A Level Physics 5.1 Energy Conservation MCQs – 30 Practice Questions with Answers (Cambridge 9702)
Free practice MCQs for Cambridge International AS & A Level Physics (9702), Section 5.1 Energy Conservation. Test yourself on work done by a force, energy transfers, power and efficiency, then reveal the worked answer for every question.
Formula recap: W = F s (force along the displacement) · W = F s cos θ (force at an angle) · P = W / t · P = F v · efficiency = useful energy (or power) out ÷ total energy (or power) in. Full list: A Level Physics Formula Sheet.
Work Done by a Force
Q1. A constant force F moves an object a distance s in the direction of the force. The work done by the force is
- (A) F ÷ s
- (B) F s
- (C) F + s
- (D) F s2
Show Answer
Answer: B – Work done = force × distance moved in the direction of the force, W = F s. Its unit is the joule (J), where 1 J = 1 N m.
Q2. A traveller pulls a suitcase with a strap that makes an angle of 30° with the horizontal. The tension in the strap is 40 N and the suitcase is pulled 25 m along level ground. The work done by the tension is
- (A) 1000 J
- (B) 500 J
- (C) 866 J
- (D) 707 J
Show Answer
Answer: C – Only the component of force along the displacement does work: W = F s cos θ = 40 × 25 × cos 30° = 1000 × 0.866 = 866 J. (Using the full 40 N gives the trap answer 1000 J.)
Q3. A crate slides 3.0 m across a rough floor. The friction force opposing the motion is a constant 18 N. The work done by the friction force is
- (A) +54 J
- (B) −54 J
- (C) −6.0 J
- (D) 0 J
Show Answer
Answer: B – Friction acts opposite to the displacement, so θ = 180° and W = F s cos 180° = 18 × 3.0 × (−1) = −54 J. The negative sign means mechanical energy is removed from the crate (transferred to internal energy).
Q4. A force applied to a trolley increases steadily from 0 N to 60 N as the trolley moves 3.0 m. On a force-displacement graph this is a straight line from the origin to (3.0 m, 60 N). The work done by the force is
- (A) 90 J
- (B) 180 J
- (C) 60 J
- (D) 20 J
Show Answer
Answer: A – Work done = area under the force-displacement graph. The area is a triangle: W = ½ × 3.0 × 60 = 90 J. (180 J would be the rectangle, which wrongly assumes the full 60 N acted throughout.)
Q5. A force decreases uniformly from 40 N to 10 N while an object moves 5.0 m. The work done by the force is
- (A) 250 J
- (B) 200 J
- (C) 50 J
- (D) 125 J
Show Answer
Answer: D – The force-displacement graph is a trapezium, so W = area = ½ × (40 + 10) × 5.0 = ½ × 50 × 5.0 = 125 J. Equivalently, average force = 25 N, and 25 × 5.0 = 125 J.
Q6. A block of mass 2.0 kg slides down a slope and drops through a vertical height of 1.5 m. Taking g = 9.8 m s−2, the work done by the weight of the block is
- (A) 59 J
- (B) 15 J
- (C) 29 J
- (D) 3.0 J
Show Answer
Answer: C – For gravity, only the vertical drop matters, whatever the slope length: W = m g Δh = 2.0 × 9.8 × 1.5 = 29.4 J ≈ 29 J.
Q7. In which situation does the force do NO work on the object?
- (A) The gravitational force on a satellite in a circular orbit around the Earth
- (B) The push of an athlete accelerating from the starting blocks
- (C) The friction force slowing a sliding box
- (D) The tension in a rope lifting a load at constant speed
Show Answer
Answer: A – In a circular orbit gravity (the centripetal force) acts perpendicular to the displacement at every instant, so cos 90° = 0 and no work is done; the satellite’s speed and kinetic energy stay constant. The same is true for the weight of a bag carried horizontally at a steady height.
Q8. A crane raises a load of mass 50 kg through a vertical height of 4.0 m at constant speed. Taking g = 9.8 m s−2, the work done by the gravitational force on the load is
- (A) +1960 J
- (B) −490 J
- (C) +490 J
- (D) −1960 J
Show Answer
Answer: D – The weight (50 × 9.8 = 490 N) acts downwards while the displacement is upwards: W = −m g h = −50 × 9.8 × 4.0 = −1960 J. The crane does +1960 J of work, which becomes gravitational potential energy.
Q9. A worker pushes a mower with a force of 100 N directed 45° below the horizontal, moving it 2.0 m along the ground. The work done by the pushing force is
- (A) 200 J
- (B) 141 J
- (C) 100 J
- (D) 283 J
Show Answer
Answer: B – Only the horizontal component of the force acts along the displacement: W = F s cos θ = 100 × 2.0 × cos 45° = 200 × 0.707 = 141 J. The vertical component does no work because there is no vertical displacement.
Q10. A box is pushed 10 m across a floor by a horizontal force of 30 N. A friction force of 24 N opposes the motion. The net (total) work done on the box is
- (A) 300 J
- (B) 240 J
- (C) 60 J
- (D) 540 J
Show Answer
Answer: C – Work by the push = 30 × 10 = +300 J. Work by friction = −24 × 10 = −240 J. Net work = 300 − 240 = 60 J, which equals the gain in kinetic energy of the box.
Energy Transfers & the Conservation Principle
Q11. Which statement expresses the principle of conservation of energy?
- (A) Energy cannot be created or destroyed; it can only be transferred from one form to another, so the total energy of an isolated system stays constant
- (B) The total kinetic energy of a system always stays constant
- (C) Energy is destroyed whenever friction acts in a system
- (D) The useful energy output of a machine equals its total energy input
Show Answer
Answer: A – Energy is never created or destroyed, only transferred between stores. Kinetic energy alone is not conserved (option B), and friction transfers energy to internal energy rather than destroying it (option C).
Q12. In a hydroelectric power station, water falls from a reservoir through turbines that drive generators. The main energy transfer chain is
- (A) electrical → kinetic → gravitational potential
- (B) kinetic → gravitational potential → electrical
- (C) gravitational potential → electrical → kinetic
- (D) gravitational potential → kinetic → electrical
Show Answer
Answer: D – Stored water has gravitational potential energy, which becomes kinetic energy as it falls, and the generators transfer this to electrical energy (with a small share wasted as internal energy in the turbines and cables).
Q13. A pendulum bob of mass 0.20 kg is released from rest at a point 0.45 m vertically above its lowest position. Neglecting air resistance and taking g = 9.8 m s−2, its kinetic energy at the lowest point is
- (A) 0.44 J
- (B) 0.88 J
- (C) 1.8 J
- (D) 8.8 J
Show Answer
Answer: B – With no friction, the loss of gravitational potential energy equals the gain in kinetic energy: Ek = m g Δh = 0.20 × 9.8 × 0.45 = 0.882 J ≈ 0.88 J.
Q14. For a swinging pendulum with negligible friction, which row is correct at the lowest point of the swing?
- (A) Kinetic energy maximum, gravitational potential energy maximum
- (B) Kinetic energy minimum, gravitational potential energy minimum
- (C) Kinetic energy maximum, gravitational potential energy minimum
- (D) Both energies are zero
Show Answer
Answer: C – The bob moves fastest at the bottom, so kinetic energy is maximum there, while its height (and so gravitational potential energy) is minimum. The sum of the two stays constant.
Q15. A block slides across a table and comes to rest because of friction. The kinetic energy of the block has been transferred mainly to
- (A) internal (thermal) energy of the block and the table surface
- (B) gravitational potential energy of the block
- (C) elastic potential energy of the table
- (D) sound energy only
Show Answer
Answer: A – Friction does negative work and the “missing” kinetic energy raises the internal energy of the rubbing surfaces (they get slightly warmer). The energy still exists – it has not been destroyed.
Q16. A machine receives 500 J of energy. It does 300 J of useful work. The energy wasted (transferred to the surroundings) is
- (A) 300 J
- (B) 800 J
- (C) 100 J
- (D) 200 J
Show Answer
Answer: D – By conservation of energy, input = useful output + wasted, so wasted = 500 − 300 = 200 J, mostly internal energy of the surroundings (with perhaps a little sound).
Q17. The mechanical energy (kinetic + gravitational potential) of a system is NOT conserved when
- (A) an object falls freely in a vacuum
- (B) a ball bounces and rises to a smaller height each time
- (C) a pendulum swings with negligible air resistance
- (D) a satellite orbits in a circle at constant speed
Show Answer
Answer: B – At each bounce some kinetic energy is transferred to internal energy (and sound) in the ball and floor, so the total mechanical energy decreases and the bounce height falls. In options A, C and D there is no friction-like transfer, so mechanical energy is (approximately) conserved – although TOTAL energy is always conserved in every case.
Q18. An object falls freely from rest through a height h, with air resistance negligible. Which statement is correct?
- (A) The gain in kinetic energy is greater than the loss of potential energy
- (B) The loss of potential energy becomes internal energy of the air
- (C) The gain in kinetic energy equals the loss of gravitational potential energy
- (D) Both kinetic and potential energy increase
Show Answer
Answer: C – With drag negligible, gravity is the only force doing work, so ΔEk = −ΔEp: ½ m v2 = m g h, giving v = √(2 g h).
Q19. When a car brakes to a stop, its kinetic energy is transferred mainly to
- (A) internal energy of the brake discs and pads (they heat up)
- (B) gravitational potential energy of the car
- (C) elastic potential energy of the tyres
- (D) chemical energy in the fuel tank
Show Answer
Answer: A – Friction between pads and discs does negative work on the car; the kinetic energy becomes internal energy of the braking parts and is then lost to the air – which is why brakes get hot and can fade.
Q20. A child slides down a playground slide. At the top the child has 400 J of gravitational potential energy (relative to the bottom). At the bottom the child’s kinetic energy is 250 J. The energy transferred to internal energy by friction is
- (A) 400 J
- (B) 250 J
- (C) 650 J
- (D) 150 J
Show Answer
Answer: D – Total energy is conserved: 400 J = 250 J (kinetic) + energy to internal energy, so the friction transfer = 400 − 250 = 150 J. This is why the child arrives slower than free fall would predict.
Power & Efficiency
Q21. A motor does 3000 J of work in 20 s. Its power is
- (A) 60 W
- (B) 300 W
- (C) 150 W
- (D) 15 W
Show Answer
Answer: C – Power = work done ÷ time taken, P = W / t = 3000 ÷ 20 = 150 W.
Q22. A machine transfers 12 kJ of energy in 2.0 minutes. Its average power is
- (A) 100 W
- (B) 200 W
- (C) 6000 W
- (D) 50 W
Show Answer
Answer: A – Convert units first: 12 kJ = 12 000 J and 2.0 min = 120 s. P = E / t = 12 000 ÷ 120 = 100 W.
Q23. A car travels at a constant speed of 25 m s−1 on a level road. The engine provides a driving force of 800 N to balance the resistive forces. The useful power of the engine is
- (A) 32 W
- (B) 200 W
- (C) 2.0 kW
- (D) 20 kW
Show Answer
Answer: D – For a force acting on a moving object, P = F v = 800 × 25 = 20 000 W = 20 kW.
Q24. An engine has a maximum power output of 30 kW. When the vehicle it drives travels at 20 m s−1, the maximum driving force available is
- (A) 600 N
- (B) 1500 N
- (C) 15 N
- (D) 600 000 N
Show Answer
Answer: B – From P = F v, F = P ÷ v = 30 000 ÷ 20 = 1500 N. This is why the available accelerating force (and acceleration) of a car falls as speed rises.
Q25. A device takes in 800 J of electrical energy and produces 560 J of useful output energy. Its efficiency is
- (A) 57%
- (B) 143%
- (C) 70%
- (D) 30%
Show Answer
Answer: C – Efficiency = useful energy out ÷ total energy in × 100% = 560 ÷ 800 × 100% = 70%. Efficiency can never exceed 100% (option B is impossible).
Q26. A motor has an electrical input power of 2.0 kW and a useful output power of 1.5 kW. Its efficiency is
- (A) 25%
- (B) 75%
- (C) 133%
- (D) 50%
Show Answer
Answer: B – Efficiency works with powers just as with energies: 1.5 ÷ 2.0 × 100% = 75%. The remaining 0.5 kW is wasted, mainly as internal energy in the windings and bearings.
Q27. A pump raises 30 kg of water through a vertical height of 12 m in 60 s. Taking g = 9.8 m s−2, the useful output power of the pump is
- (A) 59 W
- (B) 5.9 W
- (C) 588 W
- (D) 3530 W
Show Answer
Answer: A – Useful energy = m g h = 30 × 9.8 × 12 = 3528 J. Useful power = 3528 ÷ 60 = 58.8 W ≈ 59 W. (3530 W is the trap of forgetting to divide by the time; the pump’s electrical input power would be larger than 59 W because of inefficiency.)
Q28. A motor uses 400 J of electrical energy to lift a load of mass 5.0 kg through a height of 6.0 m. Taking g = 9.8 m s−2, the efficiency of the motor is
- (A) 27%
- (B) 136%
- (C) 54%
- (D) 74%
Show Answer
Answer: D – Useful output = m g h = 5.0 × 9.8 × 6.0 = 294 J. Efficiency = 294 ÷ 400 × 100% = 73.5% ≈ 74%.
Q29. The efficiency of every real machine is always less than 100%. This is because
- (A) some energy is destroyed by friction inside the machine
- (B) some input energy is always transferred to internal (thermal) energy of the machine and surroundings by friction and other resistive effects, so it is not available as useful output
- (C) machines create extra energy that is not counted in the input
- (D) the principle of conservation of energy does not apply to machines
Show Answer
Answer: B – Energy is never destroyed (option A is wrong) and conservation always applies (option D is wrong). Friction, electrical resistance and sound always divert some of the input into internal energy of the surroundings, leaving less than 100% as useful output.
Q30. A cyclist produces a useful power of 300 W. At a steady speed on a level road the total resistive force is 15 N. The cyclist’s speed is
- (A) 2.0 m s−1
- (B) 4500 m s−1
- (C) 20 m s−1
- (D) 0.05 m s−1
Show Answer
Answer: C – At steady speed the driving force equals the resistive force, so from P = F v: v = P ÷ F = 300 ÷ 15 = 20 m s−1.
Keep practising: browse all board-wise physics practice pages on the Physics MCQs hub, and keep the A Level Physics Formula Sheet open while you revise – every equation for Cambridge 9702 in one place.
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