Work and power
Learn the difference between how much energy is transferred and how quickly it is transferred. Use force, distance and time to compare machines fairly.
All material here is Core, also required for Extended. No Supplement-only kinetic or gravitational potential energy equations are needed. This is Cambridge IGCSE, not Egyptian Baccalaureate / Thanaweya Amma or Year 3.
Work is an energy transfer, not a feeling of effort
In physics, mechanical work happens when a force transfers energy through movement. Work done equals the energy transferred by that process. Electrical work is energy transferred by an electric current. Work is measured in joules (J), just like energy; it is not a store of its own.
Worked example
A motor does 500 J of work lifting a load: it transfers 500 J mechanically to the load. A lamp receiving electrical energy at 60 J each second receives 600 J in 10 seconds. Electrical work does not require the whole lamp to move across the room.
Think first: Does “the motor does 500 J of work” tell you its power?
No. You know the amount of energy transferred, but need the time to find the rate. The same 500 J could be transferred quickly or slowly.
A reliable work-and-power method
- Name the force or energy transfer being measured. Is it electrical input or useful mechanical output?
- Convert to N, m, J, W and s as appropriate. Check that movement is along the force.
- For a constant force along movement, calculate work W = Fd.
- For average power, divide work or energy transferred by the same interval’s duration.
- For energy or time, rearrange first: E = Pt or t = E/P.
- Give the correct unit and check by reversing the calculation. Compare work and power separately.
Worked example: find the force and time
A crane does 12 kJ of useful work over an 8.0 m vertical lift. Force = W/d = 12000 ÷ 8.0 = 1500 N. A different crane delivers useful output power of 0.50 kW: time for the same work = 12000 ÷ 500 = 24 s. Checks: 1500 × 8.0 = 12000 J and 500 × 24 = 12000 J. The word “useful” matters: an electrical input rating alone would not justify this time.
Worked example: motion at steady speed still needs energy
A rider travels 2.0 km in 5.0 min at steady speed against a 40 N resistance. Mechanical work against resistance = 40 × 2000 = 80000 J = 80 kJ. Average output power = 80000 ÷ 300 = 266.7 W, or 270 W to 2 significant figures. Steady speed means no kinetic-energy increase, not zero energy transfer. Energy increases internal stores of the tyres, bearings, road and air.
How fast is the same job done?
Same lift, different time
Two model hoists apply the same constant upward force over the same vertical height. Hoist A uses your selected time; B takes 10 s. We calculate work by the lifting force and average useful mechanical power, not electrical input. This is an idealised calculation, not instructions for a home lifting experiment.
For each hoist: force = 200 N; vertical distance = 3 m. Find the work first, then predict A’s average useful power.
Investigate: keep force and height fixed, then double A’s time. What changes and what stays the same? Next keep time fixed and double the height. Explain each result using the two equations. A changed control clears your previous prediction.
Practise, then explain
Four multiple-choice questions, four numerical responses and four written tasks. This lesson and all its questions are Core, and are also needed for Extended. Numerical answers are checked automatically; written answers use a self-assessment guide. The suggested marks are for this original practice, not an official Cambridge mark scheme.
Practice stays only while this page is open. Refreshing or leaving clears these responses. Copy useful written work into your notes before leaving.
A motor does 450 J of mechanical work on a load. What does this tell you?
Optional hint
Work tells you an amount of energy, not a rate or a duration.
Source clarifications and common traps
The original PDF is preserved. Its wardrobe example correctly gives 1500 J and 75 W, but “mostly thermal” needs evidence about friction and motion. We state steady speed when explaining that energy destination; work by one applied force is not automatically the net work.
Source Q6 gives a motor power without clearly stating input or output. To use t = W/P for useful lifting work, our version specifies useful mechanical output power. The zero-work wall example means no mechanical work by the push on an unmoving wall; it does not rule out heating or energy use inside the person.
Figure 16.1 measures displacement between matching points on the box before and after motion. Use distance along the force, not the box’s own length. Do not confuse W as the work symbol with W as the watt unit. A time in a question is a clue, not proof that power is what you must calculate.
Source alignment: supplied Physics-0625 Volume A, PDF pp. 113–117 / printed pp. 105–109, including Figure 16.1, and the official Cambridge 2026–2028 syllabus, sections 1.7.2 and 1.7.4. Original learning text, activity and practice. Source-download access is unchanged.