always convert to SI units (kg, m/s) before calculating.

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Extended Worked Examples

Worked Example 5 — Comparing Two Cyclists

Cyclist A (mass 70 kg) rides at 8 m/s; Cyclist B (mass 90 kg) rides at 6 m/s. Which has more kinetic energy?

1. Cyclist A: $KE = \tfrac{1}{2} \times 70 \times 8^2 = \tfrac{1}{2} \times 70 \times 64 = 2240 \text{ J}$KE=21​×70×82=21​×70×64=2240 J
2. Cyclist B: $KE = \tfrac{1}{2} \times 90 \times 6^2 = \tfrac{1}{2} \times 90 \times 36 = 1620 \text{ J}$KE=21​×90×62=21​×90×36=1620 J
3. Cyclist A has more kinetic energy even though Cyclist B has more mass — because speed is squared in the equation.

Worked Example 6 — Falling Rock

A 2 kg rock falls from a cliff and reaches 15 m/s just before landing.