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Power & Efficiency

Key idea: Define power as an energy-transfer rate, apply mechanical power as force times velocity component, and solve energy and power efficiency problems.

  • GCE A-Level H2 Physics 2027
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Learning objectives

  • Apply power, mechanical power and efficiency relationships.

1. Power

Power is the rate of energy transfer:

P = (Δ E)/(Δ t) = W/(Δ t)

Its SI unit is the watt:

1 W = 1 J s⁻¹

This quotient gives average power over the stated interval. Instantaneous power is the transfer rate at one instant.

2. Mechanical power

For a force acting on a body moving with velocity vector v:

P = vector F · vector v = Fv cos θ

where θ is the angle between the force and velocity.

  • Along the velocity: P = Fv.
  • Opposite the velocity: power delivered by that force is negative.
  • Perpendicular to the velocity: P = 0.

The F in this expression is the force whose power is required. At constant speed, the resultant force is zero, but a driving force may deliver positive power while resistance removes energy at the same rate.

3. Efficiency

Efficiency is the fraction of total input transferred to the intended useful output:

η = (E_(useful output))/(E_(total input))

For a steady process over the same interval:

η = (P_(useful output))/(P_(total input))

Efficiency is dimensionless and lies from 0 to 1, or from 0% to 100%.

Useful and dissipated outputs from an energy transferA device receives 100 joules of input energy. It transfers 72 joules to the useful output and 28 joules to internal energy stores in the device and surroundings, giving an efficiency of 72 percent.Device100 J totalInput: 100 JUseful: 72 JDissipated: 28 Jefficiency = 72 J ÷ 100 J = 0.72 = 72%
Scroll diagram horizontally to read all labels.
Efficiency compares useful output with total input. Dissipated output is still energy—it has been transferred to less useful stores.
Use like with like

Compare energy with energy or power with power. Do not divide an energy in joules by a power in watts unless a time interval is included.

4. Practical devices

No real device transfers all its input into the intended store. Motors heat due to electrical resistance and friction; vehicles transfer energy to internal stores through drag and rolling resistance.

Improving efficiency reduces the input required for a fixed useful output, but it does not remove conservation of energy: every output pathway remains part of the balance.

5. Common mistakes

  • Reversing the efficiency ratio.
  • Dividing useful output by dissipated output instead of total input.
  • Assuming constant power means constant force; from F = P/v, force falls as speed rises when power is fixed.
  • Using resultant force in P = Fv when the question asks for the engine’s power.
  • Forgetting the cosine factor when force and velocity are not parallel.

6. Worked Examples

Modelled example 1

Average power while lifting

Core

Problem

A motor raises a 120 kg load through 8.0 m in 12.0 s. Find the useful output power. Take g = 9.81 m s⁻².
Study the worked solution

  1. Identify the useful transfer

    Method

    The useful output is the load’s gravitational potential-energy increase.

    Reason

    The motor raises the load vertically through a height difference.

    Working

    Δ E_g = mgΔ h

  2. Calculate useful energy

    Method

    Δ E_g = 9.42 × 10³ J.

    Reason

    Multiply the load’s weight by its vertical displacement.

    Working

    Δ E_g = (120)(9.81)(8.0) = 9417.6 J

  3. Calculate average power

    Method

    P_useful = 785 W.

    Reason

    Average power is energy transferred per unit time.

    Working

    P_useful = 9417.6/12.0 = 784.8 W ≈ 785 W

Guided practice 2

Motor efficiency

About 4 min

Problem

The motor in the previous example supplies 785 W of useful lifting power while taking 1.10 kW of electrical input power. Find its efficiency.

Try this before viewing the solution

Unit: %

Hints

Hint 1: compare like quantities
Use power over power: η = P_useful/Pᵢₙₚᵤₜ.
View solution step by step

  1. Match the power units

    Method

    Pᵢₙₚᵤₜ = 1100 W.

    Reason

    The numerator and denominator must use consistent units.

    Working

    1.10 kW = 1100 W

  2. Calculate efficiency

    Method

    η = 0.714, or 71.4%.

    Reason

    Efficiency is useful output divided by total input.

    Working

    η = 785/1100 = 0.714 = 71.4%

Common misconception 3

Mechanical power at constant speed

Find and correct the mistake

Learner claim

A cyclist travels at 6.0 m s⁻¹ against a total opposing force of 81 N. A learner says constant speed means zero resultant force, so the cyclist’s mechanical power output is zero. Diagnose the claim and find the output.

Try this before viewing the solution

Unit: W

View solution step by step

  1. Distinguish resultant and driving force

    Method

    The cyclist’s forward driving force is 81 N.

    Reason

    At constant speed it balances, rather than removes, the opposing force.

    Working

    F_drive-81 = 0 ⇒ F_drive = 81 N

  2. Calculate driving power

    Method

    P = 486 W.

    Reason

    The driving force is parallel to velocity and transfers energy continuously.

    Working

    P = Fv = (81)(6.0) = 486 W

  3. Reconcile constant speed

    Method

    Resistance removes energy at the same rate.

    Reason

    Equal positive driving power and negative resistive power give zero net power and no kinetic-energy change.

    Working

    Pₙₑₜ = +486-486 = 0 W

Examiner practice 4

Power from an angled force

3 marks

Examination question

A 250 N force acts at 35° to the velocity of a body moving at 4.0 m s⁻¹. Find the instantaneous power delivered by this force. [3 marks]

Try this before viewing the solution

Unit: W

View solution step by step

  1. Select the parallel component

    1 mark

    Method

    F_∥ = 250 cos 35°.

    Reason

    Only the force component along velocity transfers energy instantaneously.

    Working

    F_∥ = F cos θ

  2. Apply mechanical power

    1 mark

    Method

    P = Fv cos θ.

    Reason

    Power is the scalar product of force and velocity.

    Working

    P = (250)(4.0) cos 35°

  3. Calculate the result

    1 mark

    Method

    P = 819 W, about 8.2 × 10² W.

    Reason

    The positive cosine indicates energy transfer in the direction of motion.

    Working

    P = 819 W

Challenge 5

Fixed power at two speeds

Minimal support

Independent transfer

A motor delivers a constant mechanical power of 20 kW along a vehicle’s direction of motion. Find the available driving force at 5.0 m s⁻¹ and at 10.0 m s⁻¹.

Try this before viewing the solution

Unit: N
Unit: N

Hints

Hint 1: hold power fixed
Use F = P/v separately at each speed.
View solution step by step

  1. Convert the fixed power

    Method

    P = 2.0 × 10⁴ W.

    Reason

    Watts combine directly with metres per second to give force in newtons.

    Working

    20 kW = 20,000 W

  2. Find force at the lower speed

    Method

    F = 4000 N at 5.0 m s⁻¹.

    Reason

    Rearrange the parallel-force relation P = Fv.

    Working

    F = 20,000/5.0 = 4000 N

  3. Find force at the higher speed

    Method

    F = 2000 N at 10.0 m s⁻¹.

    Reason

    Doubling speed at fixed power halves the available force.

    Working

    F = 20,000/10.0 = 2000 N

7. Mind Stretchers

Useful checks before moving on:

  • power has unit J s⁻¹ or W;
  • efficiency cannot exceed 1 or 100%;
  • at fixed power, doubling speed halves the available force component along motion.

Mind stretcher 1: Input power for an efficient liftExtension

A lift raises a 75 kg load through 10 m in 5.0 s. The motor is 60% efficient. Take g = 9.81 m s⁻².

Find the motor’s electrical input power.

Show answer

The useful lifting power is:

P_useful = (mgΔ h)/(Δ t) = (75)(9.81)(10)/5.0 = 1471.5 W

Since η = P_useful/Pᵢₙₚᵤₜ:

Pᵢₙₚᵤₜ = 1471.5/0.60 = 2452.5 W ≈ 2.45 kW

Continue with the next resource in this course.

Course and syllabus information
Course
GCE A-Level H2 Physics
Edition
GCE A-Level H2 Physics 2027