Edexcel International A Level Maths · M2 · Mechanics 2
Work, energy and power
28 free, high-quality past-paper-style questions on Work, energy and power for Edexcel IAL Maths, each with a full mark scheme. For when you have finished the past papers, or want to practise one topic at a time.IAL Maths notes and flashcards →Not sure of your level? Diagnostic test →Multiple choice or structured questions, with the mark scheme as a separate PDF.28 new this month · 245 marks · Updated 26 October 2026 · Created by Evan Kotronis with AI assistance · How questions are checked
1IALM-M2-4-B-01M2 · A2Easy6 marks
A crate of mass 250 kg is raised vertically through a height of 12 m by a crane cable. The crate moves at a constant speed of 0.40 m s−1 throughout. The crate is modelled as a particle and air resistance is ignored. Take g = 9.8 m s−2.
(a)
Find the gain in gravitational potential energy of the crate.
(2)
(b)
State the tension in the cable, giving a reason.
(1)
(c)
Find the rate at which the crane is working.
(2)
(d)
Verify your answer to (c) using your answer to (a).
(1)
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Answer
Mark
Notes
Part (a)
mgh = 250 × 9.8 × 12
M1
= 29 400 J
A1
Part (b)
constant speed so the resultant force is zero: T = 250 × 9.8 = 2450 N
B1
Part (c)
P = Fv = 2450 × 0.40
M1
= 980 W
A1
Part (d)
time = 12 ÷ 0.40 = 30 s and 29 400 ÷ 30 = 980 W ✓
B1
Exam tip: At constant speed the pulling force equals the weight, so the power is simply mgv.
A suitcase of mass 18 kg is pulled across a rough horizontal floor by its handle. The force in the handle has magnitude 30 N and acts at 40° above the horizontal. The suitcase moves in a straight line at constant speed. It is modelled as a particle. Take g = 9.8 m s−2.
Figure: the suitcase pulled by the 30 N force at 40° above the horizontal; not to scale.
(a)
Find the work done by the 30 N force as the suitcase moves 50 m.
(2)
(b)
Find the normal reaction between the suitcase and the floor.
(2)
(c)
Find the coefficient of friction between the suitcase and the floor.
(2)
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Answer
Mark
Notes
Part (a)
30 cos 40° × 50
M1
= 1150 J (1149)
A1
Part (b)
R + 30 sin 40° = 18 × 9.8
M1
R = 157 N (157.1)
A1
Part (c)
constant speed so F = 30 cos 40° = 22.98 N and F = μR
M1
μ = 22.98 ÷ 157.1 = 0.146
A1
Common error: Taking R = mg. The upward component of the pull, 30 sin 40°, reduces the normal reaction.
A stone of mass 0.3 kg is thrown from a bridge with speed 6 m s−1. The point of release is 20 m above the surface of a river. The stone is modelled as a particle moving freely under gravity. Take g = 9.8 m s−2.
(a)
Find the kinetic energy of the stone as it is released.
(1)
(b)
Use the principle of conservation of mechanical energy to find the speed of the stone as it reaches the river.
(3)
(c)
Explain why your answer to (b) does not depend on the direction in which the stone is thrown.
(1)
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Answer
Mark
Notes
Part (a)
½ × 0.3 × 62 = 5.4 J
B1
Part (b)
PE lost = 0.3 × 9.8 × 20 = 58.8 J
B1
5.4 + 58.8 = ½ × 0.3 × v2
M1
v = 20.7 m s−1
A1
Part (c)
kinetic energy depends only on speed, and the PE lost depends only on the vertical drop, so the energy equation is the same for any direction
B1
Exam tip: When no angle of projection is given, an energy method is almost certainly intended.
A car of mass 1200 kg moves along a straight horizontal road. Its engine works at a constant rate of 36 kW and the resistance to motion is constant. The maximum speed of the car on this road is 40 m s−1.
(a)
Find the magnitude of the resistance.
(2)
(b)
Find the acceleration of the car when its speed is 15 m s−1.
(3)
(c)
Explain why the acceleration of the car decreases as its speed increases.
(1)
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Answer
Mark
Notes
Part (a)
at maximum speed driving force = resistance: 36 000 ÷ 40
M1
= 900 N
A1
Part (b)
driving force = 36 000 ÷ 15 = 2400 N
M1
2400 − 900 = 1200a
M1
a = 1.25 m s−2
A1
Part (c)
the driving force P/v decreases as v increases while the resistance stays the same, so the resultant force decreases
B1
Exam tip: Maximum speed means zero acceleration, so the driving force P/v equals the total resistance.
Figure 1 shows a sledge of mass 12 kg being pulled across horizontal snow by a rope. The tension in the rope is 48 N and the rope is inclined at 30° above the horizontal throughout. The resistance to motion is constant, of magnitude 20 N. The sledge starts from rest at A and moves in a straight line to B, where AB = 25 m. The sledge is modelled as a particle.
(a)
Find the work done by the tension as the sledge moves from A to B.
(2)
(b)
Use the work–energy principle to find the speed of the sledge at B.
(3)
(c)
Find the rate at which the tension is working at the instant the sledge reaches B.
(2)
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Answer
Mark
Notes
Part (a)
48 cos 30° × 25
M1
= 1040 J (1039.2)
A1
Part (b)
work done against the resistance = 20 × 25 = 500 J
B1
½ × 12 × v2 = 1039.2 − 500 (= 539.2)
M1
v = 9.48 m s−1
A1
Part (c)
P = 48 cos 30° × 9.480 (= 41.57 × 9.480)
M1
= 394 W (ft from (b))
A1
Common error: Using the full 48 N in the work or power. Only the component of the tension along the motion, 48 cos 30°, does work.
A small bead of mass 0.2 kg is threaded on a smooth wire which is fixed in a vertical plane. Figure 1 shows the profile of the wire, with the heights of the points A, B, C, E and D above a fixed horizontal level shown on the grid. The bead is released from rest at A. Take g = 9.8 m s−2.
(a)
Find the speed of the bead at B.
(2)
(b)
Find the speed of the bead as it passes over C.
(2)
(c)
The bead is now projected from A along the wire towards B. Find the least speed of projection that allows the bead to reach D.
(2)
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Answer
Mark
Notes
Part (a)
½ × 0.2 × v2 = 0.2 × 9.8 × (5.0 − 1.0)
M1
v = 8.85 m s−1
A1
Part (b)
½v2 = 9.8 × (5.0 − 3.5)
M1
v = 5.42 m s−1
A1
Part (c)
at least enough energy to rise from 5.0 m to 6.0 m (the highest point after A): ½u2 = 9.8 × 1.0
M1
u = 4.43 m s−1
A1
Exam tip: On a smooth wire only the change in height matters: the shape of the path between two points does not affect the speed, and the mass cancels.
A motorboat of mass 600 kg moves in a straight line across a lake. Its engine works at a constant rate of P watts. Figure 1 shows the driving force of the engine plotted against the speed v of the boat, together with the resistance to motion, which is modelled as constant.
(a)
Use the point on the curve where v = 9 to find P.
(2)
(b)
Write down the magnitude of the resistance and hence find the maximum speed of the boat.
(2)
(c)
Find the acceleration of the boat at the instant when its speed is 8 m s−1.
(2)
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Answer
Mark
Notes
Part (a)
P = Fv = 2000 × 9
M1
= 18 000 W (18 kW)
A1
Part (b)
resistance = 1500 N; at maximum speed the driving force equals the resistance, 18 000 ÷ v = 1500
M1
v = 12 m s−1 (where the curve meets the line)
A1
Part (c)
driving force = 18 000 ÷ 8 = 2250 N and 2250 − 1500 = 600a
M1
a = 1.25 m s−2
A1
Exam tip: On a driving force–speed graph for constant power, every point satisfies Fv = P; the maximum speed is where the curve meets the resistance line.
Figure 1 shows a child on a rope swing. The rope has length 4.5 m and is attached to a fixed point O. The child is released from rest at A, where the rope makes an angle of 40° with the vertical, swings through C, where the rope makes an angle of 25° with the vertical, and reaches the lowest point B. The rope is modelled as light and inextensible, the child as a particle of mass 28 kg, and air resistance is ignored. Take g = 9.8 m s−2.
(a)
Show that the height h of A above B is 1.05 m, correct to 3 significant figures.
(2)
(b)
Find the speed of the child at B.
(2)
(c)
Find the speed of the child at C.
(2)
(d)
Explain why the tension in the rope does no work on the child.
(1)
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Answer
Mark
Notes
Part (a)
h = 4.5 − 4.5 cos 40°
M1
= 1.0528… = 1.05 m
A1
Part (b)
½ × 28 × v2 = 28 × 9.8 × 1.0528
M1
v = 4.54 m s−1
A1
Part (c)
height fallen from A to C = 4.5 cos 25° − 4.5 cos 40° = 0.6312 m
M1
v = √(2 × 9.8 × 0.6312) = 3.52 m s−1
A1
Part (d)
the tension acts along the rope, perpendicular to the child's circular path (direction of motion), so it does no work
B1
Exam tip: For a particle on a string, measure heights from the fixed point: the depth below O is L cos θ, so the drop between two positions is L(cos θ2 − cos θ1).
A child of mass 30 kg slides from rest down a straight playground slide of length 6 m, inclined at 35° to the horizontal. The coefficient of friction between the child and the slide is 0.3. At the bottom the child moves onto rough horizontal ground with no change in speed, and the coefficient of friction between the child and the ground is 0.4. The child is modelled as a particle and air resistance is ignored. Take g = 9.8 m s−2.
Figure: the child starts from rest at the top of the 6 m slide, then moves onto the ground; not to scale.
(a)
Use the work–energy principle to find the speed of the child at the bottom of the slide.
(5)
(b)
Find the distance the child slides along the ground before coming to rest.
(3)
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Answer
Mark
Notes
Part (a)
PE lost = 30 × 9.8 × 6 sin 35° = 1011.8 J
B1
R = 30 × 9.8 cos 35° = 240.8 N, F = 0.3R = 72.25 N
M1
work against friction = 72.25 × 6 = 433.5 J
A1
½ × 30v2 = 1011.8 − 433.5
M1
v = 6.21 m s−1
A1
Part (b)
friction on the ground = 0.4 × 30 × 9.8 = 117.6 N
B1
117.6d = ½ × 30 × 6.2092 (= 578.3)
M1
d = 4.92 m (ft from (a))
A1
Common error: Using the slide's coefficient of friction on the ground, or R = mg on the slide. Each surface needs its own normal reaction.
A cyclist and her bicycle have a total mass of 85 kg. She rides up a straight road inclined at an angle α to the horizontal, where sin α = 130. The resistance to motion from non-gravitational forces is modelled as a constant force of magnitude 25 N. Take g = 9.8 m s−2.
(a)
The cyclist rides up the road at a constant speed of 5 m s−1. Find the rate at which she is working.
(3)
(b)
She increases her rate of working to 400 W. Find her acceleration at the instant her speed is 5 m s−1.
(3)
(c)
Find the maximum speed she can reach up the road while working at 400 W.
(2)
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Answer
Mark
Notes
Part (a)
component of weight = 85 × 9.8 × 1/30 = 27.77 N
B1
driving force = 25 + 27.77 = 52.77 N
M1
P = 52.77 × 5 = 264 W
A1
Part (b)
driving force = 400 ÷ 5 = 80 N
M1
80 − 25 − 27.77 = 85a
M1
a = 0.320 m s−2
A1
Part (c)
400 ÷ v = 52.77
M1
v = 7.58 m s−1
A1
Exam tip: On a slope given as sin α = 1/n, the weight component is simply mg/n; there is no need to find α.
A box of mass 4 kg is projected with speed 7 m s−1 up a line of greatest slope of a rough ramp. The ramp is inclined at an angle α to the horizontal, where tan α = 34. The coefficient of friction between the box and the ramp is 0.5. The box is modelled as a particle. Take g = 9.8 m s−2.
Figure: the box projected up the rough ramp; tan α = 3/4; not to scale.
(a)
Use the work–energy principle to find the distance the box travels up the ramp before it comes to instantaneous rest.
(5)
(b)
Show that the box does not remain at rest.
(2)
(c)
Find the speed of the box when it returns to its point of projection.
(2)
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Answer
Mark
Notes
Part (a)
sin α = 0.6, cos α = 0.8; R = 4 × 9.8 × 0.8 = 31.36 N
B1
F = 0.5 × 31.36 = 15.68 N
M1
PE gained = 4 × 9.8 × 0.6d = 23.52d
B1
½ × 4 × 72 = 23.52d + 15.68d
M1
d = 2.5 m
A1
Part (b)
component of weight down the ramp = 23.52 N, maximum friction = 15.68 N
M1
23.52 > 15.68 so the box slides back down
A1
Part (c)
½ × 4 × v2 = 23.52 × 2.5 − 15.68 × 2.5 (= 19.6)
M1
v = 3.13 m s−1
A1
Common error: Forgetting that friction reverses direction when the box slides back, but still opposes the motion — work is done against it on both journeys.
A lorry of mass 16 000 kg is travelling at 25 m s−1 when its engine is disengaged and its brakes are applied. The total resistance to motion, including braking, is modelled as a constant force of magnitude R newtons. Take g = 9.8 m s−2.
(a)
On a straight horizontal road the lorry comes to rest in a distance of 80 m. Use the work–energy principle to find R.
(3)
(b)
The lorry is now travelling at 25 m s−1 down a straight road inclined at α to the horizontal, where sin α = 114, when the same braking begins. The resistance is unchanged. Find the distance the lorry travels before it comes to rest.
(4)
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Answer
Mark
Notes
Part (a)
KE = ½ × 16 000 × 252 = 5 000 000 J
B1
80R = 5 000 000
M1
R = 62 500
A1
Part (b)
PE lost over distance d = 16 000 × 9.8 × d/14 = 11 200d
B1
KE lost + PE lost = work done against R
M1
5 000 000 + 11 200d = 62 500d
A1
d = 97.5 m
A1
Exam tip: Downhill, the potential energy lost adds to the kinetic energy that the brakes must remove, so the stopping distance is longer.
A pump lifts water from rest at the surface of an underground tank and delivers it through a pipe whose outlet is 15 m above the surface of the water in the tank. The water leaves the outlet with speed 8 m s−1. The pump delivers 30 kg of water every minute. Assume that no energy is wasted. Take g = 9.8 m s−2.
(a)
Find the mass of water delivered each second.
(1)
(b)
Find the rate at which the pump is working.
(4)
(c)
The pump is replaced by one that works at 150 W, lifting water through the same height and delivering it at the same speed. Find the mass of water it delivers each second.
(2)
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Answer
Mark
Notes
Part (a)
30 ÷ 60 = 0.5 kg
B1
Part (b)
PE gained per second = 0.5 × 9.8 × 15 = 73.5 J
B1
KE gained per second = ½ × 0.5 × 82 = 16 J
B1
power = 73.5 + 16
M1
= 89.5 W
A1
Part (c)
energy needed per kilogram = 9.8 × 15 + ½ × 82 = 179 J, so 179m = 150
M1
m = 0.838 kg
A1
Exam tip: For a pump, work out the energy given to the water each second: that is the power. Remember both the potential energy and the kinetic energy.
A particle A of mass 3 kg lies on a rough horizontal table. It is attached to one end of a light inextensible string which passes over a small smooth pulley fixed at the edge of the table. A particle B of mass 2 kg hangs freely from the other end of the string. The coefficient of friction between A and the table is 0.25. The system is released from rest with the string taut, and B falls 1.2 m without reaching the floor while A stays on the table. Take g = 9.8 m s−2.
Figure: A on the rough table connected over the pulley to B, released from rest; not to scale.
(a)
Find the work done against friction while B falls 1.2 m.
(2)
(b)
Use the work–energy principle to find the speed of B when it has fallen 1.2 m.
(4)
(c)
Explain why the tension in the string does not appear in your energy equation in (b).
(1)
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Answer
Mark
Notes
Part (a)
F = 0.25 × 3 × 9.8 = 7.35 N
M1
work = 7.35 × 1.2 = 8.82 J
A1
Part (b)
PE lost by B = 2 × 9.8 × 1.2 = 23.52 J
B1
PE lost = KE gained by both particles + work against friction
M1
23.52 = ½ × 5 × v2 + 8.82
A1
v = 2.42 m s−1
A1
Part (c)
the tension does equal amounts of positive work on A and negative work on B (same force, same distance), so its total work on the system is zero
B1
Common error: Giving only B kinetic energy. Both particles move at the same speed, so the kinetic energy gained is ½(3 + 2)v2.
A block of mass 5 kg is pulled from rest across a rough horizontal floor by a constant horizontal force of magnitude P newtons. After 4 s the force is removed and the block slides to rest. Figure 1 shows the speed–time graph of the motion. The frictional force on the block is constant and the block is modelled as a particle. Take g = 9.8 m s−2.
(a)
Use Figure 1 to find the distance travelled by the block after the force is removed.
(1)
(b)
Use the work–energy principle to find the magnitude of the frictional force.
(2)
(c)
Find the coefficient of friction between the block and the floor.
(2)
(d)
Use the work–energy principle for the first 4 s of the motion to find P.
(3)
(e)
Find the rate at which the force is working at the instant it is removed.
(1)
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Answer
Mark
Notes
Part (a)
area = ½ × 6 × 6 = 18 m
B1
Part (b)
KE lost = ½ × 5 × 62 = 90 J = F × 18
M1
F = 5 N
A1
Part (c)
R = 5 × 9.8 = 49 N, μ = 5 ÷ 49
M1
μ = 0.102
A1
Part (d)
distance in the first 4 s = ½ × 4 × 6 = 12 m
B1
12P = 90 + 5 × 12
M1
P = 12.5
A1
Part (e)
12.5 × 6 = 75 W
B1
Exam tip: A speed–time graph gives distances (areas) and speeds; combine them with the work–energy principle instead of finding accelerations.
Figure 1 shows a straight road PQ which rises 15 m over a distance of 500 m measured along the road, followed by a straight road QR which falls 10 m over a distance of 400 m measured along the road. A lorry of mass 10 000 kg travels from P to R. The resistance to motion from non-gravitational forces is constant, of magnitude 3000 N. The lorry is modelled as a particle and no energy is lost as it passes over Q. Take g = 9.8 m s−2.
(a)
The lorry moves from P to Q at a constant speed of 12 m s−1. Find the rate at which its engine is working, giving your answer in kW.
(3)
(b)
At Q the engine is switched off. The speed of the lorry at Q is 12 m s−1. Use the work–energy principle to find the speed of the lorry at R.
(4)
(c)
Show that the lorry slows down along QR, and find the magnitude of its deceleration.
(2)
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Answer
Mark
Notes
Part (a)
component of weight down PQ = 10 000 × 9.8 × 15/500 = 2940 N
B1
driving force = 3000 + 2940 = 5940 N
M1
P = 5940 × 12 = 71 280 W = 71.3 kW
A1
Part (b)
KE at Q = ½ × 10 000 × 122 = 720 000 J
B1
PE lost = 10 000 × 9.8 × 10 = 980 000 J and work against resistance = 3000 × 400 = 1 200 000 J
B1
½ × 10 000v2 = 720 000 + 980 000 − 1 200 000
M1
v = 10 m s−1
A1
Part (c)
weight component down QR = 10 000 × 9.8 × 10/400 = 2450 N < 3000 N, so the resultant 550 N opposes the motion
M1
deceleration = 550 ÷ 10 000 = 0.055 m s−2
A1
Exam tip: A rise of h over a distance d along the road gives sin α = h/d directly; for energy you need only the vertical height.
A box of mass 8 kg rests on rough horizontal ground. The coefficient of friction between the box and the ground is 0.2. The box is pulled in a straight line by a horizontal force of magnitude P newtons. Figure 1 shows how P varies with the distance x metres moved by the box from its starting point. The box is modelled as a particle. Take g = 9.8 m s−2.
(a)
Find the total work done by the pulling force.
(2)
(b)
Find the speed of the box when x = 12.
(3)
(c)
Find the total distance moved by the box from its starting point before it comes to rest.
(2)
(d)
Explain why the speed of the box is greatest when x = 12.
(1)
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Answer
Mark
Notes
Part (a)
40 × 5 + 25 × 7
M1
= 375 J
A1
Part (b)
friction = 0.2 × 8 × 9.8 = 15.68 N
B1
½ × 8v2 = 375 − 15.68 × 12 (= 186.84)
M1
v = 6.83 m s−1
A1
Part (c)
15.68d = 186.84, d = 11.92 m
M1
total = 12 + 11.92 = 23.9 m
A1
Part (d)
for x < 12 the pulling force (40 N or 25 N) is greater than the friction 15.68 N so the box speeds up; for x > 12 only friction acts so it slows down
B1
Exam tip: When a force is constant over separate stages, its work is the area under the force–distance graph: add the work stage by stage.
Figure 1 shows a conveyor belt of length 8.0 m inclined at 20° to the horizontal. Parcels, each of mass 5 kg, are placed at rest on the belt at the bottom end, and each leaves the top end moving with the belt at 1.5 m s−1. Parcels are delivered at a steady rate of 12 per minute. Each parcel is modelled as a particle. Take g = 9.8 m s−2.
(a)
Find the total energy gained by one parcel as it moves from the bottom to the top of the belt.
(3)
(b)
Find the rate at which energy is given to the parcels.
(2)
(c)
The motor driving the belt works at a constant rate of 45 W. Find the energy, per minute, that is not transferred to the parcels as kinetic or potential energy.
(2)
(d)
Find the frictional force exerted by the belt on a parcel that is moving with the belt at constant speed.
(1)
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Answer
Mark
Notes
Part (a)
PE gained = 5 × 9.8 × 8.0 sin 20° = 134.1 J
B1
KE gained = ½ × 5 × 1.52 = 5.625 J
B1
total = 140 J (139.7)
A1
Part (b)
12 per minute = 0.2 per second; 0.2 × 139.7
M1
= 27.9 W
A1
Part (c)
45 × 60 − 12 × 139.7
M1
= 1020 J (1024)
A1
Part (d)
F = 5 × 9.8 sin 20° = 16.8 N up the belt
B1
Common error: Forgetting the kinetic energy. Each parcel starts at rest, so it gains ½mv2 as well as mgh.
Figure 1 shows a car of mass 1200 kg towing a trailer of mass 400 kg up a straight road inclined at an angle α to the horizontal, where sin α = 114. The trailer is attached to the car by a light rigid tow bar which is parallel to the road. The resistances to motion from non-gravitational forces are constant, 300 N on the car and 100 N on the trailer, as shown. The engine of the car works at a constant rate of 30 kW. Take g = 9.8 m s−2.
(a)
Find the maximum speed of the car and trailer up the road.
(3)
(b)
Find the acceleration of the car and trailer at the instant when their speed is 10 m s−1.
(3)
(c)
Find the tension in the tow bar at this instant.
(3)
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Answer
Mark
Notes
Part (a)
total weight component = 1600 × 9.8 × 1/14 = 1120 N
B1
30 000 ÷ v = 300 + 100 + 1120 (= 1520)
M1
v = 19.7 m s−1
A1
Part (b)
driving force = 30 000 ÷ 10 = 3000 N
M1
3000 − 1520 = 1600a
M1
a = 0.925 m s−2
A1
Part (c)
weight component on the trailer = 400 × 9.8 × 1/14 = 280 N
B1
trailer: T − 100 − 280 = 400 × 0.925
M1
T = 750 N
A1
Exam tip: Use the whole system for the acceleration (the tension is internal), then one vehicle on its own for the tension.
A block of mass 2.5 kg slides down a line of greatest slope of a rough plane inclined at an angle α to the horizontal, where sin α = 0.6. Figure 1 shows the kinetic energy of the block plotted against the distance it has moved down the plane, for the first 5 m of its motion. The block is modelled as a particle. Take g = 9.8 m s−2.
(a)
Find the speed of the block when it has moved 0 m.
(2)
(b)
Find the work done against friction as the block moves the first 5 m.
(3)
(c)
Find the coefficient of friction between the block and the plane.
(3)
(d)
Explain why the graph is a straight line.
(1)
Show mark scheme
Answer
Mark
Notes
Part (a)
½ × 2.5 × u2 = 10
M1
u = 2.83 m s−1 (√8)
A1
Part (b)
PE lost = 2.5 × 9.8 × 5 × 0.6 = 73.5 J
B1
KE gained = 50 − 10 = 40 J (from the graph)
B1
work against friction = 73.5 − 40 = 33.5 J
A1
Part (c)
F = 33.5 ÷ 5 = 6.7 N
M1
cos α = 0.8, R = 2.5 × 9.8 × 0.8 = 19.6 N
B1
μ = 6.7 ÷ 19.6 = 0.342
A1
Part (d)
the resultant force down the plane (weight component − friction) is constant, so the kinetic energy increases by the same amount for each metre moved
B1
Exam tip: The gradient of a kinetic energy–distance graph is the resultant force in the direction of motion (here 8 N).
A city bus has mass 9000 kg and an engine whose greatest possible power output is 120 kW. It climbs a long hill on which the road makes an angle α with the horizontal, sin α = 150. Air resistance and friction together are taken to be a fixed 3600 N throughout. Take g = 9.8 m s−2.
(a)
Find the maximum speed of the bus up the road.
(3)
(b)
Find the acceleration of the bus at the instant its speed is 12 m s−1 with the engine working at 120 kW.
(3)
(c)
On another journey the bus moves up the road from P to Q, where PQ = 600 m. Its speed increases from 12 m s−1 at P to 18 m s−1 at Q, and it takes 40 s. Find the average rate at which the engine works between P and Q.
(5)
(d)
Suggest one way in which the model for the resistance could be made more realistic.
(1)
Show mark scheme
Answer
Mark
Notes
Part (a)
mg sin α = 9000 × 9.8 ÷ 50 = 1764 N
B1
120 000 ÷ v = 3600 + 1764
M1
v = 22.4 m s−1 (22.37)
A1
Part (b)
driving force = 120 000 ÷ 12 = 10 000 N
M1
10 000 − 3600 − 1764 = 9000a
M1
a = 0.515 m s−2
A1
Part (c)
KE gained = ½ × 9000 × (182 − 122) = 810 000 J
B1
PE gained = 9000 × 9.8 × 600 ÷ 50 = 1 058 400 J
B1
work against resistance = 3600 × 600 = 2 160 000 J
B1
work by engine = 810 000 + 1 058 400 + 2 160 000 = 4 028 400 J, then ÷ 40
M1
average rate = 101 kW (100 710 W)
A1
Part (d)
let the resistance increase with speed / vary with the road surface or wind
B1
Exam tip: “Average rate of working” over a journey means total work done by the engine ÷ time. Find the work with the work–energy principle first.
A particle of mass 2 kg is pulled up a line of greatest slope of a rough plane inclined at 20° to the horizontal. The pulling force has magnitude 30 N and acts at an angle of 15° above the line of greatest slope, in the vertical plane containing that line. The coefficient of friction between the particle and the plane is 0.3. The particle starts from rest and moves 4 m up the plane. Take g = 9.8 m s−2.
Figure: the particle on the rough plane, pulled by the 30 N force at 15° to the line of greatest slope; not to scale.
(a)
Find the normal reaction between the particle and the plane.
(2)
(b)
Find the work done by the pulling force.
(2)
(c)
Find the work done against friction.
(2)
(d)
Find the gain in gravitational potential energy of the particle.
(2)
(e)
Use the work–energy principle to find the speed of the particle after it has moved 4 m.
(3)
(f)
Explain why the normal reaction does not appear in your equation in (e).
(1)
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Answer
Mark
Notes
Part (a)
R + 30 sin 15° = 2 × 9.8 cos 20°
M1
R = 10.7 N (10.65)
A1
Part (b)
30 cos 15° × 4
M1
= 116 J (115.9)
A1
Part (c)
F = 0.3 × 10.65 = 3.196 N; work = 3.196 × 4
M1
= 12.8 J
A1
Part (d)
2 × 9.8 × 4 sin 20°
M1
= 26.8 J
A1
Part (e)
work by pulling force = KE gained + PE gained + work against friction
M1
115.9 = ½ × 2 × v2 + 26.8 + 12.8 (ft)
A1
v = 8.74 m s−1
A1
Part (f)
it acts perpendicular to the plane, so perpendicular to the motion, and does no work
B1
Common error: Using R = mg cos 20°. The pulling force has a component away from the plane, 30 sin 15°, which reduces the normal reaction.
A train of mass 80 000 kg has an engine with a maximum rate of working of 600 kW. The resistance to motion from non-gravitational forces is constant, 15 000 N. The train is modelled as a particle. Take g = 9.8 m s−2.
(a)
Find the maximum speed of the train on a straight horizontal track.
(2)
(b)
Find the acceleration of the train on the horizontal track at the instant its speed is 24 m s−1, with the engine working at its maximum rate.
(3)
(c)
The train climbs a straight track inclined at α to the horizontal, where sin α = 1100. Find its maximum speed up this track.
(3)
(d)
While the train is moving up this track at the maximum speed found in (c), the engine is switched off. The resistance is unchanged. Use the work–energy principle to find the distance the train travels before it comes to rest.
(3)
Show mark scheme
Answer
Mark
Notes
Part (a)
600 000 ÷ v = 15 000
M1
v = 40 m s−1
A1
Part (b)
driving force = 600 000 ÷ 24 = 25 000 N
M1
25 000 − 15 000 = 80 000a
M1
a = 0.125 m s−2
A1
Part (c)
mg sin α = 80 000 × 9.8 ÷ 100 = 7840 N
B1
600 000 ÷ v = 15 000 + 7840
M1
v = 26.3 m s−1 (26.27)
A1
Part (d)
KE lost = PE gained + work against resistance: ½ × 80 000 × 26.272 = 7840d + 15 000d
M1 A1
d = 1210 m (1208.6)
A1
Exam tip: With the engine off on an upward slope, both the resistance and the weight component take kinetic energy away, so they add in the energy equation.
A trolley of mass 120 kg runs on a fixed track. It is released from rest at a point A, 18 m vertically above the lowest point B. The section AB is smooth. From B the track is horizontal and rough for 30 m to a point C, where a constant resistance acts on the trolley. Beyond C the track is smooth and rises steadily. The trolley is modelled as a particle and air resistance is ignored. Take g = 9.8 m s−2.
Figure: the track from A down to B, along the rough section BC and up beyond C; not to scale.
(a)
Find the speed of the trolley at B.
(3)
(b)
The trolley passes C with speed 12 m s−1. Find the magnitude of the constant resistance on BC.
(3)
(c)
Find the greatest height above BC reached by the trolley beyond C.
(2)
(d)
The trolley then runs back down to C and along CB. Show that it comes to rest before reaching B, and find its distance from C when it stops.
(4)
(e)
State one assumption, other than those given in the question, that you have used in answering parts (c) and (d).
(1)
Show mark scheme
Answer
Mark
Notes
Part (a)
mechanical energy conserved on the smooth section
M1
½ × 120 × v2 = 120 × 9.8 × 18
A1
v = 18.8 m s−1
A1
Part (b)
KE at B = 21 168 J, KE at C = ½ × 120 × 122 = 8640 J
B1
30R = 21 168 − 8640
M1
R = 418 N (417.6)
A1
Part (c)
120 × 9.8 × h = 8640
M1
h = 7.35 m
A1
Part (d)
it returns to C with KE 8640 J (smooth section)
B1
work needed to cross CB = 417.6 × 30 = 12 528 J > 8640 J, so it stops before B
B1
417.6d = 8640
M1
d = 20.7 m from C
A1
Part (e)
no energy is lost where the track sections join at C / the trolley stays in contact with the track / the resistance on CB is the same when the trolley moves from C to B as from B to C
B1
Exam tip: Split a journey at every change of surface. On each smooth part KE + PE is constant; on each rough part include the work done against the resistance.
Figure 1 shows a fixed wedge whose faces are inclined at 30° and 60° to the horizontal. Particle A, of mass 3 kg, lies on the face inclined at 30°, which is rough; the coefficient of friction between A and this face is 0.25. Particle B, of mass 4 kg, lies on the face inclined at 60°, which is smooth. The particles are attached to the ends of a light inextensible string which passes over a small smooth pulley at the top of the wedge, each section of the string being parallel to a line of greatest slope of its face. B is 1.5 m from the foot of its face, measured along the face. The system is released from rest with the string taut. A does not reach the pulley. Take g = 9.8 m s−2.
(a)
Show that the system starts to move.
(3)
(b)
Use the work–energy principle to find the speed of B as it reaches the foot of its face.
(5)
(c)
When B reaches the foot of its face it stops and the string becomes slack. Find the further distance that A moves up the plane before it comes to instantaneous rest.
(3)
(d)
Show that A then slides back down the plane.
(1)
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Answer
Mark
Notes
Part (a)
component of the weight of B down its face = 4 × 9.8 sin 60° = 33.95 N
B1
for A: 3 × 9.8 sin 30° = 14.7 N and limiting friction 0.25 × 3 × 9.8 cos 30° = 6.365 N
B1
33.95 > 14.7 + 6.365 = 21.07, so B moves down and A moves up
A1
Part (b)
PE lost by B = 4 × 9.8 × 1.5 sin 60° = 50.92 J
B1
PE gained by A = 3 × 9.8 × 1.5 sin 30° = 22.05 J
B1
work against friction = 6.365 × 1.5 = 9.548 J
B1
½ × (3 + 4)v2 = 50.92 − 22.05 − 9.548 (= 19.32)
M1
v = 2.35 m s−1
A1
Part (c)
KE of A = ½ × 3 × 2.3502 = 8.282 J
B1
8.282 = (14.7 + 6.365)d
M1
d = 0.393 m
A1
Part (d)
weight component 14.7 N > limiting friction 6.365 N, so A cannot stay at rest
B1
Common error: Giving the kinetic energy to B only, or forgetting the potential energy gained by A. Both particles move with the same speed through the same distance.
A van of mass 2000 kg has an engine whose maximum rate of working is P watts. The resistance to motion from non-gravitational forces is a constant R newtons. The van is driven with the engine working at its maximum rate up several straight roads, each inclined at an angle α to the horizontal, and its maximum speed v m s−1 up each road is found. Figure 1 shows 1v plotted against sin α. Take g = 9.8 m s−2.
(a)
Show that 1v = RP + 2000gP sin α.
(3)
(b)
Use Figure 1 to find P and R.
(4)
(c)
Write down the maximum speed of the van on a horizontal road.
(1)
(d)
The van moves up a road with sin α = 0.05, with the engine working at its maximum rate. Find the acceleration of the van at the instant when its speed is 15 m s−1.
(3)
(e)
The van later moves down a road with sin α = 0.05 with its engine switched off. Explain why, according to the model, it moves at constant speed.
(2)
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Answer
Mark
Notes
Part (a)
at maximum speed the acceleration is zero, so driving force = resistance + weight component
M1
P/v = R + 2000g sin α
M1
dividing by P gives the result
A1
Part (b)
gradient = (0.060 − 0.020) ÷ 0.10 = 0.4
M1
2000 × 9.8 ÷ P = 0.4, P = 49 000
A1
intercept R/P = 0.020
M1
R = 0.020 × 49 000 = 980
A1
Part (c)
1 ÷ 0.020 = 50 m s−1
B1
Part (d)
driving force = 49 000 ÷ 15 = 3266.7 N
M1
3266.7 − 980 − 2000 × 9.8 × 0.05 = 2000a
M1
a = 0.653 m s−2
A1
Part (e)
component of the weight down the road = 2000 × 9.8 × 0.05 = 980 N
B1
this equals the resistance of 980 N, so the resultant force is zero and the speed is constant
B1
Exam tip: Rearranging a model into the form y = c + mx lets you read two unknowns from a straight-line graph: one from the gradient and one from the intercept.
A car of mass 1250 kg is driven 400 m up a straight road AB inclined at an angle α to the horizontal. Its speed at A is 10 m s−1 and the journey from A to B takes 20 s. The resistance to motion from non-gravitational forces is constant. Figure 1 shows, for the journey from A to B, the work done by the engine, the gain in gravitational potential energy of the car and the work done against the resistance. The gain in kinetic energy is not shown. The car is modelled as a particle. Take g = 9.8 m s−2.
(a)
Find the height of B above A, and hence find sin α.
(3)
(b)
Find the magnitude of the resistance.
(2)
(c)
Use the work–energy principle to find the speed of the car at B.
(4)
(d)
Find the average rate at which the engine works between A and B.
(2)
(e)
At B the engine is working at 60 kW. Find the acceleration of the car at B.
(3)
Show mark scheme
Answer
Mark
Notes
Part (a)
1250 × 9.8 × h = 294 000
M1
h = 24 m
A1
sin α = 24 ÷ 400 = 0.06
A1
Part (b)
400R = 320 000
M1
R = 800 N
A1
Part (c)
gain in KE = 1000 − 294 − 320
M1
= 386 kJ
A1
½ × 1250v2 = ½ × 1250 × 102 + 386 000
M1
v = 26.8 m s−1
A1
Part (d)
1 000 000 ÷ 20
M1
= 50 kW
A1
Part (e)
driving force = 60 000 ÷ 26.79 = 2240 N
M1
2240 − 800 − 1250 × 9.8 × 0.06 (= 735) = 1250a
M1
a = 0.564 m s−2
A1
Exam tip: Work done by the engine = gain in KE + gain in PE + work done against resistance. If any three are known, the fourth follows.
A lift of mass 800 kg is raised vertically from rest by a cable attached to a motor at the top of the shaft. Figure 1 shows the speed–time graph of the lift for one journey, which ends with the lift at rest. A constant resistance of 300 N opposes the motion throughout. The lift is modelled as a particle and the cable as light and inextensible. Take g = 9.8 m s−2.
(a)
Use Figure 1 to find the height through which the lift rises.
(2)
(b)
Use the work–energy principle to find the total work done by the cable on the lift during the journey.
(3)
(c)
Find the rate at which the cable does work while the lift moves at constant speed.
(2)
(d)
Find the greatest rate at which the cable does work during the journey, and state when it occurs.
(3)
(e)
Find the tension in the cable during the last 5 s of the journey.
(1)
Show mark scheme
Answer
Mark
Notes
Part (a)
½ × 4 × 2 + 10 × 2 + ½ × 5 × 2
M1
= 29 m
A1
Part (b)
the lift starts and ends at rest, so there is no overall change in KE
B1
work = 800 × 9.8 × 29 + 300 × 29
M1
= 236 060 J = 236 kJ
A1
Part (c)
T = 800 × 9.8 + 300 = 8140 N
M1
P = 8140 × 2 = 16 280 W = 16.3 kW
A1
Part (d)
first 4 s: a = 0.5 m s−2, T − 7840 − 300 = 800 × 0.5, T = 8540 N
M1
greatest at the largest speed in this stage: 8540 × 2 = 17 080 W = 17.1 kW, which exceeds 16.3 kW and the values in the last stage
A1
it occurs at t = 4 s, just before the lift stops accelerating
B1
Part (e)
a = −0.4 m s−2: T = 7840 + 300 − 800 × 0.4 = 7820 N
B1
Common error: Taking the power as mgv throughout. The cable tension changes from stage to stage, so find T for each stage before using P = Tv.
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