Cambridge IGCSE Physics · topic 1.7.4
Power
Connect the ideas, make your working visible, and test the result. Use the hints when you need a starting point; open the full solution after your attempt.
Extra challenging questions
Work through the linked stages and explain what each result means.
Choosing a hoist from an energy audit
Optional enrichment: this challenge may extend beyond the usual syllabus depth and assumes confidence with algebra and graph interpretation. A hoist raises a 20 kg load through 5.0 m in 10 s at constant speed. Use g = 10 N kg⁻¹. A force sensor measures an upward cable force of 240 N throughout the rise, including a constant resisting force from guide rails. The electrical input power is the supplied 300 W throughout the lift. Supplied data (Time / s; Input power / W): (0, 300), (5, 300), (10, 300). (a) Calculate cable work, useful gravitational-energy gain, guide-rail dissipation and overall efficiency. [3 marks] (b) A second hoist performs the same lift in 5 s with 50% overall efficiency. Find its mean input power and explain the trade-off. [3 marks]
Electrical input during the rise
Supplied question data. Read the labelled axes and use the observations to test the physical model.
Vertical axis (y): Input power / W
Horizontal axis (x): Time / s
Read the exact data as a table
| Time / s | Input power / W |
|---|---|
| 0 | 300 |
| 5 | 300 |
| 10 | 300 |
These are the supplied values, not a worked answer. Lines join the supplied values; they are not a fitted model.
Show a hint
Begin with part (a): Calculate cable work, useful gravitational-energy gain, guide-rail dissipation and overall efficiency.
Reveal the worked solution
- Cable work is Fh = 240(5.0) = 1200 J. Gravitational energy increases by mgh = 20(10)(5) = 1000 J.
- At constant speed the load's kinetic energy does not change; 200 J of cable work is transferred by rail friction.
- Electrical energy is the power–time area, 300(10) = 3000 J. Useful efficiency is 1000/3000 = 1/3, while 1800 J is also dissipated within the drive.
- The useful gravitational-energy gain remains 1000 J because mass and height do not change.
- At 50% overall efficiency, input energy is 1000/0.50 = 2000 J; mean input power is 2000/5 = 400 W.
- The faster hoist draws greater mean power yet uses less energy per lift. A supply must meet its power demand; energy cost follows energy use rather than power alone.
Answer (a) Cable work = 1200 J; useful gain = 1000 J; rail dissipation = 200 J; overall efficiency = 33.3%. (b) Mean input power = 400 W; total electrical energy = 2000 J.
Compare the assumptions, method and units with your attempt. Explain any correction before trying another question.
A community energy plan that survives the demand peak
Optional enrichment: this challenge may extend beyond the usual syllabus depth and assumes confidence with algebra and graph interpretation. A community uses 120 kWh of electricity per day, with the supplied six-hour average demand levels. A solar array generates 24 kWh per installed kW per week, and a wind installation generates 60 kWh per installed kW per week. A proposed plan uses 14 kW of solar plus 8.4 kW of wind. Treat the quoted outputs as weekly averages; neither source is dispatchable on demand. Supplied data (Period start / h; Average demand / kW): (0, 2), (6, 5), (12, 4), (18, 9). (a) Check daily demand and weekly energy balance. Explain why average energy balance is insufficient to establish reliability. [3 marks] (b) An evening backup battery must supply the full 9 kW for 6 h. Its usable depth of discharge is 80% and discharge efficiency is 90%. Find nominal energy capacity and state two further decision criteria. [3 marks]
Six-hour average demand
Supplied question data. Read the labelled axes and use the observations to test the physical model.
Vertical axis (y): Average demand / kW
Horizontal axis (x): Period start / h
Read the exact data as a table
| Period start / h | Average demand / kW |
|---|---|
| 0 | 2 |
| 6 | 5 |
| 12 | 4 |
| 18 | 9 |
These are the supplied values, not a worked answer. No fitted trend is drawn.
Show a hint
Begin with part (a): Check daily demand and weekly energy balance. Explain why average energy balance is insufficient to establish reliability.
Reveal the worked solution
- The four six-hour blocks consume 6(2+5+4+9) = 120 kWh per day, or 840 kWh per week.
- Proposed weekly generation is 14(24)+8.4(60) = 336+504 = 840 kWh, matching the average weekly energy.
- Matching totals does not align generation with the evening 9 kW demand or calm cloudy intervals. Storage, flexible demand or reliable backup is required, with losses and power limits included.
- Required delivered energy is 9(6) = 54 kWh. Usable delivered fraction is 0.80×0.90 = 0.72.
- Nominal energy capacity must be at least 54/0.72 = 75 kWh. This capacity calculation is separate from the required 9 kW output-power rating.
- Compare capital and lifecycle costs, resource variability, maintenance, local noise or land impacts, material supply and recycling. Stakeholder priorities and environmental effects should be stated explicitly rather than hidden in one energy total.
Answer (a) Daily demand = 120 kWh; generation = 840 kWh per week, equal to average demand, but peak reliability is unproven. (b) Nominal battery energy≥75 kWh; output power rating≥9 kW.
Compare the assumptions, method and units with your attempt. Explain any correction before trying another question.