Power and Efficiency

Key idea: Calculate O Level power and efficiency using energy, work and time; compare devices, convert units and solve linked lifting problems with clear working.

  • SEC G3 Physics 2027
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Learning objectives

  • Recognise kinetic, potential, nuclear and internal energy stores
  • Describe mechanical energy transfer by a force acting over a distance
  • Describe electrical energy transfer by an electric current
  • Describe energy transfer by heating due to a temperature difference
  • Describe energy transfer by electromagnetic and mechanical waves
  • Recall and apply Ek = ½mv² in new situations
  • Recall and apply Ep = mgh near the Earth's surface in new situations
  • State and apply the principle of conservation of energy
  • Recall and apply work done = force × distance moved in the force direction
  • Recall and apply power = energy transfer / time taken
  • Calculate efficiency as useful energy output / total energy input
  • Evaluate prescribed electricity-generation resources by efficiency, cost, reliability and environmental impact

1. Definition

Power

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

P = W/t = E/t

  • P is power in watts (W), W is work done in joules (J), E is energy transferred in joules (J), and t is time in seconds (s).
  • 1 W = 1 J s⁻¹
  • Power is a scalar quantity.

Review Energy calculations first if finding Eₖ, Δ Eₚ or dissipated energy still feels difficult.

Power is the gradient of an energy–time graph

Two straight lines on an energy-time graph: the steeper line represents higher power (faster energy transfer).

Scroll across the graph to read all labels.

Two straight lines on an energy-time graph: the steeper line represents higher power (faster energy transfer).Two straight lines on an energy-time graph: the steeper line represents higher power (faster energy transfer).
For constant power, energy transferred increases linearly with time: E = Pt. The gradient (rise/run) equals the power.
Open full-size graph
View figure data
Values for Power is the gradient of an energy–time graph
Time (s)100 W device200 W device
000
55001000

For the steeper line, gradient = (1000 - 0)/(5.0 - 0) = 200 J s⁻¹ = 200 W

The gradient unit confirms that the graph gives a rate of energy transfer.

2. Key Ideas

  • Always convert time to seconds before using P = E/t.
  • A device has higher power if it transfers the same energy in less time, or more energy in the same time.
  • If a load is lifted at constant speed, the useful work done is the gain in GPE: W_useful = mgh (or Fh if weight F is given).
  • Efficiency, η, compares the useful output with the total input. Use either energy values, or power values measured over the same interval, but do not mix them: η = E_useful/Eᵢₙₚᵤₜ × 100% η = P_useful/Pᵢₙₚᵤₜ × 100%
  • Efficiency cannot exceed 100%.
  • High power does not necessarily mean high efficiency: power describes how fast energy is transferred, while efficiency describes the useful fraction.

3. Detailed Explanations

Calculating power

To find power:

  1. Find the energy transferred E (or work done W).
  2. Convert time to seconds.
  3. Use P = E/t (or P = W/t).

For example, if 600 J is transferred in 3.0 s, P = 600/3.0 = 200 W

Efficiency

Energy is conserved, but not all input energy is transferred usefully. For a device:

total input energy = useful energy output + energy dissipated to the surroundings

The dissipated share commonly increases internal energy stores in the device and surroundings. Sound waves may also carry energy away, often because of friction, vibration or electrical resistance.

Efficiency of an energy transfer is:

η = E_useful/Eᵢₙₚᵤₜ × 100%

Sankey diagram for a sixty-percent efficient deviceA 100 joule input arrow splits into a 60 joule useful output arrow and a narrower 40 joule arrow for energy dissipated to the surroundings. The labelled widths are proportional to the energy transferred.Energy flow through a deviceArrow width represents the amount of energy transferred.100 J inputelectrical input60 J usefuloutput40 J dissipatedEfficiency = 60 J ÷ 100 J × 100% = 60%
Scroll diagram horizontally to read all labels.
The output arrows add to the input: 60 J useful + 40 J dissipated = 100 J input. Efficiency is the useful share of the input, so this device is 60% efficient.

4. Common Mistakes

  • Not converting minutes/hours to seconds.
  • Mixing up power (W) and energy (J).
  • Writing P = W/t but using W as “watt” instead of work done (J).
  • Getting efficiency > 100% (usually input/output swapped).
  • Forgetting to multiply by 100% when the answer is required as a percentage.
  • Assuming the device with the greatest power is automatically the most efficient.
  • Using useful energy over input power, or useful power over input energy. Both parts of an efficiency ratio must be the same type of quantity.

5. Exam Tips

  • Start with the definition: power = energy transfer per unit time.
  • Show unit conversions clearly (e.g. 10 min = 600 s).
  • When comparing two power outputs, check what is the same:
    • same work but different time → compare time
    • same time but different work → compare work
  • For efficiency questions, write input = useful + dissipated before substituting numbers.
  • Convert a percentage efficiency to a decimal before rearranging an equation: 20% = 0.20.
  • Sense-check the result: useful output cannot exceed total input, and efficiency cannot exceed 100%.

6. Worked Examples

Modelled example 1

Two runners climb the same hill

Core

Problem

Runner A has twice the mass of runner B. They climb the same vertical height in the same time. Whose average useful mechanical power output is higher? Explain.

Study the worked solution
  1. Relate climbing work to GPE

    Method

    Model the useful work as the gain in gravitational potential energy.

    Reason

    Each runner raises their mass through vertical height h.

    Working

    W = Δ Eₚ = mgh
  2. Compare the average powers

    Method

    Divide each energy transfer by the common time.

    Reason

    With g, h and t fixed, P = mgh/t is proportional to mass.

    Working

    P = mgh/t ⇒ P_A = 2P_B

Guided practice 2

Power of a motor

About 4 min

Problem

A motor does 60 000 J of work in 10 min. Calculate its average power.

Convert time before calculating the rate

Unit: s
Unit: W

Hints

Hint 1: convert the time
10 min = 10 × 60 s.
Hint 2: calculate the rate
Use P = W/t after the conversion.
View solution step by step
  1. Convert minutes to seconds

    Method

    Express the time in SI units.

    Reason

    A watt is a joule per second.

    Working

    10 min = 10 × 60 = 600 s
  2. Calculate average power

    Reason

    Average power is work done per unit time.

    Working

    P = W/t = (60 000)/600 = 100 W

Common misconception 3

Useful energy from input power

Find and correct the mistake

Learner response

An electric motor has input power 1.0 kW. In half an hour, 60% of the input energy is dissipated. A student states that the efficiency is therefore 60%. Locate the error, find the useful energy transferred, and state the actual efficiency.

Separate useful and dissipated fractions

Unit: J
Unit: %

View solution step by step
  1. Convert power and time

    Method

    Use watts and seconds.

    Reason

    E = Pt requires consistent SI units.

    Working

    1.0 kW = 1000 W, 0.5 h = 1800 s
  2. Find the total input energy

    Reason

    Power is the rate of input energy transfer.

    Working

    Eᵢₙ = Pt = (1000)(1800) = 1.8 × 10⁶ J
  3. Use the complementary fraction

    Method

    Subtract the dissipated fraction from 100%.

    Reason

    Input energy is split into useful and dissipated shares.

    Working

    E_useful = 0.40(1.8 × 10⁶) = 7.2 × 10⁵ J

Examiner practice 4

Energy dissipated by a lifting motor

4 marks

Examination question

An electric motor lifts a 10 N load through 5.0 m. The electrical energy input is 65 J. Calculate the energy dissipated by the motor and its efficiency. [4 marks]

Show useful energy, dissipation and efficiency

View solution step by step
  1. Calculate useful output energy

    1 mark

    Method

    Find the work done lifting the load.

    Reason

    The load’s gain in GPE is the useful output.

    Working

    E_useful = Fh = (10)(5.0) = 50 J
  2. Find dissipated energy

    1 mark

    Reason

    Input energy equals useful output plus dissipated energy.

    Working

    E_dissipated = 65-50 = 15 J
  3. Form the efficiency ratio

    1 mark

    Method

    Divide useful output by total input.

    Reason

    Efficiency measures the useful fraction of the input.

    Working

    η = 50/65 × 100%
  4. State the efficiency

    1 mark

    Reason

    The result must be below 100% because some energy is dissipated.

    Working

    η = 76.9% ≈ 77%

Challenge 5

Input power from efficiency

Minimal support

Reverse efficiency calculation

A crane raises a weight of 200 N through a vertical height of 8.0 m in 4.0 s at constant speed. The efficiency of the crane is 20%. Find the electrical input power.

Work from useful output back to total input

Unit: W
Unit: W

Hints

Hint 1: find useful output power
Use Pₒᵤₜ = Fh/t.
Hint 2: reverse the efficiency ratio
Write 0.20 = Pₒᵤₜ/Pᵢₙ.
View solution step by step
  1. Find useful output energy

    Method

    Calculate the work done on the load.

    Reason

    The vertical lift is the crane’s useful transfer.

    Working

    E_useful = Fh = (200)(8.0) = 1600 J
  2. Find useful output power

    Reason

    Power is energy transferred per unit time.

    Working

    Pₒᵤₜ = 1600/4.0 = 400 W
  3. Infer the input power

    Method

    Rearrange the efficiency equation for input power.

    Reason

    The 400 W useful output is only 20% of the electrical input.

    Working

    Pᵢₙ = 400/0.20 = 2000 W = 2.0 kW

7. Mind Stretchers

Mind stretcher 1: Finding efficiency from power and workExtension

A motor takes in power 500 W for 20 s. It lifts a 100 N load through 6.0 m at constant speed. Find the efficiency of the motor.

Show Answer

Total input energy: Eᵢₙ = Pt = (500)(20) = 10 000 J

Useful output energy: E_useful = Fh = (100)(6.0) = 600 J

efficiency = 600/(10 000) × 100% = 6.0%

Mind stretcher 2: Same work, different powerExtension

Two machines each do 3000 J of work. Machine A takes 12 s, machine B takes 7.5 s. Find the power of each machine and state which is more powerful.

Show Answer

P_A = 3000/12 = 250 W P_B = 3000/7.5 = 400 W

Machine B has the higher power because it does the same work in less time.

Continue with the next resource in this course.

Course and syllabus information
Course
SEC G3 Physics
Edition
SEC G3 Physics 2027