ExaminerPrep › Question bank › IAL Physics › 1.6 Power and efficiency
Edexcel International A Level Physics · Unit 1 · Mechanics and Materials
32 free, high-quality past-paper-style questions on Power and efficiency for Edexcel IAL Physics, each with a full mark scheme. For when you have finished the past papers, or want to practise one topic at a time.IAL Physics notes and flashcards →Not sure of your level? Diagnostic test →Multiple choice or structured questions, with the mark scheme as a separate PDF.
Which of the following is equivalent to the watt, expressed in SI base units?
| Answer | Mark | Notes |
|---|---|---|
| 1 W = 1 J s−1. A joule is a newton metre: (kg m s−2) × m = kg m2 s−2. Dividing by a second gives kg m2 s−3. | ✓ 1 | Answer A |
Answer: A
Exam tip: Build base units step by step from defining equations: force from F = ma, energy from W = FΔs, power from P = W/t.
A squirrel of mass 0.46 kg runs 9.0 m vertically up a tree trunk in 3.0 s. g = 9.81 N kg−1.
What is the average useful power output of the squirrel in gaining gravitational potential energy?
| Answer | Mark | Notes |
|---|---|---|
| Gain in gravitational potential energy = mgΔh = 0.46 × 9.81 × 9.0 = 40.6 J. Power = 40.6 ÷ 3.0 = 14 W (13.5 W). | ✓ 1 | Answer B |
Answer: B
Common error: Stopping at the energy. The question asks for a power, so divide the energy transferred by the time taken.
A heat-pump water heater uses 2.5 kWh of energy during one heating cycle.
What is this energy in joules?
| Answer | Mark | Notes |
|---|---|---|
| 1 kWh = 1000 W × 3600 s = 3.6 × 106 J, so 2.5 kWh = 2.5 × 3.6 × 106 = 9.0 × 106 J. | ✓ 1 | Answer D |
Answer: D
Exam tip: Learn 1 kWh = 3.6 × 106 J by deriving it: 1000 W for 3600 s. Both factors are needed.
A weightlifter raises a barbell of mass 140 kg vertically through 1.9 m. The barbell starts and finishes at rest, and the lift takes 1.4 s. g = 9.81 N kg−1.
State what is meant by power.
Calculate the gain in gravitational potential energy of the barbell.
Calculate the average useful power output of the weightlifter during the lift.
| Answer | Mark | Notes |
|---|---|---|
| Part (a) | ||
| the rate of doing work / the rate of transferring energy | ✓ 1 | |
| Part (b) | ||
| ΔEgrav = mgΔh = 140 × 9.81 × 1.9 | ✓ 1 | |
| = 2.6 × 103 J (2609 J) | ✓ 1 | |
| Part (c) | ||
| 2609 ÷ 1.4 = 1.9 kW (1864 W) (allow error carried forward) | ✓ 1 | |
Exam tip: A definition of power must mention a rate: “energy per second” or “rate of doing work”. “Energy transferred” alone describes work, not power.
On a building site a worker uses a rope over a single fixed pulley to raise a bucket of cement of mass 35 kg through a height of 6.0 m at constant speed. She pulls down on the rope with a constant force of 420 N, and her end of the rope moves 6.0 m. g = 9.81 N kg−1.
Calculate the work done by the worker.
Calculate the useful energy transferred to the bucket.
Calculate the efficiency of the pulley system.
Suggest why the efficiency is less than 100%.
| Answer | Mark | Notes |
|---|---|---|
| Part (a) | ||
| 420 × 6.0 = 2.5 × 103 J (2520 J) | ✓ 1 | |
| Part (b) | ||
| 35 × 9.81 × 6.0 = 2.1 × 103 J (2060 J) | ✓ 1 | |
| Part (c) | ||
| 2060 ÷ 2520 = 0.82 / 82% (allow error carried forward) | ✓ 1 | |
| Part (d) | ||
| friction in the pulley bearings / between the rope and the pulley transfers some energy to thermal energy in the surroundings | ✓ 1 | |
Exam tip: Decide which energy is useful before you calculate an efficiency: here it is the gain in gravitational potential energy of the bucket, not the work done by the worker.
The graph shows how the input power to a garden shredder varies with time while it shreds one bag of branches.
What is the total energy transferred to the shredder during the 70 s shown?
| Answer | Mark | Notes |
|---|---|---|
| The energy transferred is the area under the power–time graph: 2.0 kW × 40 s + 0.5 kW × 20 s = 80 kJ + 10 kJ = 90 kJ. | ✓ 1 | Answer B |
Answer: B
Exam tip: The area under a power–time graph is the energy transferred. With power in kW and time in s, the area comes out directly in kJ.
The graph shows the total energy supplied to a dehumidifier against time. The dehumidifier operates at constant power.
What is the input power of the dehumidifier?
| Answer | Mark | Notes |
|---|---|---|
| Power = gradient of the energy–time graph = 360 kJ ÷ 20 min = 360 000 J ÷ 1200 s = 300 W. | ✓ 1 | Answer B |
Answer: B
Common error: Taking the gradient straight from the axis units in kJ per minute. Convert to joules and seconds first, or the answer is out by a factor of 60 or 1000.
The Sankey diagram shows the energy transfers in an electric drill each second. The band widths are drawn to scale.
What is the efficiency of the drill?
| Answer | Mark | Notes |
|---|---|---|
| Input = sum of outputs, so the useful output = 750 − 180 − 45 = 525 W. Efficiency = 525 ÷ 750 = 0.70 = 70%. | ✓ 1 | Answer D |
Answer: D
Exam tip: In a Sankey diagram the input equals the sum of all the outputs. Find an unlabelled output by subtraction, then divide the useful output by the input.
A heater in a greenhouse is left switched on overnight. The figure shows the readings on its electricity meter at the start and at the end of the night.
Determine the energy transferred to the heater, in kWh.
Calculate the average power of the heater, in kW.
Calculate the energy transferred to the heater in joules.
| Answer | Mark | Notes |
|---|---|---|
| Part (a) | ||
| 18 451.9 − 18 432.7 = 19.2 kWh | ✓ 1 | |
| Part (b) | ||
| time from 18:00 to 06:00 = 12 h; 19.2 ÷ 12 = 1.6 kW (allow error carried forward) | ✓ 1 | |
| Part (c) | ||
| 1 kWh = 1000 W × 3600 s = 3.6 × 106 J | ✓ 1 | |
| 19.2 × 3.6 × 106 = 6.9 × 107 J (6.91 × 107 J) | ✓ 1 | |
Common error: Taking the time from 18:00 to 06:00 as 6 hours. Count forward through midnight: 6 h before it and 6 h after it.
The Sankey diagram shows the energy transfers each second in the engine of a car travelling at a steady speed. It is drawn on a grid in which each small square represents 4.0 kW.
Use the diagram to determine the useful output power of the engine.
Calculate the efficiency of the engine.
The car is driven at this steady speed for 45 minutes. Calculate the energy transferred as thermal energy in the exhaust gases during this time, in kWh.
| Answer | Mark | Notes |
|---|---|---|
| Part (a) | ||
| useful band is 5 squares wide: 5 × 4.0 = 20 kW | ✓ 1 | |
| Part (b) | ||
| 20 ÷ 80 = 0.25 / 25% (allow error carried forward) | ✓ 1 | |
| Part (c) | ||
| exhaust band is 11 squares wide: 11 × 4.0 = 44 kW | ✓ 1 | |
| 44 kW × 0.75 h = 33 kWh | ✓ 1 | |
Exam tip: Count squares across a band where its edges are straight, and check that the output widths add up to the input width before you use them.
A tractor pulls a plough across a field at a constant speed of 1.8 m s−1. The horizontal force exerted by the tractor on the plough is 14 kN.
What is the rate at which the tractor does work on the plough?
| Answer | Mark | Notes |
|---|---|---|
| The force moves its point of application at constant velocity, so P = Fv = 14 000 × 1.8 = 25 kW (25.2 kW). | ✓ 1 | Answer C |
Answer: C
Exam tip: P = Fv follows from P = W/t = FΔs/t. Check the units: N × m s−1 = J s−1 = W.
A small pump in a village well raises 45 kg of water through a vertical height of 8.0 m every minute. The input power to the pump is 95 W. g = 9.81 N kg−1.
What is the efficiency of the pump?
| Answer | Mark | Notes |
|---|---|---|
| Useful energy per minute = 45 × 9.81 × 8.0 = 3532 J, so useful power = 3532 ÷ 60 = 58.9 W. Efficiency = 58.9 ÷ 95 = 0.62 = 62%. | ✓ 1 | Answer C |
Answer: C
Common error: Dividing an energy by a power. Convert the useful energy per minute into a useful power (÷ 60 s) before comparing it with the input power.
An outboard engine on a small boat has an efficiency of 0.32. Its useful output power is 18 kW.
At what rate does the engine waste energy?
| Answer | Mark | Notes |
|---|---|---|
| Input power = useful ÷ efficiency = 18 ÷ 0.32 = 56.25 kW. Wasted power = 56.25 − 18 = 38 kW. | ✓ 1 | Answer B |
Answer: B
Exam tip: Input = useful + wasted. Find the input from the efficiency first, then subtract the useful output.
The battery of an electric motorbike stores 9.6 kWh of energy. On a level road at a constant speed of 22 m s−1 the useful output power of the motor is 7.5 kW. Assume that all the stored energy is transferred as useful output.
Show that the energy stored in the battery is about 3.5 × 107 J.
Calculate the time for which the motorbike can travel at 22 m s−1.
Calculate the distance travelled in this time.
Calculate the total resistive force on the motorbike at this speed.
| Answer | Mark | Notes |
|---|---|---|
| Part (a) | ||
| 9.6 × 1000 × 3600 = 3.46 × 107 J | ✓ 1 | |
| Part (b) | ||
| t = E ÷ P = 3.456 × 107 ÷ 7500 | ✓ 1 | |
| = 4.6 × 103 s (4608 s) | ✓ 1 | |
| Part (c) | ||
| 22 × 4608 = 1.0 × 105 m (101 km) (allow error carried forward) | ✓ 1 | |
| Part (d) | ||
| constant speed, so driving force = resistive force and P = Fv | ✓ 1 | |
| F = 7500 ÷ 22 = 340 N (341 N) | ✓ 1 | |
Exam tip: Energy in kWh = power in kW × time in hours. To get a time in seconds, convert the energy to joules and the power to watts first.
A planetary rover of mass 900 kg drives up a straight slope on Mars at a constant speed of 0.040 m s−1. The slope is inclined at 12° to the horizontal. The gravitational field strength on Mars is 3.71 N kg−1. The total resistive force on the rover, acting down the slope, is 260 N.
Calculate the component of the rover's weight acting down the slope.
Show that the driving force on the rover is about 950 N.
Calculate the useful output power of the rover's motors.
The motors have an efficiency of 0.55. Calculate the input power to the motors.
| Answer | Mark | Notes |
|---|---|---|
| Part (a) | ||
| W = 900 × 3.71 = 3339 N; component = W sin 12° | ✓ 1 | |
| = 694 N | ✓ 1 | |
| Part (b) | ||
| constant speed, so driving force = 694 + 260 = 954 N | ✓ 1 | |
| Part (c) | ||
| P = Fv = 954 × 0.040 = 38 W (38.2 W) | ✓ 1 | |
| Part (d) | ||
| input = useful ÷ efficiency = 38.2 ÷ 0.55 | ✓ 1 | |
| = 69 W (69.4 W) (allow error carried forward) | ✓ 1 | |
Common error: Using the Earth value of g by habit. Use the field strength given for the place in the question.
A cyclist rides along a level road at a constant speed of 9.0 m s−1. The total resistive force on the cyclist and bicycle is 32 N.
Starting from the definition of work done, show that the power of a constant force F moving its point of application at constant velocity v is given by P = Fv.
Calculate the useful power output of the cyclist at 9.0 m s−1.
At a constant speed of 12.0 m s−1 the total resistive force is 54 N. Calculate the extra useful power the cyclist must produce to ride at this speed.
Explain why there is a maximum speed at which the cyclist can travel on a level road.
| Answer | Mark | Notes |
|---|---|---|
| Part (a) | ||
| P = W/t and W = FΔs, so P = FΔs/t | ✓ 1 | |
| Δs/t = v at constant velocity, so P = Fv | ✓ 1 | |
| Part (b) | ||
| driving force = 32 N; 32 × 9.0 = 290 W (288 W) | ✓ 1 | |
| Part (c) | ||
| 54 × 12.0 = 648 W | ✓ 1 | |
| extra power = 648 − 288 = 360 W | ✓ 1 | |
| Part (d) | ||
| the resistive force, and so the power needed (Fv), increases with speed; at the speed where the power needed equals the cyclist's maximum power output there is no resultant force, so no further acceleration | ✓ 1 | |
Exam tip: In a “show that” derivation, write each equation you start from and state the substitution Δs/t = v in words; the final line alone earns nothing.
A student estimates her useful power output by running up the steps of a sports stadium. g = 9.81 N kg−1.
Describe how she should determine the vertical height she climbs and the time she takes. Name the measuring instruments.
Her mass is 71.5 kg and the vertical height is 6.40 m. Her five times are 5.1 s, 4.8 s, 5.0 s, 4.7 s and 5.0 s. Calculate the mean time and its uncertainty.
Calculate her useful power output.
The uncertainties in her mass and in the height are negligible. Calculate the percentage uncertainty in her power.
| Answer | Mark | Notes |
|---|---|---|
| Part (a) | ||
| measure the rise (vertical height) of one step with a metre rule / tape measure | ✓ 1 | |
| multiply by the number of steps, after checking that several steps have the same rise / measure the total height directly | ✓ 1 | |
| a partner times the run with a stopwatch from the bottom step to the top step, repeated several times and a mean taken | ✓ 1 | |
| Part (b) | ||
| mean = 24.6 ÷ 5 = 4.92 s (allow 4.9 s) | ✓ 1 | |
| uncertainty = half the range = (5.1 − 4.7) ÷ 2 = ± 0.2 s | ✓ 1 | |
| Part (c) | ||
| 71.5 × 9.81 × 6.40 ÷ 4.92 = 910 W (912 W; allow 920 W from a mean of 4.9 s) | ✓ 1 | |
| Part (d) | ||
| 0.2 ÷ 4.92 × 100 = 4.1% | ✓ 1 | |
Exam tip: Only the vertical height counts for the gain in gravitational potential energy. Say so in a method description, and do not measure along the slope of the stairs.
A kayaker paddles in a straight line at a constant speed of 4.0 m s−1. The graph shows how the total resistive force on the kayak varies with its speed.
What is the useful power output of the kayaker?
| Answer | Mark | Notes |
|---|---|---|
| At constant speed the driving force equals the resistive force. From the graph, F = 80 N at 4.0 m s−1. P = Fv = 80 × 4.0 = 320 W. | ✓ 1 | Answer C |
Answer: C
Exam tip: Read the force at the speed in the question. Because the force changes with speed, neither the gradient nor the area of this graph gives the power.
The graph shows how the efficiency of an electric motor varies with its useful output power. The motor is operated at a useful output power of 6.0 kW.
At what rate is energy wasted in the motor?
| Answer | Mark | Notes |
|---|---|---|
| From the graph the efficiency at 6.0 kW is 80% (0.80). Input power = 6.0 ÷ 0.80 = 7.5 kW. Wasted power = 7.5 − 6.0 = 1.5 kW. | ✓ 1 | Answer A |
Answer: A
Common error: Multiplying the output power by the efficiency. Efficiency = useful output ÷ input, so the input is the useful output divided by the efficiency.
A delivery drone of mass 2.4 kg rises vertically. The graph shows its height above the ground against time. g = 9.81 N kg−1.
What is the useful power output of the drone in gaining gravitational potential energy while it is rising?
| Answer | Mark | Notes |
|---|---|---|
| Between 2 s and 12 s the height increases steadily by 30 m in 10 s, so the vertical speed is 3.0 m s−1 and the kinetic energy is constant. P = mg × (vertical speed) = 2.4 × 9.81 × 3.0 = 71 W (70.6 W). | ✓ 1 | Answer C |
Answer: C
Exam tip: The gradient of a height–time graph is the vertical speed. Use only the interval in which the height is changing.
A sprinter of mass 72 kg starts from rest. The graph shows how her velocity varies with time for the first 6.0 s. Assume that resistive forces on her are negligible during the first 3.0 s.
Determine her acceleration during the first 3.0 s.
Calculate the resultant force on her during the first 3.0 s.
Calculate her average useful power output during the first 3.0 s.
Calculate her useful power output at t = 3.0 s, just before her velocity becomes constant. Explain why this is greater than her average power output during the first 3.0 s.
| Answer | Mark | Notes |
|---|---|---|
| Part (a) | ||
| gradient = 9.0 ÷ 3.0 = 3.0 m s−2 | ✓ 1 | |
| Part (b) | ||
| F = ma = 72 × 3.0 = 220 N (216 N) | ✓ 1 | |
| Part (c) | ||
| gain in kinetic energy = ½ × 72 × 9.02 = 2916 J | ✓ 1 | |
| P = 2916 ÷ 3.0 = 970 W (972 W) | ✓ 1 | |
| Part (d) | ||
| P = Fv = 216 × 9.0 = 1900 W (1944 W) | ✓ 1 | |
| the force is constant but the velocity increases steadily, so Fv increases; her average velocity over the 3.0 s is only 4.5 m s−1, half the final value, so the average power is half the final power | ✓ 1 | |
Exam tip: P = Fv gives the power at one instant. With a constant force and a steadily increasing speed, the power rises in proportion to v.
A horse walking along a towpath pulls a narrowboat along a straight canal at a constant velocity of 1.1 m s−1. The figure is a view from above. The rope makes an angle of 22° with the direction of motion of the boat, and the tension in the rope is 650 N.
Calculate the component of the tension in the direction of motion of the boat.
Calculate the rate at which the horse does useful work on the boat.
Explain why the component of the tension at right angles to the direction of motion does no work on the boat.
On another day the boat moves at the same constant velocity driven by its own engine, which pushes it directly forwards. The resistive force on the boat is the same as before. The engine has an efficiency of 0.28. Calculate the rate at which the engine transfers energy from its fuel.
| Answer | Mark | Notes |
|---|---|---|
| Part (a) | ||
| 650 cos 22° | ✓ 1 | |
| = 603 N (allow 600 N) | ✓ 1 | |
| Part (b) | ||
| P = Fv = 603 × 1.1 = 660 W (663 W) (allow error carried forward) | ✓ 1 | |
| Part (c) | ||
| the boat has no displacement in that direction, and work done = force × displacement in the direction of the force | ✓ 1 | |
| Part (d) | ||
| useful power = 663 W, since the forward force again equals the resistive force of 603 N | ✓ 1 | |
| input = 663 ÷ 0.28 = 2.4 kW (2370 W) (allow error carried forward) | ✓ 1 | |
Common error: Using the full tension in P = Fv. Only the component of the force along the velocity transfers energy to the boat.
A student compares three ways of heating the same mass of water from the same starting temperature to the same final temperature. The bar chart shows the energy supplied by each method. The useful energy gained by the water is 300 kJ in each case.
Use the bar chart to calculate the efficiency of the kettle.
The kettle has a power of 2.2 kW. Calculate the time it takes to heat the water.
The microwave oven takes 4.0 minutes to heat the water. Calculate the input power of the microwave oven.
Suggest why the gas hob has the lowest efficiency.
| Answer | Mark | Notes |
|---|---|---|
| Part (a) | ||
| energy supplied to the kettle = 400 kJ | ✓ 1 | |
| 300 ÷ 400 = 0.75 / 75% | ✓ 1 | |
| Part (b) | ||
| t = E ÷ P = 400 000 ÷ 2200 | ✓ 1 | |
| = 180 s (182 s) | ✓ 1 | |
| Part (c) | ||
| 500 000 ÷ 240 = 2.1 kW (2080 W) | ✓ 1 | |
| Part (d) | ||
| much of the energy from the flame heats the surrounding air, as hot gases flow up around the sides of the pan / heats the pan and the hob rather than the water | ✓ 1 | |
Exam tip: When the useful energy is the same for every method, the method that needs the least input energy is the most efficient.
The graph shows how the electrical output power of a wind turbine depends on the wind speed.
State the output power of the turbine when the wind speed is 10 m s−1.
At a wind speed of 10 m s−1, the kinetic energy of the air passing through the area swept by the blades is 3.1 MJ each second. Calculate the efficiency of the turbine at this wind speed.
On one day the wind speed is 10 m s−1 for 9.0 hours and 8.0 m s−1 for the other 15 hours. Calculate the electrical energy output of the turbine for the day, in kWh.
Convert your answer to the previous part into joules.
Suggest why the efficiency of a wind turbine must be less than 1 even if there were no friction or electrical losses.
| Answer | Mark | Notes |
|---|---|---|
| Part (a) | ||
| 1200 kW (allow 1150–1250 kW) | ✓ 1 | |
| Part (b) | ||
| 1200 ÷ 3100 = 0.39 (0.387) (allow error carried forward) | ✓ 1 | |
| Part (c) | ||
| at 8.0 m s−1 the output is 600 kW (allow 580–620 kW) | ✓ 1 | |
| 1200 × 9.0 + 600 × 15 = 10 800 + 9000 | ✓ 1 | |
| = 2.0 × 104 kWh (19 800 kWh) | ✓ 1 | |
| Part (d) | ||
| 19 800 × 3.6 × 106 = 7.1 × 1010 J (allow error carried forward) | ✓ 1 | |
| Part (e) | ||
| the air must still be moving after it passes the blades so that it can flow away, so it keeps some of its kinetic energy | ✓ 1 | |
Exam tip: With power in kW and time in hours, kW × h gives kWh directly. Convert to joules only when the question asks for it.
A high-speed train travels on a level track at constant speed. The total resistive force on the train is proportional to the square of its speed.
The train's speed is doubled and it again travels at constant speed. By what factor does the useful power output of the train increase?
| Answer | Mark | Notes |
|---|---|---|
| At constant speed, driving force = resistive force ∝ v2. Power = Fv ∝ v2 × v = v3. Doubling v multiplies the power by 23 = 8. | ✓ 1 | Answer D |
Answer: D
Exam tip: In a proportion question write the power in terms of the speed alone (P = Fv with F ∝ v2) before putting in the factor.
A chairlift carries skiers up a straight slope inclined at 20° to the horizontal. At any instant the skiers and chairs moving up the slope have a total mass of 4800 kg, and they move at a constant 2.5 m s−1 along the slope. Ignore friction and the returning chairs. g = 9.81 N kg−1.
What is the minimum power needed to raise the skiers and chairs?
| Answer | Mark | Notes |
|---|---|---|
| The rate of gain of height is the vertical component of the velocity, 2.5 sin 20° = 0.855 m s−1. Power = mg × (vertical speed) = 4800 × 9.81 × 0.855 = 40.26 kW, so 40.3 kW (rounded up: 40 kW would be too little). Equivalently, the force along the slope is mg sin 20° and P = Fv. | ✓ 1 | Answer A |
Answer: A
Common error: Using the speed along the slope with the full weight. Only the vertical component of the velocity raises the gravitational potential energy — or use the component of weight along the slope with the speed along the slope.
A student investigates how the efficiency of a small motor depends on the load it lifts. The input power to the motor is kept constant at 3.20 W, measured with a power meter. The motor winds up a string to raise a mass through 0.800 m at constant speed, and she times each lift. g = 9.81 N kg−1.
Results: 0.100 kg — 2.60 s; 0.200 kg — 3.40 s; 0.300 kg — 4.60 s; 0.400 kg — 7.00 s.
Calculate the efficiency of the motor when it lifts the 0.300 kg mass.
The efficiencies for the 0.100 kg, 0.200 kg and 0.400 kg masses are 9.4%, 14.4% and 14.0%. Describe how the efficiency varies with the load, and estimate the load at which it is greatest.
Suggest one improvement to the method that would locate the load for maximum efficiency more precisely.
| Answer | Mark | Notes |
|---|---|---|
| Part (a) | ||
| useful energy = 0.300 × 9.81 × 0.800 = 2.354 J | ✓ 1 | |
| useful power = 2.354 ÷ 4.60 = 0.512 W | ✓ 1 | |
| efficiency = 0.512 ÷ 3.20 = 0.160 / 16.0% | ✓ 1 | |
| Part (b) | ||
| efficiency increases with load and then decreases | ✓ 1 | |
| greatest at a load of about 0.3 kg (between 0.2 kg and 0.4 kg) | ✓ 1 | |
| Part (c) | ||
| take more readings at loads between 0.2 kg and 0.4 kg / use a greater lifting height or light gates so that the percentage uncertainty in each time is smaller / repeat each timing and use the mean | ✓ 1 | |
Model sketch
Exam tip: With a constant input power, efficiency = useful power ÷ input power. Convert the useful energy for each lift into a power by dividing by the time for that lift.
In a hydroelectric power station, water flows through the turbines at a rate of 85 m3 s−1 after falling through a vertical height of 120 m. The density of water is 1000 kg m−3. The overall efficiency of the station is 0.88. g = 9.81 N kg−1.
Calculate the mass of water passing through the turbines each second.
Calculate the rate at which the water loses gravitational potential energy.
Calculate the useful output power of the station.
Calculate the useful energy output of the station in one day, in kWh.
| Answer | Mark | Notes |
|---|---|---|
| Part (a) | ||
| 85 × 1000 = 8.5 × 104 kg | ✓ 1 | |
| Part (b) | ||
| energy lost per second = mgΔh for one second's mass of water: 8.5 × 104 × 9.81 × 120 | ✓ 1 | |
| = 1.0 × 108 W (1.0006 × 108 W) | ✓ 1 | |
| Part (c) | ||
| 0.88 × 1.0006 × 108 = 8.8 × 107 W (88 MW) | ✓ 1 | |
| Part (d) | ||
| 8.805 × 104 kW × 24 h | ✓ 1 | |
| = 2.1 × 106 kWh (allow error carried forward) | ✓ 1 | |
Exam tip: When a flow rate is given, work per second: the mass arriving each second gives the energy transferred each second, which is the power.
The diagram shows the energy chain for a solar-powered water pump. Each block passes its useful output to the next block. The useful output power of the motor is 54 W.
What is the power of the sunlight incident on the solar panel?
| Answer | Mark | Notes |
|---|---|---|
| Overall efficiency = 0.20 × 0.90 × 0.75 = 0.135. Input power = 54 ÷ 0.135 = 400 W. | ✓ 1 | Answer D |
Answer: D
Exam tip: For devices that pass energy one to the next, multiply the efficiencies to find the overall efficiency. Never add them.
For a car on a level road, the graph shows the power needed to overcome resistive forces at each constant speed, and the maximum power the engine can deliver at that speed. The top speed of the car is where the two curves meet.
What is the total resistive force on the car at its top speed?
| Answer | Mark | Notes |
|---|---|---|
| The curves meet at 40 m s−1 and 60 kW. At constant speed the driving force equals the resistive force, so F = P/v = 60 000 ÷ 40 = 1500 N = 1.5 kN. | ✓ 1 | Answer A |
Answer: A
Exam tip: Above the crossing point the power needed is greater than the power available, so there is no spare power to accelerate: that is why the crossing gives the top speed.
A student uses a small electric motor to lift a load of mass 0.400 kg at constant speed through different heights h. For each height she measures the energy E supplied to the motor using a joulemeter. The graph shows her results and her line of best fit. g = 9.81 N kg−1.
Determine the gradient of the line of best fit.
Explain why the efficiency with which the motor lifts the load through each additional metre is equal to mg ÷ gradient, and calculate this efficiency.
Suggest a reason why the line of best fit does not pass through the origin.
Calculate the efficiency of a single lift through 0.30 m, and explain why it is less than the efficiency you calculated for each additional metre.
| Answer | Mark | Notes |
|---|---|---|
| Part (a) | ||
| (6.40 − 0.80) ÷ 1.00 = 5.6 J m−1 (allow 5.5–5.7) | ✓ 1 | |
| Part (b) | ||
| each extra metre gives the load mg × 1 m of extra gravitational potential energy (useful), while the gradient is the extra energy supplied per metre | ✓ 1 | |
| 0.400 × 9.81 ÷ 5.6 = 3.92 ÷ 5.6 = 0.70 (allow error carried forward) | ✓ 1 | |
| Part (c) | ||
| energy is supplied before the load starts to rise, e.g. to take up slack in the string / to accelerate the motor and load / to overcome friction when starting | ✓ 1 | |
| Part (d) | ||
| E = 0.80 + 5.6 × 0.30 = 2.48 J; useful energy = 0.400 × 9.81 × 0.30 = 1.18 J; efficiency = 1.18 ÷ 2.48 = 0.47 | ✓ 1 | |
| the fixed 0.80 J supplied on every lift is wasted, and it is a larger fraction of the total for a small height | ✓ 1 | |
Exam tip: Find a gradient from a large triangle drawn on the line of best fit, using points read from the line rather than individual data points.
The figure shows a lift (elevator) system. The lift car and its passengers rise at a constant speed of 1.5 m s−1 while the counterweight descends at the same speed. A motor drives the pulley. g = 9.81 N kg−1.
Calculate the rate at which the car and passengers gain gravitational potential energy.
Show that the useful power output of the motor is about 3.7 kW.
The motor has an efficiency of 0.65. Calculate its input power.
Calculate the energy supplied to the motor while the car rises through 36 m. Give your answer in kWh.
| Answer | Mark | Notes |
|---|---|---|
| Part (a) | ||
| (1200 + 450) × 9.81 × 1.5 | ✓ 1 | |
| = 2.4 × 104 W (24 300 W) | ✓ 1 | |
| Part (b) | ||
| the counterweight loses gravitational potential energy at 1400 × 9.81 × 1.5 = 20 600 W | ✓ 1 | |
| useful power = 24 280 − 20 600 = 3680 W ≈ 3.7 kW | ✓ 1 | |
| Part (c) | ||
| 3680 ÷ 0.65 = 5.7 kW (5660 W) | ✓ 1 | |
| Part (d) | ||
| time = 36 ÷ 1.5 = 24 s; 5660 × 24 = 1.36 × 105 J = 0.038 kWh (allow error carried forward) | ✓ 1 | |
Exam tip: With a counterweight, the motor supplies only the difference between the rate at which the car side gains gravitational potential energy and the rate at which the counterweight loses it.
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