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What do work done, power and efficiency measure, and how are they calculated?

Work done as energy transferred (W = F s), power as the rate of doing work or transferring energy (P = E / t), and efficiency as the fraction of energy transferred usefully.

A CCEA GCSE Double Award Science (Physics Unit P1) answer on work done as energy transferred by a force, power as the rate of transferring energy, and efficiency as the fraction of energy transferred usefully, with the equations and worked calculations.

Generated by Claude Opus 4.810 min answer

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  1. What this dot point is asking
  2. Work done
  3. Power
  4. Efficiency
  5. Examples in context
  6. Try this

What this dot point is asking

CCEA Double Award wants you to use work done (W = F s), power (P = E / t) and efficiency as the fraction of energy transferred usefully. Power and efficiency calculations appear regularly, and you must give efficiency as a fraction or percentage.

Work done

If nothing moves, no work is done, however large the force. Holding a heavy bag still does no work in the physics sense.

Power

A more powerful device transfers the same energy in less time, or more energy in the same time.

Efficiency

Examples in context

Example 1. Comparing kettles
A 3 kW3\ \text{kW} kettle boils water faster than a 2 kW2\ \text{kW} one because it transfers energy at a higher rate (more joules per second). The total energy needed to boil the same water is the same; the more powerful kettle just delivers it in less time, so t=E/Pt = E/P is shorter.
Example 2. LED versus filament bulbs
An LED bulb is far more efficient than an old filament bulb: a larger fraction of the electrical energy becomes light rather than waste heat, so it costs less to run. An old filament bulb might be only about 5 percent efficient (most energy wasted as heat), while an LED can be over 80 percent efficient, which is why filament bulbs have been phased out.
Example 3. A power station
A typical fossil-fuel power station is only about 40 percent efficient, because much of the chemical energy in the fuel is dissipated as heat in the cooling towers and exhaust gases rather than being transferred usefully to electricity. Improving efficiency means wasting less energy for the same useful output, which saves fuel and reduces emissions.

A useful exam tip is that work done and energy transferred are two names for the same quantity in joules, so a question asking for "the energy transferred when a force moves an object" is asking for W=FsW = Fs. Once you have the energy, dividing by the time gives the power, and comparing useful energy with total energy gives the efficiency. These three ideas often appear together in one structured question.

Try this

Q1. State the equation for power and its unit. [2 marks]

  • Cue. P=E/tP = E/t (or W/tW/t); the unit is the watt (W).

Q2. A crane does 24000 J24000\ \text{J} of work in 8.0 s8.0\ \text{s}. Find its power. [2 marks]

  • Cue. P=24000/8.0=3000 WP = 24000/8.0 = 3000\ \text{W}.

Q3. A motor takes in 500 J500\ \text{J} and transfers 300 J300\ \text{J} usefully. Find its efficiency. [2 marks]

  • Cue. efficiency=300/500×100=60%\text{efficiency} = 300/500 \times 100 = 60\%.

Exam-style practice questions

Practice questions written in the style of CCEA exam questions on this dot point, with worked answer explainers. The year tag is the paper they imitate, not the source.

CCEA-style3 marksA motor transfers 4800 J of energy in 6.0 s. Calculate its power output and state the unit.
Show worked answer →

Power is the rate of transferring energy, energy divided by time.

P=Et=48006.0=800 W.P = \dfrac{E}{t} = \dfrac{4800}{6.0} = 800\ \text{W}.

So the power is 800 W (watts).

Markers reward P=E/tP = E/t, the substitution, the value 800, and the unit watts.

CCEA-style3 marksA lamp supplied with 60 J of energy transfers 9 J as light. Calculate its efficiency as a percentage.
Show worked answer →

Efficiency is the useful energy transferred divided by the total energy supplied, times 100.

efficiency=960×100=15%.\text{efficiency} = \dfrac{9}{60} \times 100 = 15\%.

So the lamp is 15 percent efficient (the other 85 percent is wasted as heat).

Markers reward useful over total, times 100, and the value 15 percent.

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