Power and efficiency

Key idea: Power tells you how quickly energy is transferred. Efficiency compares useful output with total input and helps explain why real devices need cooling, fuel or larger power supplies.

GCE A-Level H1 Physics 2027

H1 Physics 8867 · Lesson 3 of 3

Check your understanding

By the end of this lesson, you should be able to

Define power as energy transferred per unit time.
Use mechanical power from the force component along velocity.
Calculate efficiency and explain practical consequences of dissipated energy.

Learn the idea

Big question: Why can a machine be powerful without being efficient, and efficient without being powerful?

Keep rate separate from fraction

Power is the rate of energy transfer: one watt is one joule per second. Efficiency is the fraction of input transferred to the intended useful output. They answer different questions.

A 2 kW device may transfer energy quickly but waste a large fraction; a small LED may be efficient while using little power. Always compare like with like: energy ratio or power ratio.

Check your understanding: Can an 80% efficient machine have lower useful power than a 40% efficient one?

Yes. The less efficient machine may have a much larger input power.

Find mechanical power from motion

During a short displacement, work is F cosθ times distance. Dividing by time gives P = Fv cosθ, using the force component along velocity.

Energy transferred to less useful stores is still conserved. Naming its effect—such as heating that requires cooling—turns a percentage into a physical explanation.

Check your understanding: A force is perpendicular to velocity. What instantaneous power does it transfer?

Zero, because the force component along velocity is zero.

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.

Key ideas

Efficiency has no unit and cannot exceed 1 or 100%.
Use energy ratios or power ratios consistently.
A device can be powerful but inefficient, or efficient but low-powered.

Relationships to know

P = ΔE/Δt
Pmechanical = Fv cos θ
efficiency = useful output / total input

Follow the reasoning

Worked example

Combine angled force, speed and efficiency

Question: A tractor pulls a load at constant 2.5 m s⁻¹ with a 6.0 kN force at 20° to its motion. Its engine input power is 20 kW. Find useful mechanical power and efficiency.

Step 1: Resolve along the motion
Why: Only the parallel component transfers mechanical energy to the moving load.
Working: Fparallel = 6000 cos20° = 5638 N.
Step 2: Find useful power
Why: Power is force component multiplied by speed.
Working: Puseful = Fv cosθ = 5638(2.5) = 14.1 kW.
Step 3: Form the correct ratio
Why: Efficiency is useful output divided by total input.
Working: η = 14.1/20.0 = 0.705.

Answer: Useful power is 14 kW and efficiency is about 70%.

Check: The useful output is below the input, so the efficiency is physically possible.

Now try it with support

Practise with support

A winch lifts a 75 kg load through 6.0 m in 12 s while taking 500 W. Find its useful power and efficiency. Use g = 9.81 m s⁻².

Hints

Useful energy gained is mgh.
Divide by time before forming the power ratio.

View the guided answer

Useful power = 75(9.81)(6.0)/12 = 368 W. Efficiency = 368/500 = 0.736 = 73.6%.

Your turn

Practise independently

A pump raises 600 kg of water by 8.0 m each minute while drawing 1.0 kW. Calculate its useful power and efficiency.

Check your answer

Useful energy each minute is mgh = 600(9.81)(8.0) = 47.1 kJ, so useful power = 47.1 kJ/60 s = 785 W. Efficiency = 785/1000 = 0.785, or 78.5%.

Common mistakes and exam guidance

Watch out for

Dividing total input by useful output and obtaining more than 100%.
Using Fv when force is not along the direction of motion without resolving it.

In an exam

If your efficiency exceeds 100%, reverse the ratio and check the stated input/output quantities.
Explain a practical loss by naming both the energy transfer and its consequence.

Put the ideas together

Exam-style practice [6 marks]

A pump raises 900 kg of water through 12 m every 5.0 minutes. Its electrical input is 420 W. Find its useful output power, efficiency and energy transferred to less useful stores during the 5.0 minutes. Use g = 9.81 m s⁻².

Plan before you answer

Convert minutes to seconds.
Find gravitational-energy gain per unit time.
Use both power difference and time for the final transfer.

View the marking points and model answer

Marking points

Uses 300 s.
Finds useful energy 1.059 × 10⁵ J.
Finds useful power about 353 W.
Finds less useful power about 67 W.
Finds efficiency about 0.841 or 84.1%.
Obtains about 2.01 × 10⁴ J transferred to less useful stores.

Model answer

Useful energy = 900(9.81)(12) = 1.059 × 10⁵ J, so useful power = 1.059 × 10⁵/300 = 353 W. Efficiency = 353/420 = 0.841 = 84.1%. Less useful energy = (420 − 353)(300) ≈ 2.01 × 10⁴ J.

Finish from memory

Three-question recap

Define one watt.
Check
One joule of energy transferred per second.
Write mechanical power for force at angle θ to velocity.
Check
P = Fv cosθ.
Why can efficiency not exceed 100%?
Check
Useful output cannot exceed total input because energy is conserved.

Try this next

Compare two devices using both power and efficiency rather than treating the terms as interchangeable.

Check the whole topic Return to Energy and Fields Review Fields and potential energy

Continue with the next resource in this course.

Course and syllabus information

Course: GCE A-Level H1 Physics
Edition: GCE A-Level H1 Physics 2027