CSEC PHYSICS · SECTION A · LESSON A14

Power and Efficiency

Defining power as the rate of doing work, calculating power and efficiency, and explaining why no machine is 100% efficient and how its efficiency can be improved.

MechanicsA 5.11A 5.12A 5.13≈ 35 min🎮 3 games
Before you start

By the end of this lesson you can…

Step 1 · Watch

Explainer video

4 narrated scenes. Press play, listen and read along — the points appear as they are explained. Use ‹ Prev / Next › or the scene bar to jump to any scene.

LESSON A14 · EXPLAINER

Power and Efficiency

A crane and a person can both lift a load to the same height. The crane does it faster — it has more power.

Narration uses your device’s built-in voice. No sound? Switch to “Captions only” — the captions advance on their own.

Step 2 · Read

Key notes

Copy these into your notebook in your own words — that is how they stick.

Power

Rate of doing work / transferring energy; P = W/t = E/t.

Watt

1 W = 1 J/s; 1 kW = 1 000 W.

Energy from power

E = Pt.

Efficiency

(useful energy output ÷ energy input) × 100%.

Efficiency from power

(useful power output ÷ power input) × 100%.

Never 100%

Energy is wasted as thermal energy and sound.

Improve

Lubricate, insulate, better design.

Check

Useful output can never be larger than the input.

EXAM TIPS & COMMON MISTAKES

  • Power is a RATE: always divide by time.
  • Efficiency has no unit (it is a percentage); do not write J or W after it.
  • If the efficiency in your answer is above 100%, you have reversed input and output.
  • Convert time to seconds before using P = E ÷ t (hours × 3 600).
Play & practise

Games

3 quick games on this lesson. Earn XP and badges as you play — your best score for each game is saved. Games are for practice and do not change your exam readiness. More games in the Physics Arcade.

Step 3 · Practise

Paper 1 practice — multiple choice

8 items modelled on real CSEC Paper 1 questions. Choose one answer for each, then check. Aim for 80% or better. Work each item on paper first.

Syllabus A 5.11

1. The unit of power is the

Answer: D. 1 W = 1 J/s.
Syllabus A 5.11

2. A machine does 600 J of work in 12 s. Its power is

Answer: A. P = 600 ÷ 12 = 50 W.
Syllabus A 5.11

3. A 2.0 kW motor runs for 30 s. The energy transferred is

Answer: B. E = Pt = 2 000 × 30 = 60 000 J.
Syllabus A 5.13

4. A motor takes in 800 J and does 600 J of useful work. Its efficiency is

Answer: C. 600 ÷ 800 × 100 = 75%.
Syllabus A 5.12

5. The energy wasted by a machine is usually

Answer: C. Mostly heat (and sound).
Syllabus A 5.12

6. Which is TRUE about efficiency?

Answer: D. Some energy is always wasted.
Syllabus A 5.11

7. A 60 W lamp is on for 5.0 s. The energy supplied is

Answer: A. E = 60 × 5.0 = 300 J.
Syllabus A 5.13

8. A device has an input power of 40 W and an efficiency of 25%. The useful power output is

Answer: B. 25% of 40 W = 10 W.
Step 4 · Apply

Paper 2 practice — structured questions

Write full answers, then reveal the mark scheme and tick the points you earned (14 marks in total). Show all your working — method marks count. Your answers are saved on this device.

Paper 2 style · Power and efficiency

Question 1 [14 marks]

A small electric motor lifts a suitcase of weight 80 N through a height of 1.5 m in 5.0 s. The motor takes 200 J of electrical energy.

(a) (a) Define power. [1]

MARK SCHEME — (a) (max 1)

(b) (b) Calculate the useful work done in lifting the suitcase. [3]

MARK SCHEME — (b) (max 3)

(c) (c) Calculate the useful power output of the motor. [3]

MARK SCHEME — (c) (max 3)

(d) (d) Calculate the efficiency of the motor. [3]

MARK SCHEME — (d) (max 3)

(e) (e) State what happens to the 80 J of energy that is not useful. [2]

MARK SCHEME — (e) (max 2)

(f) (f) Suggest two ways to increase the efficiency of the motor. [2]

MARK SCHEME — (f) (max 2)

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Sources

Where this topic appears in CSEC papers

Find these in your own past-paper booklets for extra timed practice. Items above are written in the style of these papers; any adapted from a specific question say so.