Cambridge IGCSE Physics · topic 1.7.1
Energy
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 braking model tested at two speeds
Optional enrichment: this challenge may extend beyond the usual syllabus depth and assumes confidence with algebra and graph interpretation. A 2.0 kg trolley reaches a level rough strip after descending from rest through height 1.8 m on a smooth ramp. Use g = 10 m s⁻². On the strip the opposing force depends on distance as shown. It remains 8 N beyond x = 4 m. Neglect rotational energy. Supplied data (Distance along strip / m; Resisting force / N): (0, 4), (4, 8), (6, 8). (a) Find entry speed and stopping distance, using the force–distance graph rather than a constant deceleration formula. [3 marks] (b) If the entry speed doubles, determine the new stopping distance and explain the energy scaling. [3 marks]
Resisting force on the rough strip
Supplied question data. Read the labelled axes and use the observations to test the physical model.
Vertical axis (y): Resisting force / N
Horizontal axis (x): Distance along strip / m
Read the exact data as a table
| Distance along strip / m | Resisting force / N |
|---|---|
| 0 | 4 |
| 4 | 8 |
| 6 | 8 |
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): Find entry speed and stopping distance, using the force–distance graph rather than a constant deceleration formula.
Reveal the worked solution
- The trolley gains mgh = 36 J, so ½(2)v² = 36 and v = 6.0 m s⁻¹.
- Work against resistance over the first 4 m is trapezium area ½(4+8)(4) = 24 J, leaving 12 J of kinetic energy.
- At 8 N it travels another 12/8 = 1.5 m, stopping at 5.5 m. Resistance is not constant on the first section, so neither is deceleration.
- Kinetic energy varies as v², so doubling speed multiplies it by four, giving 144 J.
- The first 4 m still dissipates 24 J. The remaining 120 J requires 120/8 = 15 m beyond that point.
- Total distance is 19 m. It is not four times 5.5 m because the spatial resistance profile has a lower-force entry section.
Answer (a) Entry speed = 6.0 m s⁻¹; stopping distance = 5.5 m. (b) Kinetic energy becomes 144 J and stopping distance becomes 19 m.
Compare the assumptions, method and units with your attempt. Explain any correction before trying another question.