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Edexcel International A Level Physics · Unit 1 · Mechanics and Materials
40 free, high-quality past-paper-style questions on Work and energy 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 joule?
| Answer | Mark | Notes |
|---|---|---|
| Work done = force × distance, so 1 J = 1 N × 1 m. Since 1 N = 1 kg m s−2, 1 J = 1 kg m2 s−2. | ✓ 1 | Answer D |
Answer: D
Exam tip: To find base units, start from a defining equation (here work = force × distance) and replace each unit by its base units.
A child pulls a toy wagon 15 m along level ground. The handle makes an angle of 30° with the horizontal and the pull along the handle is 24 N.
How much work does the pull do on the wagon?
| Answer | Mark | Notes |
|---|---|---|
| Only the component along the displacement does work: 24 cos 30° × 15 = 310 J (312 J). | ✓ 1 | Answer B |
Answer: B
Common error: The vertical component of the pull, F sin θ, does no work because the wagon does not move vertically.
A sprinter of mass 68 kg runs at 9.5 m s−1.
What is her kinetic energy?
| Answer | Mark | Notes |
|---|---|---|
| Ek = ½mv2 = ½ × 68 × 9.52 = 3070 J = 3.1 kJ. | ✓ 1 | Answer C |
Answer: C
Common error: Forgetting to square the speed, or forgetting the ½. Write ½ × m × v2 in full before using your calculator.
A hiker of mass 72 kg walks from a village 450 m above sea level to a summit 1230 m above sea level. The path is 9.6 km long. g = 9.81 N kg−1.
What is the increase in the hiker's gravitational potential energy?
| Answer | Mark | Notes |
|---|---|---|
| Only the vertical height gained counts: Δh = 1230 − 450 = 780 m. ΔEgrav = 72 × 9.81 × 780 = 5.5 × 105 J. | ✓ 1 | Answer D |
Answer: D
Common error: Using the length of the path. In mgΔh, Δh is the change in vertical height, whatever route is taken.
A hospital porter pushes a patient's bed 35 m along a level corridor. The porter's push is horizontal and constant at 140 N, and the bed moves at constant velocity.
Calculate the work done by the porter on the bed.
Explain why the weight of the bed does no work on it.
State the work done against resistive forces. Give a reason for your answer.
| Answer | Mark | Notes |
|---|---|---|
| Part (a) | ||
| ΔW = FΔs = 140 × 35 | ✓ 1 | |
| = 4.9 × 103 J (4900 J) | ✓ 1 | |
| Part (b) | ||
| the weight acts vertically and the bed moves horizontally / there is no displacement in the direction of the weight | ✓ 1 | |
| Part (c) | ||
| 4.9 × 103 J, because the velocity is constant so the kinetic energy does not change / the resistive force equals the push | ✓ 1 | |
Exam tip: A force does work only when its point of application moves in the direction of the force. Check the direction of every force against the direction of motion.
On a building site, a crane lifts a steel beam of mass 850 kg from the ground to the top of a building 36 m high. g = 9.81 N kg−1.
Calculate the increase in gravitational potential energy of the beam.
The beam is then lowered to a floor 12 m above the ground. Calculate the change in its gravitational potential energy during this lowering.
State an assumption made when using ΔEgrav = mgΔh in this question.
| Answer | Mark | Notes |
|---|---|---|
| Part (a) | ||
| ΔEgrav = 850 × 9.81 × 36 | ✓ 1 | |
| = 3.0 × 105 J (3.00 × 105 J) | ✓ 1 | |
| Part (b) | ||
| 850 × 9.81 × (36 − 12) = 2.0 × 105 J decrease (−2.0 × 105 J) | ✓ 1 | |
| Part (c) | ||
| g is constant over the height change / the heights are small compared with the radius of the Earth | ✓ 1 | |
Common error: Using the beam's final height above the ground in the second part. The change is from 36 m to 12 m, a drop of 24 m.
A gardener pushes a roller across a level lawn. The figure shows the push on the roller, which acts along the handle. The push does 2.2 × 103 J of work on the roller.
How far does the roller move?
| Answer | Mark | Notes |
|---|---|---|
| Only the horizontal component of the push does work: ΔW = F cos θ × Δs. Δs = 2200 ÷ (120 × cos 35°) = 2200 ÷ 98.3 = 22 m. | ✓ 1 | Answer D |
Answer: D
Common error: Multiplying by cos θ when rearranging. The component along the motion is F cos θ, so you divide the work by F cos θ.
A cyclist and bicycle have a total mass of 80 kg. The graph shows how the velocity of the cyclist varies with time.
What is the increase in kinetic energy of the cyclist and bicycle between t = 0 and t = 12 s?
| Answer | Mark | Notes |
|---|---|---|
| From the graph, u = 4.0 m s−1 and v = 10.0 m s−1. ΔEk = ½ × 80 × (10.02 − 4.02) = 40 × 84 = 3360 J = 3.4 kJ. | ✓ 1 | Answer D |
Answer: D
Common error: Squaring the change in speed. ΔEk = ½m(v2 − u2), which is not the same as ½m(v − u)2.
A drone lifts a parcel vertically from the ground. The graph shows how the gravitational potential energy of the parcel varies with its height above the ground. g = 9.81 N kg−1.
What is the mass of the parcel?
| Answer | Mark | Notes |
|---|---|---|
| The gradient of the graph is ΔEgrav ÷ Δh = mg. Using P: mg = 98 ÷ 4.0 = 24.5 N, so m = 24.5 ÷ 9.81 = 2.5 kg. | ✓ 1 | Answer B |
Answer: B
Exam tip: For a straight-line graph, ask what the gradient means physically. Here ΔEgrav = mgΔh, so the gradient is the weight mg.
A pine cone falls from rest from a branch 20 m above the ground. The graph shows how its gravitational potential energy, its kinetic energy and its total energy change with the distance it has fallen.
How much work is done against air resistance during the whole fall?
| Answer | Mark | Notes |
|---|---|---|
| At the start the total energy is 50 J (all gravitational potential). At the ground it is 38 J (all kinetic). The energy transferred by work against air resistance is 50 − 38 = 12 J, the fall in the total-energy line. | ✓ 1 | Answer A |
Answer: A
Exam tip: On an energy–distance graph, the drop in the total-energy line is the work done against resistive forces.
Two people pull a small boat along a level beach on rollers, using two ropes attached to its bow. The figure shows the ropes from above. The boat moves 6.0 m in a straight line at constant velocity.
Calculate the component of the tension in one rope along the direction of motion.
Calculate the total work done on the boat by the two ropes as it moves 6.0 m.
Explain why the components of the tensions at right angles to the direction of motion do no work on the boat.
| Answer | Mark | Notes |
|---|---|---|
| Part (a) | ||
| 180 × cos 25° = 163 N | ✓ 1 | |
| Part (b) | ||
| ΔW = 2 × 163 × 6.0 | ✓ 1 | |
| = 2.0 × 103 J (1960 J) — allow error carried forward from (a) | ✓ 1 | |
| Part (c) | ||
| they are perpendicular to the displacement / the boat does not move sideways (the two sideways components cancel) | ✓ 1 | |
Exam tip: With several forces, find the work done by each one (component along the displacement × displacement) and then add.
A lift and its passengers have a total mass of 900 kg. The graph shows how the velocity of the lift varies with time as it moves upwards from one floor to another. g = 9.81 N kg−1.
Calculate the maximum kinetic energy of the lift.
Use the graph to show that the lift rises 16 m.
Calculate the increase in gravitational potential energy of the lift.
Friction is negligible. State the total work done on the lift by the cable between t = 0 and t = 10 s. Give a reason for your answer.
| Answer | Mark | Notes |
|---|---|---|
| Part (a) | ||
| ½ × 900 × 2.02 = 1.8 × 103 J | ✓ 1 | |
| Part (b) | ||
| distance = area under the velocity–time graph | ✓ 1 | |
| ½ × 2.0 × 2.0 + 6.0 × 2.0 + ½ × 2.0 × 2.0 = 16 m | ✓ 1 | |
| Part (c) | ||
| 900 × 9.81 × 16 = 1.4 × 105 J (1.41 × 105 J) | ✓ 1 | |
| Part (d) | ||
| 1.4 × 105 J, because the lift is at rest at the start and the end, so there is no overall change in kinetic energy and all the work done becomes gravitational potential energy | ✓ 1 | |
Exam tip: The area under a velocity–time graph gives the distance, which you can then use as Δh in mgΔh.
A resultant force does 3.6 J of work on a trolley of mass 0.80 kg that starts from rest on a level track.
What is the final speed of the trolley?
| Answer | Mark | Notes |
|---|---|---|
| The work done by the resultant force becomes kinetic energy: 3.6 = ½ × 0.80 × v2, so v2 = 9.0 and v = 3.0 m s−1. | ✓ 1 | Answer B |
Answer: B
Exam tip: Rearrange Ek = ½mv2 to v = √(2Ek/m) before substituting, and remember the square root at the end.
A child starts from rest at the top of a water slide and slides to the bottom, 4.5 m lower. Resistive forces are negligible. g = 9.81 N kg−1.
What is the child's speed at the bottom of the slide?
| Answer | Mark | Notes |
|---|---|---|
| Loss of GPE = gain in KE: mgΔh = ½mv2, so v = √(2gΔh) = √(2 × 9.81 × 4.5) = 9.4 m s−1. The mass cancels, and the shape of the slide does not matter. | ✓ 1 | Answer B |
Answer: B
Exam tip: With no resistive forces the mass cancels. You do not need it, and the answer is the same for a child or an adult.
In a sorting office, a parcel of mass 2.5 kg slides from rest down a curved chute. The bottom of the chute is 3.0 m lower than the top. The parcel leaves the chute at 5.5 m s−1. g = 9.81 N kg−1.
How much energy is transferred to the surroundings as the parcel slides down the chute?
| Answer | Mark | Notes |
|---|---|---|
| Loss of GPE = 2.5 × 9.81 × 3.0 = 73.6 J. Gain in KE = ½ × 2.5 × 5.52 = 37.8 J. Energy transferred to the surroundings = 73.6 − 37.8 = 36 J. | ✓ 1 | Answer A |
Answer: A
Exam tip: Write the energy balance in words first: loss of GPE = gain in KE + energy transferred to the surroundings.
A gardener pulls the starter cord of a lawnmower. The force on the cord, along the direction the cord moves, increases uniformly from 0 to 150 N over the first 0.40 m and then stays at 150 N for a further 0.60 m.
How much work is done on the cord?
| Answer | Mark | Notes |
|---|---|---|
| Work done = area under the force–displacement graph: ½ × 0.40 × 150 + 0.60 × 150 = 30 + 90 = 120 J. | ✓ 1 | Answer C |
Answer: C
Exam tip: Sketch the force–displacement graph and split the area into a triangle and a rectangle.
A tug tows a cargo ship 250 m into a harbour at constant velocity. The towing cable makes an angle of 15° with the ship's direction of motion. The tension in the cable is 3.8 × 104 N.
Calculate the component of the tension along the ship's direction of motion.
Calculate the work done by the tension as the ship moves 250 m.
State the total work done on the ship by all the forces acting on it. Explain your answer.
| Answer | Mark | Notes |
|---|---|---|
| Part (a) | ||
| 3.8 × 104 × cos 15° = 3.7 × 104 N (3.67 × 104 N) | ✓ 1 | |
| Part (b) | ||
| ΔW = 3.67 × 104 × 250 | ✓ 1 | |
| = 9.2 × 106 J — allow error carried forward from (a) | ✓ 1 | |
| Part (c) | ||
| zero | ✓ 1 | |
| the velocity is constant, so the kinetic energy does not change / the resultant force is zero | ✓ 1 | |
Common error: Using sin 15° for the component along the motion. The component next to the angle uses cos.
A tram of mass 3.2 × 104 kg travels along a straight, level track at 14.0 m s−1. The driver applies the brakes, which exert a constant braking force of 48 kN until the tram stops.
Calculate the kinetic energy of the tram before braking.
Calculate the braking distance.
Explain why the braking distance from 28.0 m s−1, with the same braking force, would be about 260 m.
State what happens to the kinetic energy of the tram as it brakes.
| Answer | Mark | Notes |
|---|---|---|
| Part (a) | ||
| Ek = ½ × 3.2 × 104 × 14.02 | ✓ 1 | |
| = 3.1 × 106 J (3.14 × 106 J) | ✓ 1 | |
| Part (b) | ||
| work done by braking force = kinetic energy: 48 × 103 × d = 3.14 × 106 | ✓ 1 | |
| d = 65 m (65.3 m) | ✓ 1 | |
| Part (c) | ||
| kinetic energy ∝ v2, so doubling the speed gives four times the kinetic energy | ✓ 1 | |
| the same force must do four times the work, so the distance is 4 × 65.3 = 261 m | ✓ 1 | |
| Part (d) | ||
| it is transferred to thermal energy (in the brakes, wheels and rails) and then to the surroundings | ✓ 1 | |
Exam tip: A braking problem is a work–energy problem: braking force × braking distance = kinetic energy removed. You do not need the deceleration.
A skateboarder of mass 58 kg (including the board) starts from rest at the rim of a half-pipe, 3.4 m above its lowest point, and rolls down without pushing. g = 9.81 N kg−1.
Calculate the speed at the lowest point if resistive forces are negligible.
The skateboarder comes momentarily to rest on the opposite side 2.9 m above the lowest point. Calculate the energy transferred to the surroundings.
The distance travelled along the half-pipe between the two rest positions is 9.0 m. Calculate the average resistive force.
State the main form in which the energy in (b) ends up.
| Answer | Mark | Notes |
|---|---|---|
| Part (a) | ||
| ½mv2 = mgΔh, so v = √(2 × 9.81 × 3.4) | ✓ 1 | |
| = 8.2 m s−1 (8.17 m s−1) | ✓ 1 | |
| Part (b) | ||
| loss of GPE between the two rest positions = 58 × 9.81 × (3.4 − 2.9) | ✓ 1 | |
| = 2.8 × 102 J (284 J) | ✓ 1 | |
| Part (c) | ||
| 284 ÷ 9.0 = 32 N — allow error carried forward from (b) | ✓ 1 | |
| Part (d) | ||
| thermal energy (internal energy) of the wheels, ramp and air | ✓ 1 | |
Exam tip: Between two moments at rest the kinetic energy is zero at both ends, so the energy transferred to the surroundings is simply the loss of gravitational potential energy.
A student investigates energy transfers for a trolley rolling down a ramp. The trolley has a mass of 0.850 kg and carries a card of length 50.0 mm. It is released from rest, and when it has descended a vertical height of 0.150 m the card passes through a light gate. The card blocks the light beam for 0.0305 s. g = 9.81 N kg−1.
Calculate the speed of the trolley at the light gate.
Calculate the kinetic energy of the trolley at the light gate.
Calculate the loss of gravitational potential energy of the trolley.
Determine the energy transferred to the surroundings.
The height was measured with a metre rule as 0.150 m ± 0.001 m. Calculate the percentage uncertainty in the height.
Explain why the speed calculated in (a) is a mean speed.
| Answer | Mark | Notes |
|---|---|---|
| Part (a) | ||
| 0.0500 ÷ 0.0305 = 1.64 m s−1 | ✓ 1 | |
| Part (b) | ||
| ½ × 0.850 × 1.642 = 1.14 J | ✓ 1 | |
| Part (c) | ||
| 0.850 × 9.81 × 0.150 = 1.25 J | ✓ 1 | |
| Part (d) | ||
| 1.25 − 1.14 = 0.11 J | ✓ 1 | |
| Part (e) | ||
| 0.001 ÷ 0.150 × 100 = 0.67% | ✓ 1 | |
| Part (f) | ||
| it is the card length divided by the time the card takes to pass, so it is the average over that time; the trolley is still accelerating as the card passes | ✓ 1 | |
Exam tip: With a light gate, speed = length of the card ÷ time the beam is blocked. Convert the card length to metres first.
During an eruption, a volcano throws out a lump of rock of mass 2.4 kg at 45 m s−1. The rock lands on the side of the volcano 120 m below the vent. g = 9.81 N kg−1.
Calculate the kinetic energy of the rock as it leaves the vent.
Calculate the speed at which the rock lands if air resistance is negligible.
Explain why the answer to (b) does not depend on the direction in which the rock is thrown out.
The rock actually lands at 52 m s−1. Calculate the energy transferred to the surroundings during the flight.
| Answer | Mark | Notes |
|---|---|---|
| Part (a) | ||
| ½ × 2.4 × 452 = 2.4 × 103 J (2430 J) | ✓ 1 | |
| Part (b) | ||
| ½v2 = ½ × 452 + 9.81 × 120 | ✓ 1 | |
| v = 66 m s−1 (66.2) | ✓ 1 | |
| Part (c) | ||
| kinetic energy is a scalar / depends only on the speed, not the direction | ✓ 1 | |
| the loss of GPE depends only on the vertical drop of 120 m, not on the path | ✓ 1 | |
| Part (d) | ||
| (2430 + 2.4 × 9.81 × 120) − ½ × 2.4 × 522 = 5255 − 3245 | ✓ 1 | |
| = 2.0 × 103 J (2010 J) | ✓ 1 | |
Exam tip: Energy gives a projectile's speed, but not its direction or time of flight. For those, use the horizontal and vertical components separately.
The graph shows how the force on the handle of a rowing machine varies with the displacement of the handle during one stroke. The force acts in the direction in which the handle moves.
How much work is done on the handle during the stroke?
| Answer | Mark | Notes |
|---|---|---|
| Work done = area under the force–displacement graph. Triangle 0–0.40 m: ½ × 0.40 × 400 = 80 J. Rectangle 0.40–1.20 m: 0.80 × 400 = 320 J. Triangle 1.20–1.40 m: ½ × 0.20 × 400 = 40 J. Total = 440 J. | ✓ 1 | Answer C |
Answer: C
Exam tip: Split the area under a force–displacement graph into triangles and rectangles, find each area with the axis units, then add.
A marble is released from rest at A on the curved track shown in the figure. Resistive forces are negligible. g = 9.81 N kg−1.
What is the speed of the marble at C?
| Answer | Mark | Notes |
|---|---|---|
| The marble falls a vertical height of 0.80 − 0.35 = 0.45 m from A to C. ½mv2 = mgΔh, so v = √(2 × 9.81 × 0.45) = 3.0 m s−1. The shape of the track between A and C does not matter. | ✓ 1 | Answer C |
Answer: C
Common error: Using the height of C above the bench. The energy change depends on the height fallen from the release point, here 0.80 − 0.35 m.
A toy car is launched along a level track by a spring launcher at five different settings. For each launch, the energy given to the car (its kinetic energy Ek) is known from the launcher setting, and the speed v of the car is measured with a light gate. The graph shows Ek plotted against v2.
What is the mass of the toy car?
| Answer | Mark | Notes |
|---|---|---|
| Ek = ½mv2, so the gradient of the graph is ½m. Gradient = 12 ÷ 16 = 0.75 kg, so m = 2 × 0.75 = 1.5 kg. | ✓ 1 | Answer C |
Answer: C
Exam tip: Compare the graph with y = mx: here y = Ek and x = v2, so the gradient is ½m, not m.
A tennis ball of mass 0.058 kg is dropped from rest and bounces on a hard floor. The graph shows how the height of the bottom of the ball above the floor varies with time. Air resistance is negligible. g = 9.81 N kg−1.
Calculate the speed of the ball just before it hits the floor for the first time.
Calculate the energy transferred from the ball to other forms during the first bounce.
Show that the ball keeps about the same fraction of its energy at each bounce. Use this to estimate the height the ball reaches after its third bounce.
| Answer | Mark | Notes |
|---|---|---|
| Part (a) | ||
| ½mv2 = mgh, so v = √(2 × 9.81 × 1.20) | ✓ 1 | |
| = 4.85 m s−1 | ✓ 1 | |
| Part (b) | ||
| heights from the graph: 1.20 m before and 0.80 m after | ✓ 1 | |
| 0.058 × 9.81 × (1.20 − 0.80) = 0.23 J (0.228 J) | ✓ 1 | |
| Part (c) | ||
| 0.80 ÷ 1.20 = 0.67 and 0.55 ÷ 0.80 = 0.69, so roughly the same fraction | ✓ 1 | |
| height ≈ 0.55 × 0.68 = 0.37 m (allow 0.36–0.38 m) | ✓ 1 | |
Exam tip: At the top of each bounce the ball is momentarily at rest, so its energy there is all gravitational potential. Comparing peak heights compares total energies.
A funicular railway car of total mass 4.5 × 103 kg is hauled at constant speed up a straight track by a cable, as shown in the figure. The tension in the cable is 1.80 × 104 N and acts parallel to the track. The car travels the full length of the track. g = 9.81 N kg−1.
Calculate the vertical height gained by the car.
Calculate the increase in gravitational potential energy of the car.
Calculate the work done on the car by the cable.
Use ΔW = FΔs cos θ and the angle shown in the figure to show that the work done on the car by its weight is equal in size and opposite in sign to your answer to (b).
Calculate the average resistive force on the car.
| Answer | Mark | Notes |
|---|---|---|
| Part (a) | ||
| 380 × sin 22° = 142 m | ✓ 1 | |
| Part (b) | ||
| 4.5 × 103 × 9.81 × 142 = 6.3 × 106 J (6.28 × 106 J) | ✓ 1 | |
| Part (c) | ||
| 1.80 × 104 × 380 = 6.84 × 106 J | ✓ 1 | |
| Part (d) | ||
| weight = 4.5 × 103 × 9.81 = 4.41 × 104 N | ✓ 1 | |
| 4.41 × 104 × 380 × cos 112° = −6.28 × 106 J, the negative of (b) | ✓ 1 | |
| Part (e) | ||
| (6.84 − 6.28) × 106 ÷ 380 = 1.5 × 103 N (1.46 × 103 N) — allow error carried forward | ✓ 1 | |
Exam tip: A force at more than 90° to the displacement does negative work. Lifting an object means its weight does negative work, which shows up as a gain in gravitational potential energy.
In a pumped-storage scheme, water is pumped at night from a lower reservoir to an upper reservoir. The figure shows the water levels in both reservoirs before and after one night of pumping. Both reservoirs have vertical sides. During the night 1.5 × 106 m3 of water is pumped. Density of water = 1000 kg m−3. g = 9.81 N kg−1.
Calculate the mass of water pumped.
Use the figure to show that the centre of mass of the water that is moved rises by 320 m.
Calculate the increase in gravitational potential energy of the water.
A student uses 352 m − 19 m as the height change. Explain why this gives too large a value for the energy stored.
| Answer | Mark | Notes |
|---|---|---|
| Part (a) | ||
| 1000 × 1.5 × 106 = 1.5 × 109 kg | ✓ 1 | |
| Part (b) | ||
| centre of the layer removed from the lower reservoir = (31 + 19) ÷ 2 = 25 m; centre of the layer added to the upper reservoir = (338 + 352) ÷ 2 = 345 m | ✓ 1 | |
| 345 − 25 = 320 m | ✓ 1 | |
| Part (c) | ||
| 1.5 × 109 × 9.81 × 320 = 4.7 × 1012 J | ✓ 1 | |
| Part (d) | ||
| 333 m is the rise of only the water that ends at the very top, from the very bottom; on average the water rises by the rise of its centre of mass, 320 m, so 333 m overestimates Δh | ✓ 1 | |
Common error: Using surface levels for a large body of water. Δh in mgΔh is the rise of the centre of mass of the water that is actually moved.
A pole vaulter of mass 70 kg runs up and takes off. The figure shows the height of her centre of mass at take-off and at the top of the vault. At take-off she is moving horizontally at 9.2 m s−1. Treat the vaulter as a point at her centre of mass, and assume that the only energy transfers are between kinetic energy and gravitational potential energy. g = 9.81 N kg−1.
Calculate her kinetic energy at take-off.
Calculate the greatest height her centre of mass could reach if all of this kinetic energy were transferred to gravitational potential energy.
At the top of the vault she is still moving horizontally. Use the figure to calculate her speed at the top.
Suggest why the vaulter needs some horizontal speed at the top of the vault.
| Answer | Mark | Notes |
|---|---|---|
| Part (a) | ||
| ½ × 70 × 9.22 = 3.0 × 103 J (2960 J) | ✓ 1 | |
| Part (b) | ||
| Δh = 9.22 ÷ (2 × 9.81) = 4.31 m | ✓ 1 | |
| height = 1.05 + 4.31 = 5.36 m | ✓ 1 | |
| Part (c) | ||
| ½v2 = ½ × 9.22 − 9.81 × (4.90 − 1.05) | ✓ 1 | |
| v = 3.0 m s−1 (3.02 m s−1) | ✓ 1 | |
| Part (d) | ||
| so that she keeps moving forward over the bar and lands beyond it, rather than dropping back onto the bar | ✓ 1 | |
Exam tip: Energy conservation can be applied between any two points. Write ½mu2 = ½mv2 + mgΔh and cancel m before substituting.
An electric go-kart and its driver have a total mass of 160 kg. The kart starts from rest on a level track. The graph shows how the resultant force on the kart varies with its displacement.
Calculate the work done by the resultant force in the first 20 m.
Calculate the speed of the kart at a displacement of 20 m.
Determine the speed of the kart at a displacement of 60 m.
Describe the motion of the kart between 60 m and 70 m. Give a reason for your answer.
| Answer | Mark | Notes |
|---|---|---|
| Part (a) | ||
| 500 × 20 = 1.0 × 104 J | ✓ 1 | |
| Part (b) | ||
| work done by the resultant force = gain in kinetic energy: 1.0 × 104 = ½ × 160 × v2 | ✓ 1 | |
| v = 11 m s−1 (11.2 m s−1) | ✓ 1 | |
| Part (c) | ||
| area from 20 m to 60 m = ½ × 40 × 500 = 1.0 × 104 J, so total work = 2.0 × 104 J | ✓ 1 | |
| v = √(2 × 2.0 × 104 ÷ 160) = 16 m s−1 (15.8 m s−1) | ✓ 1 | |
| Part (d) | ||
| constant speed (about 16 m s−1), because the resultant force is zero, so no work is done on the kart and its kinetic energy does not change | ✓ 1 | |
Exam tip: The area under a resultant force–displacement graph equals the change in kinetic energy. A falling force still does positive work: the kart keeps speeding up until the force reaches zero.
A car travelling at 18 m s−1 brakes on a level road and stops in 24 m. The braking force is constant.
With the same braking force, what would the braking distance be from 27 m s−1?
| Answer | Mark | Notes |
|---|---|---|
| The work done by the braking force equals the kinetic energy removed: Fd = ½mv2, so d ∝ v2. d = 24 × (27/18)2 = 54 m. | ✓ 1 | Answer C |
Answer: C
Exam tip: In a ratio question, write the proportionality first (d ∝ v2) and scale; you do not need the mass or the force.
A playground swing has chains 2.4 m long. A child on the swing is pulled back until the chains make 40° with the vertical, and is then released from rest. Resistive forces are negligible. g = 9.81 N kg−1.
What is the speed of the child at the lowest point of the swing?
| Answer | Mark | Notes |
|---|---|---|
| The vertical drop is h = L(1 − cos 40°) = 2.4 × (1 − 0.766) = 0.56 m. Then v = √(2gh) = √(2 × 9.81 × 0.562) = 3.3 m s−1. | ✓ 1 | Answer A |
Answer: A
Common error: L cos θ is the height of the pivot above the released child, not the drop. The drop is L − L cos θ.
At a polar research station, a loaded cargo sledge of total mass 75 kg is released from rest at the top of a straight snow slope. The slope is 120 m long and makes an angle of 18° with the horizontal. The average resistive force on the sledge is 95 N. g = 9.81 N kg−1.
Show that the sledge descends a vertical height of about 37 m.
Calculate the loss of gravitational potential energy of the sledge.
Calculate the work done against the resistive force.
Calculate the speed of the sledge at the bottom of the slope.
Calculate the speed at the bottom if the resistive force were negligible, and explain why this speed does not depend on the mass of the sledge.
| Answer | Mark | Notes |
|---|---|---|
| Part (a) | ||
| 120 × sin 18° = 37.1 m | ✓ 1 | |
| Part (b) | ||
| 75 × 9.81 × 37.1 = 2.73 × 104 J | ✓ 1 | |
| Part (c) | ||
| 95 × 120 = 1.14 × 104 J | ✓ 1 | |
| Part (d) | ||
| gain in KE = 2.73 × 104 − 1.14 × 104 = 1.59 × 104 J | ✓ 1 | |
| v = √(2 × 1.59 × 104 ÷ 75) = 20.6 m s−1 | ✓ 1 | |
| Part (e) | ||
| v = √(2 × 9.81 × 37.1) = 27.0 m s−1 | ✓ 1 | |
| both the loss of GPE (mgΔh) and the gain in KE (½mv2) are proportional to m, so m cancels | ✓ 1 | |
Common error: Multiplying the resistive force by the vertical height. Work done against a resistive force uses the distance moved along the path, here 120 m.
A student pulls a wooden block of mass 0.94 kg along a horizontal bench using a force sensor connected to a data logger. The pull is always horizontal and the frictional force on the moving block is constant. The block is at rest at displacement s = 0, where the pull has just reached 6.0 N and the block starts to move. The data logger's force–displacement graph is made of straight-line sections: the force falls from (0, 6.0 N) to (0.10 m, 4.5 N) and then stays at 4.5 N until s = 0.80 m. Between 0.10 m and 0.80 m the block moves at a constant speed of 0.40 m s−1.
State what is represented by the area under a force–displacement graph.
Determine the total work done by the student over the 0.80 m.
State the frictional force on the block, giving a reason.
At 0.80 m the student lets go and the block slides to rest. Calculate the further distance the block slides.
| Answer | Mark | Notes |
|---|---|---|
| Part (a) | ||
| work done (by the force) | ✓ 1 | |
| Part (b) | ||
| trapezium 0 to 0.10 m: ½ × (6.0 + 4.5) × 0.10 = 0.525 J | ✓ 1 | |
| rectangle 0.10 to 0.80 m: 0.70 × 4.5 = 3.15 J | ✓ 1 | |
| total = 3.7 J (3.675 J) | ✓ 1 | |
| Part (c) | ||
| 4.5 N, because between 0.10 m and 0.80 m the block moves at constant speed, so the resultant force is zero | ✓ 1 | |
| Part (d) | ||
| Ek = ½ × 0.94 × 0.402 = 0.075 J | ✓ 1 | |
| d = 0.075 ÷ 4.5 = 0.017 m (1.7 cm) | ✓ 1 | |
Exam tip: When a graph is described in words, sketch it with the key coordinates before you split the area into shapes.
A mobile phone of mass 0.190 kg slides off a table and falls 0.750 m from rest onto a carpet. The carpet compresses by 4.0 mm as it brings the phone to rest. Air resistance is negligible. g = 9.81 N kg−1.
Calculate the speed of the phone as it reaches the carpet.
Calculate the average force exerted by the carpet on the phone as it stops.
Suggest how a soft phone case reduces the force on the phone in a fall like this.
| Answer | Mark | Notes |
|---|---|---|
| Part (a) | ||
| ½mv2 = mgh, so v = √(2 × 9.81 × 0.750) | ✓ 1 | |
| = 3.84 m s−1 | ✓ 1 | |
| Part (b) | ||
| energy removed by the carpet = loss of GPE over the whole fall, 0.750 m + 0.0040 m | ✓ 1 | |
| 0.190 × 9.81 × 0.754 = 1.41 J | ✓ 1 | |
| F = 1.41 ÷ 0.0040 = 3.5 × 102 N (351 N) | ✓ 1 | |
| allow the equivalent method, one mark per step: ½ × 0.190 × 3.842 = 1.40 J; ÷ 0.0040 = 349 N (+ mg = 351 N); = 3.5 × 102 N — allow error carried forward from (a) | — | |
| Part (c) | ||
| the case deforms, increasing the stopping distance, so the same energy is removed by a smaller average force | ✓ 1 | |
Exam tip: In a “stopping” problem, energy to be removed ÷ stopping distance = average stopping force. Include the drop during the stopping distance if it is not negligible.
A worker pushes a crate up a ramp with a constant horizontal force, as shown in the figure.
How much work does the horizontal force do on the crate as the crate moves 4.0 m up the slope?
| Answer | Mark | Notes |
|---|---|---|
| The angle between the horizontal force and the displacement (along the slope) is 20°. ΔW = FΔs cos 20° = 250 × 4.0 × cos 20° = 940 J = 0.94 kJ. | ✓ 1 | Answer B |
Answer: B
Exam tip: In ΔW = FΔs cos θ, θ is the angle between the force and the displacement. Find it from the diagram before choosing sin or cos.
A car-park barrier is a uniform arm of mass 12 kg, hinged at one end. The figure shows the arm lowered and raised. g = 9.81 N kg−1.
What is the increase in gravitational potential energy of the arm when it is raised?
| Answer | Mark | Notes |
|---|---|---|
| The centre of mass is at the middle of the arm, 1.5 m from the hinge. It rises 1.5 × sin 50° = 1.15 m. ΔEgrav = 12 × 9.81 × 1.15 = 140 J (135 J). | ✓ 1 | Answer B |
Answer: B
Common error: Using the rise of the free end. For an extended object, Δh in mgΔh is the rise of its centre of mass.
A puck of mass 0.50 kg is sent up a rough ramp inclined at 30° to the horizontal. The graph shows how its kinetic energy varies with the distance it moves up the ramp until it stops. The frictional force on the puck is constant. g = 9.81 N kg−1.
What is the frictional force on the puck?
| Answer | Mark | Notes |
|---|---|---|
| The kinetic energy lost per metre is 6.0 ÷ 1.6 = 3.75 N. This equals the component of the weight down the slope plus the friction: 3.75 = 0.50 × 9.81 × sin 30° + F = 2.45 + F, so F = 1.3 N. | ✓ 1 | Answer A |
Answer: A
Exam tip: The gradient of a kinetic energy–distance graph is the resultant force along the path. Split it into the separate forces acting along the slope.
A student investigates energy transfers using a trolley on a level track that has been compensated for friction. The trolley is attached by a light string, passing over a pulley, to a hanging mass of 0.200 kg. The system is released from rest, and a light gate measures the speed v of the trolley when the hanging mass has fallen through a height h. The graph shows the results. g = 9.81 N kg−1.
Show that v2 = 2mgh ÷ (M + m), where M is the mass of the trolley and m is the hanging mass.
Determine the gradient of the graph.
Use your answer to (b) to determine the mass M of the trolley.
Explain how the gradient would change if the track were not compensated for friction.
| Answer | Mark | Notes |
|---|---|---|
| Part (a) | ||
| loss of gravitational potential energy of the hanging mass, mgh = gain in kinetic energy of the trolley and the hanging mass | ✓ 1 | |
| mgh = ½(M + m)v2, which rearranges to the equation given | ✓ 1 | |
| Part (b) | ||
| large triangle on the line of best fit, e.g. 2.75 ÷ 0.70 | ✓ 1 | |
| gradient = 3.9 m s−2 (allow 3.8–4.0) | ✓ 1 | |
| Part (c) | ||
| M + m = 2 × 0.200 × 9.81 ÷ 3.9 = 1.0 kg | ✓ 1 | |
| M = 1.0 − 0.200 = 0.80 kg (allow 0.78–0.84 kg) — allow error carried forward from (b) | ✓ 1 | |
| Part (d) | ||
| the gradient would be smaller, because some of the gravitational potential energy lost would be used to do work against friction, so less becomes kinetic energy for each height fallen | ✓ 1 | |
Exam tip: Both masses speed up together, so the kinetic energy gained is ½(M + m)v2, while only the hanging mass loses gravitational potential energy.
An archer draws back the string of a bow. The graph shows how the force needed to hold the string varies with the draw distance.
State what is represented by the area under the graph.
Estimate the work done in drawing the bow to 0.50 m.
A student estimates the work as ½ × 200 × 0.50 = 50 J. Explain why this is an underestimate.
The arrow has a mass of 0.025 kg and leaves the bow with 44 J of kinetic energy. Calculate the speed of the arrow and suggest what happens to the rest of the work done in drawing the bow.
| Answer | Mark | Notes |
|---|---|---|
| Part (a) | ||
| the work done (energy transferred) in drawing the bow | ✓ 1 | |
| Part (b) | ||
| count the squares under the curve (or split the area into trapezia) | ✓ 1 | |
| each grid square represents 0.05 m × 20 N = 1.0 J | ✓ 1 | |
| work ≈ 56 J (allow 52–59 J) | ✓ 1 | |
| Part (c) | ||
| the curve lies above the straight line from the origin to (0.50 m, 200 N), so the area under the curve is greater than the area of the triangle | ✓ 1 | |
| Part (d) | ||
| v = √(2 × 44 ÷ 0.025) = 59 m s−1 (59.3 m s−1) | ✓ 1 | |
| the rest becomes kinetic energy of the bow limbs and string, and thermal energy and sound | ✓ 1 | |
Exam tip: When counting squares, first work out the energy one square represents (width × height in the axis units), then count whole squares and combine the part squares.
A ski jumper of mass 68 kg starts from rest at A and travels down the straight in-run to the take-off point B, as shown in the figure. She leaves B at 25.0 m s−1 and lands at C. g = 9.81 N kg−1.
Calculate the speed she would have at B if resistive forces were negligible.
Calculate the work done against resistive forces between A and B.
Calculate the average resistive force on her along the in-run.
Air resistance is negligible during the flight from B to C. Calculate her speed when she lands at C.
| Answer | Mark | Notes |
|---|---|---|
| Part (a) | ||
| ½mv2 = mgΔh with Δh = 105 − 63 = 42 m | ✓ 1 | |
| v = √(2 × 9.81 × 42) = 28.7 m s−1 | ✓ 1 | |
| Part (b) | ||
| loss of GPE = 68 × 9.81 × 42 = 2.80 × 104 J and kinetic energy at B = ½ × 68 × 25.02 = 2.125 × 104 J | ✓ 1 | |
| work against resistive forces = 6.8 × 103 J (6770 J) | ✓ 1 | |
| Part (c) | ||
| 6770 ÷ 98 = 69 N — allow error carried forward from (b) | ✓ 1 | |
| Part (d) | ||
| ½v2 = ½ × 25.02 + 9.81 × (63 − 26) | ✓ 1 | |
| v = 36.8 m s−1 | ✓ 1 | |
Common error: Using the vertical drop as the distance for the resistive force. Work done against a resistive force uses the distance along the path, here 98 m along the in-run.
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