Power and Efficiency
Key idea: Calculate G3 Physics and O-Level Physics power and efficiency using energy, work and time; compare devices, convert units and solve linked lifting problems with clear working.
By the end, you can
- Calculate power as energy transferred or work done per unit time.
- Calculate efficiency from matched energy or power input and useful output values.
1. Definition
Power
Power is the rate of doing work, or the rate of energy transfer.
P = W/t = E/t
- P is power in watts (W), W is work done in joules (J), E is energy transferred in joules (J), and t is time in seconds (s).
- 1 W = 1 J s⁻¹
- Power is a scalar quantity.
Review Energy calculations first if finding Eₖ, Δ Eₚ or dissipated energy is not yet secure.
Data table
| 100 W device | |
|---|---|
| Time (s) | Energy transferred (J) |
| 0 | 0 |
| 5 | 500 |
| 200 W device | |
|---|---|
| Time (s) | Energy transferred (J) |
| 0 | 0 |
| 5 | 1000 |
For the steeper line, gradient = 1000 - 0/5.0 - 0 = 200 J s⁻¹ = 200 W
The gradient unit confirms that the graph gives a rate of energy transfer.
2. Key Ideas
- Always convert time to seconds before using P = E/t.
- A device has higher power if it transfers the same energy in less time, or more energy in the same time.
- If a load is lifted at constant speed, the useful work done is the gain in GPE: Wᵤₛₑfᵤₗ = mgh (or Fh if weight F is given).
- Efficiency, eta, compares the useful output with the total input. Use either energy values, or power values measured over the same interval, but do not mix them: eta = Eᵤₛₑfᵤₗ/Eᵢₙₚᵤₜ × 100\% eta = Pᵤₛₑfᵤₗ/Pᵢₙₚᵤₜ × 100\%
- Efficiency cannot exceed 100\%.
- High power does not necessarily mean high efficiency: power describes how fast energy is transferred, while efficiency describes the useful fraction.
3. Detailed Explanations
Calculating power
To find power:
- Find the energy transferred E (or work done W).
- Convert time to seconds.
- Use P = E/t (or P = W/t).
For example, if 600 J is transferred in 3.0 s, P = 600/3.0 = 200 W
Efficiency
Energy is conserved, but not all input energy is transferred usefully. For a device:
total input energy = useful energy output + energy dissipated to the surroundings
The dissipated share commonly increases internal energy stores in the device and surroundings. Sound waves may also carry energy away, often because of friction, vibration or electrical resistance.
Efficiency of an energy transfer is:
eta = Eᵤₛₑfᵤₗ/Eᵢₙₚᵤₜ × 100\%
The relationship between instantaneous power and force/speed is beyond O Level. See A Level Work, Energy and Power.
4. Common Mistakes
- Not converting minutes/hours to seconds.
- Mixing up power (W) and energy (J).
- Writing P = W/t but using W as “watt” instead of work done (J).
- Getting efficiency >100\% (usually input/output swapped).
- Forgetting to multiply by 100\% when the answer is required as a percentage.
- Assuming the device with the greatest power is automatically the most efficient.
- Using useful energy over input power, or useful power over input energy. Both parts of an efficiency ratio must be the same type of quantity.
5. Exam Tips
- Start with the definition: power = energy transfer per unit time.
- Show unit conversions clearly (e.g. 10 min = 600 s).
- When comparing two power outputs, check what is the same:
- same work but different time → compare time
- same time but different work → compare work
- For efficiency questions, write input = useful + dissipated before substituting numbers.
- Convert a percentage efficiency to a decimal before rearranging an equation: 20\% = 0.20.
- Sense-check the result: useful output cannot exceed total input, and efficiency cannot exceed 100\%.
6. Worked Examples
Example 1: Two runners climb the same hillCore
Runner A has twice the mass of runner B. They climb the same vertical height in the same time. Whose average power output is higher? Explain.
Show Answer
For the useful mechanical transfer modelled in the question, climbing increases the GPE of the runner–Earth system: W = Δ Eₚ = mgh
Average power: P = W/t = mgh/t
Both runners climb the same height h in the same time t, so this useful mechanical power is proportional to mass m. Runner A has twice the mass, so Runner A gains twice the GPE in the same time and has twice the useful mechanical power output in this model.
Example 2: Power of a motorCore
A motor does 60 000 J of work in 10 min. Calculate its average power.
Show Answer
Convert time: 10 min = 10 × 60 = 600 s
P = W/t = 60 000/600 = 100 W
Example 3: Useful energy from input powerCore
An electric motor has an electrical input power of 1.0 kW. During operation, 60\% of the input energy is dissipated to the surroundings. Find the useful energy transferred in half an hour. Hence, state the efficiency of the motor.
Show Answer
Input power: 1.0 kW = 1000 W
Time: 0.5 h = 30 min = 1800 s
Total input energy: Eᵢₙ = Pt = (1000)(1800) = 1.8 × 10⁶ J
If 60\% is dissipated, 40\% is useful: Eᵤₛₑfᵤₗ = 0.40Eᵢₙ; = 0.40(1.8 × 10⁶); = 7.2 × 10⁵ J
The useful fraction is 0.40, so the efficiency is 40\%.
Example 4: Energy dissipated by a lifting motorCore
An electric motor lifts a 10 N load through 5.0 m. The electrical energy input is 65 J. Calculate the energy dissipated by the motor and its efficiency.
Show Answer
Useful energy output is the work done lifting the load: Eᵤₛₑfᵤₗ = Fh = (10)(5.0) = 50 J
Energy dissipated: Edᵢₛₛᵢₚₐₜₑd = Eᵢₙ - Eᵤₛₑfᵤₗ = 65 - 50 = 15 J
Efficiency: efficiency = 50/65 × 100\% = 76.9\% ≈ 77\%
Example 5: Input power from efficiencyCore
A crane raises a weight of 200 N through a vertical height of 8.0 m in 4.0 s at constant speed. The efficiency of the crane is 20\%. Find the electrical power input.
Show Answer
Useful output energy (work done lifting): Eᵤₛₑfᵤₗ = Fh = (200)(8.0) = 1600 J
Useful output power: Pₒᵤₜ = Eᵤₛₑfᵤₗ/t = 1600/4.0 = 400 W
Efficiency: 0.20 = Pₒᵤₜ/Pᵢₙ Rightarrow Pᵢₙ = 400/0.20 = 2000 W = 2.0 kW
7. Mind Stretchers
Mind stretcher 1: Finding efficiency from power and workExtension
A motor takes in power 500 W for 20 s. It lifts a 100 N load through 6.0 m at constant speed. Find the efficiency of the motor.
Show Answer
Total input energy: Eᵢₙ = Pt = (500)(20) = 10 000 J
Useful output energy: Eᵤₛₑfᵤₗ = Fh = (100)(6.0) = 600 J
efficiency = 600/10 000 × 100\% = 6.0\%
Mind stretcher 2: Same work, different powerExtension
Two machines each do 3000 J of work. Machine A takes 12 s, machine B takes 7.5 s. Find the power of each machine and state which is more powerful.
Show Answer
PA = 3000/12 = 250 W PB = 3000/7.5 = 400 W
Machine B has the higher power because it does the same work in less time.
8. Practice, Quiz and Next Step
Close your notes and use Power and Efficiency in the supplied context below. This requires a constructed explanation or working, not recognition of an option.
Fresh context: Two lifts each raise a 6000 J load: lift A takes 15 s and lift B takes 10 s.
- Retrieve: define power as rate of energy transfer in your own words, including units, sign or conditions where relevant.
- Represent: Draw equal energy-transfer arrows with different time labels.
- Apply: Calculate each power and explain why the faster lift is more powerful without saying it transfers more energy in this event.
Check the response before looking back
- The chosen system and transfer pathway are stated before applying conservation.
- Energy and power are not used interchangeably, and all quantities use compatible units.
- The numerical result is checked against the physical situation and any efficiency limit.
If one check fails, name that exact gap, revisit the matching explanation or worked example, and redo the task with different values or a different situation. Then use theO-Level topic checks orpractice browser for an independent re-test.
Recommended next step
work-energy-power
Why this will help: Use one focused question set to check that you can apply the lesson without prompts.
About 10 minutes