AQA GCSE Physics (8463), Higher tier · Energy › Conservation and dissipation of energy
Practise Efficiency. 14 exam-style questions plus unlimited generated ones on this subtopic, at up to four difficulty levels, with full mark schemes and a progress tracker. Free, no account needed.
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Calculate the total energy dissipated from the moment the ball is released until the ball reaches the top of its rise after the 4th bounce.
0.86 J
Write down the equation that links efficiency, useful output energy transfer and total input energy transfer.
efficiency = useful output energy transfer ÷ total input energy transfer
Sample questions
Written for this site in the style of AQA exam questions. They are not taken from real past papers.
Question 1Easy4 marks
An electric motor is used in a toy car.
(a) The motor is supplied with 500 J of energy. 350 J is transferred usefully to the kinetic energy store of the car. Calculate the efficiency of the motor. Use the equation: efficiency = useful output energy transfer ÷ total input energy transfer Give your answer as a decimal.[2]
(b) Calculate the energy wasted by the motor.[1]
(c) What happens to the energy wasted by the motor? Tick (✓) one box.[1]
It is destroyed.
It is dissipated to the surroundings.
It is stored in the battery.
It is transferred to the kinetic energy store of the car.
Show the answer and mark scheme
(a)Answer: 0.70
efficiency = 350 ÷ 500
efficiency = 0.70
(b)Answer: 150 J
150 (J)
(c)Answer: It is dissipated to the surroundings.
Question 2Medium6 marks
An electric motor in a fan has an efficiency of 0.65.
(a) The useful power output of the motor is 390 W. Calculate the total power input to the motor.[3]
(b) Calculate the power wasted by the motor.[1]
(c) The fan runs for 5.0 minutes. Calculate the energy wasted by the motor in this time.[2]
Show the answer and mark scheme
(a)Answer: 600 W
0.65 = 390 ÷ Pin
Pin = 390 ÷ 0.65
Pin = 600 (W)
(b)Answer: 210 W
600 − 390 = 210 (W)
(c)Answer: 63 000 J
E = 210 × 300
E = 63 000 (J)
Question 3Hard8 marks
A factory uses an electric winch to lift crates. The winch lifts a crate of mass 45 kg vertically through 8.0 m. The electrical energy supplied to the winch during the lift is 6000 J. Gravitational field strength = 9.8 N/kg
(a) Calculate the efficiency of the winch. Give your answer as a decimal to 2 significant figures.[3]
(b) Suggest two ways the efficiency of the winch could be increased. Explain how each change increases the efficiency.[4]
(c) The factory manager says that the winch could be improved until it is 100% efficient. Explain why this is not possible.[1]
Show the answer and mark scheme
(a)Answer: 0.59
useful energy = 45 × 9.8 × 8.0 (= 3528 J)
efficiency = 3528 ÷ 6000 (= 0.588)
efficiency = 0.59 given to 2 significant figures
(b)Answer: Lubricate the gears and pulley to reduce friction, so less energy is dissipated by heating; use lower-resistance (thicker) wires in the motor so less energy is dissipated in the wires; use a lighter hook and cable so less energy is transferred to them.
lubricate the moving parts / gears / pulley
reduces friction so less energy is dissipated (to the thermal store of the surroundings)
use wires with a lower resistance / thicker wires in the motor
less energy is dissipated by heating in the wires
use a lighter hook / cable
less energy is (wastefully) transferred to the gravitational potential store of the hook and cable
(c)Answer: There will always be some friction / resistance, so some energy is always dissipated to the surroundings.
there will always be some friction / electrical resistance so some energy is always dissipated (to the surroundings)