Kinetic and Gravitational Potential Energy

This lesson covers the equations for kinetic energy (KE) and gravitational potential energy (GPE) as required by the Edexcel GCSE Physics specification (1PH0), Topic 3: Conservation of Energy. You need to be able to use, rearrange and apply these equations, and understand how energy is converted between these two stores.

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Kinetic Energy (KE)

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

The Equation

$KE = \frac{1}{2} m v^2$KE=21​mv2

Where:

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

Key Points About the KE Equation

• KE depends on speed squared — if you double the speed, KE increases by a factor of four (2² = 4).
• If you treble the speed, KE increases by a factor of nine (3² = 9).
• KE depends linearly on mass — double the mass, double the KE.

Exam Tip: The v² relationship is very commonly tested. If a question says "the car doubles its speed," the kinetic energy quadruples. Many students forget to square the speed — always check.

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Worked Examples — Kinetic Energy

Example 1: Calculating KE

Question: A car of mass 1200 kg is travelling at 15 m/s. Calculate its kinetic energy.

Solution:

$KE = \frac{1}{2} \times 1200 \times 15^2$KE=21​×1200×152 $KE = \frac{1}{2} \times 1200 \times 225$KE=21​×1200×225 $KE = 135{,}000 \text{ J} = 135 \text{ kJ}$KE=135,000 J=135 kJ

Example 2: Finding Speed from KE

Question: A 0.5 kg ball has 10 J of kinetic energy. Calculate its speed.

Solution:

Rearranging $KE = \frac{1}{2}mv^2$KE=21​mv2:

$v^2 = \frac{2 \times KE}{m} = \frac{2 \times 10}{0.5} = 40$v2=m2×KE​=0.52×10​=40

$v = \sqrt{40} = 6.3 \text{ m/s (2 s.f.)}$v=40​=6.3 m/s (2 s.f.)

Example 3: Finding Mass from KE

Question: A toy car has 0.36 J of kinetic energy and is moving at 3 m/s. What is its mass?

Solution:

$m = \frac{2 \times KE}{v^2} = \frac{2 \times 0.36}{3^2} = \frac{0.72}{9} = 0.08 \text{ kg}$m=v22×KE​=322×0.36​=90.72​=0.08 kg

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Gravitational Potential Energy (GPE)

Gravitational potential energy is the energy stored in an object because of its height above the ground (or some reference point). The higher the object and the greater its mass, the more GPE it has.

The Equation

$GPE = m g h$GPE=mgh

Where:

• GPE = gravitational potential energy (joules, J)
• m = mass (kilograms, kg)
• g = gravitational field strength (newtons per kilogram, N/kg) — on Earth, g = 9.8 N/kg (often approximated as 10 N/kg)
• h = height (metres, m)

Exam Tip: In Edexcel exams, g = 9.8 N/kg is usually given on the formula sheet, but you may sometimes be told to use g = 10 N/kg. Always read the question carefully and use the value given.

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Worked Examples — GPE

Example 1: Calculating GPE

Question: A book of mass 2 kg is placed on a shelf 1.5 m above the floor. Calculate the gravitational potential energy stored. (Use g = 9.8 N/kg)

Solution:

$GPE = m g h = 2 \times 9.8 \times 1.5 = 29.4 \text{ J}$GPE=mgh=2×9.8×1.5=29.4 J

Example 2: Finding Height from GPE

Question: A 5 kg object has 245 J of gravitational potential energy. How high is it above the ground? (g = 9.8 N/kg)

Solution:

$h = \frac{GPE}{m g} = \frac{245}{5 \times 9.8} = \frac{245}{49} = 5.0 \text{ m}$h=mgGPE​=5×9.8245​=49245​=5.0 m

Example 3: Finding Mass from GPE

Question: An object 4 m above the ground stores 156.8 J of GPE. What is its mass? (g = 9.8 N/kg)

Solution:

$m = \frac{GPE}{g h} = \frac{156.8}{9.8 \times 4} = \frac{156.8}{39.2} = 4.0 \text{ kg}$m=ghGPE​=9.8×4156.8​=39.2156.8​=4.0 kg

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Converting Between GPE and KE

When an object falls freely (ignoring air resistance), its GPE decreases and its KE increases. By the principle of conservation of energy, the decrease in GPE equals the increase in KE (if no energy is dissipated).

The Key Relationship

$\text{GPE lost} = \text{KE gained}$GPE lost=KE gained $m g h = \frac{1}{2} m v^2$mgh=21​mv2

flowchart TD
    A["Object at top\nMax GPE, zero KE"] -->|"Falls — GPE converts to KE"| B["Object halfway\nSome GPE, some KE"]
    B -->|"Continues falling"| C["Object at bottom\nZero GPE, max KE"]

Important Points

• At the top: maximum GPE, zero KE (if released from rest).
• At the bottom: zero GPE (taking the bottom as the reference point), maximum KE.
• At any point in between: the total energy (GPE + KE) stays the same (ignoring air resistance).
• In reality, some energy is always transferred to the thermal store via air resistance, so the speed at the bottom is slightly less than the calculated value.

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Multi-Step Problem: Finding Speed at the Bottom of a Ramp

Question: A 2 kg ball rolls down a ramp from a height of 3 m. Assuming no energy is dissipated, calculate the speed of the ball at the bottom of the ramp. (g = 9.8 N/kg)

Solution:

Step 1: Calculate the GPE at the top.

$GPE = m g h = 2 \times 9.8 \times 3 = 58.8 \text{ J}$GPE=mgh=2×9.8×3=58.8 J

Step 2: By conservation of energy, all GPE converts to KE at the bottom.

$KE = 58.8 \text{ J}$KE=58.8 J

Step 3: Use the KE equation to find speed.

$\frac{1}{2} m v^2 = 58.8$21​mv2=58.8 $v^2 = \frac{2 \times 58.8}{2} = 58.8$v2=22×58.8​=58.8 $v = \sqrt{58.8} = 7.7 \text{ m/s (2 s.f.)}$v=58.8​=7.7 m/s (2 s.f.)

Exam Tip: Notice that when you set mgh = ½mv², the mass cancels out. This means the speed at the bottom does not depend on mass (when air resistance is ignored). This is a very common exam trick — students often think heavier objects fall faster, but they reach the same speed.

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Rearranging the Equations — Summary

Find this	From KE equation	From GPE equation
Energy	$KE = \frac{1}{2}mv^2$KE=21​mv2	$GPE = mgh$GPE=mgh
Mass	$m = \frac{2 \times KE}{v^2}$m=v22×KE​	$m = \frac{GPE}{gh}$m=ghGPE​
Speed	$v = \sqrt{\frac{2 \times KE}{m}}$v=m2×KE​​	—
Height	—	$h = \frac{GPE}{mg}$h=mgGPE​

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Practice Problems