Work, Energy and Power

This lesson covers work done, kinetic energy, gravitational potential energy, power, and efficiency — as required by the Edexcel GCSE Physics specification (1PH0), Topic 1: Key Concepts of Physics. You need to be able to calculate each of these quantities and understand how energy is transferred in different situations.

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Work Done

When a force moves an object through a distance, work is done. Work done is a measure of the energy transferred.

work done = force × distance (in the direction of the force)

$W = Fs$W=Fs

Where:

• W = work done (J)
• F = force (N)
• s = distance moved in the direction of the force (m)

Key Points

• Work done is measured in joules (J). 1 J = 1 N m (1 joule = 1 newton metre).
• Work done = energy transferred. If 500 J of work is done, 500 J of energy is transferred.
• If the force and the movement are in different directions, only the component of the force in the direction of movement does work.
• If an object is not moving, no work is done — even if a force is applied.

Worked Examples

Example 1: A person pushes a box with a force of 50 N across the floor for 4 m. Calculate the work done.

W = Fs = 50 × 4 = 200 J

Example 2: A crane lifts a 200 kg load to a height of 15 m. Calculate the work done (g = 9.8 N/kg).

First find the force (weight): F = mg = 200 × 9.8 = 1960 N W = Fs = 1960 × 15 = 29 400 J (or 29.4 kJ)

Exam Tip: When calculating work done against gravity, the force is the weight (mg) and the distance is the vertical height. Do not use the distance along a ramp unless the question specifically asks about a force along the ramp.

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Kinetic Energy

Kinetic energy (KE) is the energy an object has because of its motion.

$KE = \frac{1}{2}mv^2$KE=21​mv2

Where:

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

Key Points

• Kinetic energy depends on speed squared — if the speed doubles, the kinetic energy quadruples.
• Kinetic energy is a scalar quantity.
• A stationary object has zero kinetic energy.

Worked Examples

Example 1: A 1500 kg car travels at 20 m/s. Calculate its kinetic energy.

KE = ½mv² = ½ × 1500 × 20² = ½ × 1500 × 400 = 300 000 J (300 kJ)

Example 2: The same car doubles its speed to 40 m/s. Calculate the new kinetic energy.

KE = ½mv² = ½ × 1500 × 40² = ½ × 1500 × 1600 = 1 200 000 J (1200 kJ)

Notice: doubling the speed (×2) quadrupled the kinetic energy (×4), from 300 kJ to 1200 kJ.

Example 3: A ball of mass 0.4 kg has 20 J of kinetic energy. What is its speed?

Rearrange: v² = 2 × KE ÷ m = 2 × 20 ÷ 0.4 = 100 v = √100 = 10 m/s

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Gravitational Potential Energy

Gravitational potential energy (GPE) is the energy an object has because of its height above the ground.

$GPE = mgh$GPE=mgh

Where:

• GPE = gravitational potential energy (J)
• m = mass (kg)
• g = gravitational field strength (N/kg) — 9.8 N/kg on Earth
• h = height (m)

Worked Examples

Example 1: A 5 kg ball is lifted to a height of 3 m. Calculate the GPE gained (g = 9.8 N/kg).

GPE = mgh = 5 × 9.8 × 3 = 147 J

Example 2: A 60 kg person climbs a 12 m high staircase. How much GPE do they gain?

GPE = mgh = 60 × 9.8 × 12 = 7056 J (7.1 kJ to 2 s.f.)

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Energy Transfers — Combining KE and GPE

In many situations, energy is transferred between KE and GPE.

Falling Objects (Ignoring Air Resistance)

When an object falls:

• GPE decreases (height decreases).
• KE increases (speed increases).
• GPE lost = KE gained (conservation of energy, no energy is lost to friction).

$mgh = \frac{1}{2}mv^2$mgh=21​mv2

Worked Example

A 2 kg ball is dropped from a height of 5 m. Calculate its speed just before it hits the ground (g = 9.8 N/kg, ignore air resistance).

GPE lost = KE gained: mgh = ½mv² 2 × 9.8 × 5 = ½ × 2 × v² 98 = v² v = √98 = 9.9 m/s

Exam Tip: When you set mgh = ½mv², notice that mass (m) cancels from both sides. This means the speed of a falling object does not depend on its mass (ignoring air resistance). This is a useful shortcut.

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Power

Power is the rate of doing work or the rate of energy transfer.

$P = \frac{W}{t} = \frac{E}{t}$P=tW​=tE​

Where:

• P = power (W)
• W = work done (J) or E = energy transferred (J)
• t = time (s)

Key Points

• Power is measured in watts (W). 1 watt = 1 joule per second (1 W = 1 J/s).
• 1 kilowatt (kW) = 1000 W.
• 1 megawatt (MW) = 1 000 000 W.
• A more powerful device transfers energy faster, not necessarily more total energy.

Worked Examples

Example 1: A motor does 6000 J of work in 30 s. Calculate its power.

P = W ÷ t = 6000 ÷ 30 = 200 W

Example 2: A 1500 W kettle boils water in 3 minutes. How much energy does it transfer?

Convert time: t = 3 × 60 = 180 s E = P × t = 1500 × 180 = 270 000 J (270 kJ)

Example 3: A person of mass 75 kg runs up a 10 m staircase in 12 s. Calculate their power output (g = 9.8 N/kg).

Work done = GPE gained = mgh = 75 × 9.8 × 10 = 7350 J Power = W ÷ t = 7350 ÷ 12 = 612.5 W

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Efficiency

Efficiency is the proportion of energy that is usefully transferred. No device is 100% efficient — some energy is always dissipated (wasted), usually as thermal energy.

$\text{efficiency} = \frac{\text{useful output energy transfer}}{\text{total input energy transfer}} \times 100\%$efficiency=total input energy transferuseful output energy transfer​×100%

Or in terms of power:

$\text{efficiency} = \frac{\text{useful output power}}{\text{total input power}} \times 100\%$efficiency=total input poweruseful output power​×100%

Worked Examples

Example 1: A motor has an input energy of 500 J and does 350 J of useful work. Calculate its efficiency.

Efficiency = (350 ÷ 500) × 100% = 70%

Example 2: A light bulb has a power rating of 60 W but only 9 W is emitted as light. Calculate the efficiency.

Efficiency = (9 ÷ 60) × 100% = 15%

The remaining 51 W is wasted as thermal energy (heat).