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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-01 M2 · 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)
Show mark scheme
AnswerMarkNotes
Part (a)
mgh = 250 × 9.8 × 12M1
= 29 400 JA1
Part (b)
constant speed so the resultant force is zero: T = 250 × 9.8 = 2450 NB1
Part (c)
P = Fv = 2450 × 0.40M1
= 980 WA1
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.

2IALM-M2-4-B-02 M2 · A2Easy6 marks

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.

18 kg30 N40°
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)
Show mark scheme
AnswerMarkNotes
Part (a)
30 cos 40° × 50M1
= 1150 J (1149)A1
Part (b)
R + 30 sin 40° = 18 × 9.8M1
R = 157 N (157.1)A1
Part (c)
constant speed so F = 30 cos 40° = 22.98 N and F = μRM1
μ = 22.98 ÷ 157.1 = 0.146A1

Common error: Taking R = mg. The upward component of the pull, 30 sin 40°, reduces the normal reaction.

3IALM-M2-4-B-03 M2 · A2Easy5 marks

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)
Show mark scheme
AnswerMarkNotes
Part (a)
½ × 0.3 × 62 = 5.4 JB1
Part (b)
PE lost = 0.3 × 9.8 × 20 = 58.8 JB1
5.4 + 58.8 = ½ × 0.3 × v2M1
v = 20.7 m s−1A1
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 directionB1

Exam tip: When no angle of projection is given, an energy method is almost certainly intended.

4IALM-M2-4-B-04 M2 · A2Easy6 marks

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)
Show mark scheme
AnswerMarkNotes
Part (a)
at maximum speed driving force = resistance: 36 000 ÷ 40M1
= 900 NA1
Part (b)
driving force = 36 000 ÷ 15 = 2400 NM1
2400 − 900 = 1200aM1
a = 1.25 m s−2A1
Part (c)
the driving force P/v decreases as v increases while the resistance stays the same, so the resultant force decreasesB1

Exam tip: Maximum speed means zero acceleration, so the driving force P/v equals the total resistance.

5IALM-M2-4-FB-01 M2 · A2Easy7 marks

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.

AB25 m12 kg48 N30°resistance 20 Ndirection of motionFigure 1
(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)
Show mark scheme
AnswerMarkNotes
Part (a)
48 cos 30° × 25M1
= 1040 J (1039.2)A1
Part (b)
work done against the resistance = 20 × 25 = 500 JB1
½ × 12 × v2 = 1039.2 − 500 (= 539.2)M1
v = 9.48 m s−1A1
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.

6IALM-M2-4-FB-02 M2 · A2Easy6 marks

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.

01234567height h / mhorizontal distanceABCEDbead released from rest at AFigure 1
(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)
Show mark scheme
AnswerMarkNotes
Part (a)
½ × 0.2 × v2 = 0.2 × 9.8 × (5.0 − 1.0)M1
v = 8.85 m s−1A1
Part (b)
½v2 = 9.8 × (5.0 − 3.5)M1
v = 5.42 m s−1A1
Part (c)
at least enough energy to rise from 5.0 m to 6.0 m (the highest point after A): ½u2 = 9.8 × 1.0M1
u = 4.43 m s−1A1

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.

7IALM-M2-4-FB-03 M2 · A2Easy6 marks

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.

03691215182105001000150020002500300035004000force / Nspeed v / m s⁻¹resistancedriving forceFigure 1
(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)
Show mark scheme
AnswerMarkNotes
Part (a)
P = Fv = 2000 × 9M1
= 18 000 W (18 kW)A1
Part (b)
resistance = 1500 N; at maximum speed the driving force equals the resistance, 18 000 ÷ v = 1500M1
v = 12 m s−1 (where the curve meets the line)A1
Part (c)
driving force = 18 000 ÷ 8 = 2250 N and 2250 − 1500 = 600aM1
a = 1.25 m s−2A1

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.

8IALM-M2-4-FB-04 M2 · A2Easy7 marks

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.

OACB40°25°4.5 mhFigure 1
(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)
Show mark scheme
AnswerMarkNotes
Part (a)
h = 4.5 − 4.5 cos 40°M1
= 1.0528… = 1.05 mA1
Part (b)
½ × 28 × v2 = 28 × 9.8 × 1.0528M1
v = 4.54 m s−1A1
Part (c)
height fallen from A to C = 4.5 cos 25° − 4.5 cos 40° = 0.6312 mM1
v = √(2 × 9.8 × 0.6312) = 3.52 m s−1A1
Part (d)
the tension acts along the rope, perpendicular to the child's circular path (direction of motion), so it does no workB1

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).

9IALM-M2-4-B-05 M2 · A2Medium8 marks

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.

35°30 kg6 mrough horizontal ground
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)
Show mark scheme
AnswerMarkNotes
Part (a)
PE lost = 30 × 9.8 × 6 sin 35° = 1011.8 JB1
R = 30 × 9.8 cos 35° = 240.8 N, F = 0.3R = 72.25 NM1
work against friction = 72.25 × 6 = 433.5 JA1
½ × 30v2 = 1011.8 − 433.5M1
v = 6.21 m s−1A1
Part (b)
friction on the ground = 0.4 × 30 × 9.8 = 117.6 NB1
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.

10IALM-M2-4-B-06 M2 · A2Medium8 marks

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)
Show mark scheme
AnswerMarkNotes
Part (a)
component of weight = 85 × 9.8 × 1/30 = 27.77 NB1
driving force = 25 + 27.77 = 52.77 NM1
P = 52.77 × 5 = 264 WA1
Part (b)
driving force = 400 ÷ 5 = 80 NM1
80 − 25 − 27.77 = 85aM1
a = 0.320 m s−2A1
Part (c)
400 ÷ v = 52.77M1
v = 7.58 m s−1A1

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.

11IALM-M2-4-B-07 M2 · A2Medium9 marks

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.

α4 kg7 m s⁻¹
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)
Show mark scheme
AnswerMarkNotes
Part (a)
sin α = 0.6, cos α = 0.8; R = 4 × 9.8 × 0.8 = 31.36 NB1
F = 0.5 × 31.36 = 15.68 NM1
PE gained = 4 × 9.8 × 0.6d = 23.52dB1
½ × 4 × 72 = 23.52d + 15.68dM1
d = 2.5 mA1
Part (b)
component of weight down the ramp = 23.52 N, maximum friction = 15.68 NM1
23.52 > 15.68 so the box slides back downA1
Part (c)
½ × 4 × v2 = 23.52 × 2.5 − 15.68 × 2.5 (= 19.6)M1
v = 3.13 m s−1A1

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.

12IALM-M2-4-B-08 M2 · A2Medium7 marks

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)
Show mark scheme
AnswerMarkNotes
Part (a)
KE = ½ × 16 000 × 252 = 5 000 000 JB1
80R = 5 000 000M1
R = 62 500A1
Part (b)
PE lost over distance d = 16 000 × 9.8 × d/14 = 11 200dB1
KE lost + PE lost = work done against RM1
5 000 000 + 11 200d = 62 500dA1
d = 97.5 mA1

Exam tip: Downhill, the potential energy lost adds to the kinetic energy that the brakes must remove, so the stopping distance is longer.

13IALM-M2-4-B-09 M2 · A2Medium7 marks

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)
Show mark scheme
AnswerMarkNotes
Part (a)
30 ÷ 60 = 0.5 kgB1
Part (b)
PE gained per second = 0.5 × 9.8 × 15 = 73.5 JB1
KE gained per second = ½ × 0.5 × 82 = 16 JB1
power = 73.5 + 16M1
= 89.5 WA1
Part (c)
energy needed per kilogram = 9.8 × 15 + ½ × 82 = 179 J, so 179m = 150M1
m = 0.838 kgA1

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.

14IALM-M2-4-B-10 M2 · A2Medium7 marks

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.

A3 kgB2 kg
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)
Show mark scheme
AnswerMarkNotes
Part (a)
F = 0.25 × 3 × 9.8 = 7.35 NM1
work = 7.35 × 1.2 = 8.82 JA1
Part (b)
PE lost by B = 2 × 9.8 × 1.2 = 23.52 JB1
PE lost = KE gained by both particles + work against frictionM1
23.52 = ½ × 5 × v2 + 8.82A1
v = 2.42 m s−1A1
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 zeroB1

Common error: Giving only B kinetic energy. Both particles move at the same speed, so the kinetic energy gained is ½(3 + 2)v2.

15IALM-M2-4-FB-05 M2 · A2Medium9 marks

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.

0123456789101101234567v / m s⁻¹time t / sforce removedFigure 1
(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)
Show mark scheme
AnswerMarkNotes
Part (a)
area = ½ × 6 × 6 = 18 mB1
Part (b)
KE lost = ½ × 5 × 62 = 90 J = F × 18M1
F = 5 NA1
Part (c)
R = 5 × 9.8 = 49 N, μ = 5 ÷ 49M1
μ = 0.102A1
Part (d)
distance in the first 4 s = ½ × 4 × 6 = 12 mB1
12P = 90 + 5 × 12M1
P = 12.5A1
Part (e)
12.5 × 6 = 75 WB1

Exam tip: A speed–time graph gives distances (areas) and speeds; combine them with the work–energy principle instead of finding accelerations.

16IALM-M2-4-FB-06 M2 · A2Medium9 marks

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.

PQR500 m400 m15 m10 mFigure 1 (vertical scale exaggerated)
(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)
Show mark scheme
AnswerMarkNotes
Part (a)
component of weight down PQ = 10 000 × 9.8 × 15/500 = 2940 NB1
driving force = 3000 + 2940 = 5940 NM1
P = 5940 × 12 = 71 280 W = 71.3 kWA1
Part (b)
KE at Q = ½ × 10 000 × 122 = 720 000 JB1
PE lost = 10 000 × 9.8 × 10 = 980 000 J and work against resistance = 3000 × 400 = 1 200 000 JB1
½ × 10 000v2 = 720 000 + 980 000 − 1 200 000M1
v = 10 m s−1A1
Part (c)
weight component down QR = 10 000 × 9.8 × 10/400 = 2450 N < 3000 N, so the resultant 550 N opposes the motionM1
deceleration = 550 ÷ 10 000 = 0.055 m s−2A1

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.

17IALM-M2-4-FB-07 M2 · A2Medium8 marks

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.

0246810121416051015202530354045P / Ndistance x / mFigure 1
(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)
Show mark scheme
AnswerMarkNotes
Part (a)
40 × 5 + 25 × 7M1
= 375 JA1
Part (b)
friction = 0.2 × 8 × 9.8 = 15.68 NB1
½ × 8v2 = 375 − 15.68 × 12 (= 186.84)M1
v = 6.83 m s−1A1
Part (c)
15.68d = 186.84, d = 11.92 mM1
total = 12 + 11.92 = 23.9 mA1
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 downB1

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.

18IALM-M2-4-FB-08 M2 · A2Medium8 marks

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.

belt speed 1.5 m s⁻¹8.0 m20°bottomtopFigure 1
(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)
Show mark scheme
AnswerMarkNotes
Part (a)
PE gained = 5 × 9.8 × 8.0 sin 20° = 134.1 JB1
KE gained = ½ × 5 × 1.52 = 5.625 JB1
total = 140 J (139.7)A1
Part (b)
12 per minute = 0.2 per second; 0.2 × 139.7M1
= 27.9 WA1
Part (c)
45 × 60 − 12 × 139.7M1
= 1020 J (1024)A1
Part (d)
F = 5 × 9.8 sin 20° = 16.8 N up the beltB1

Common error: Forgetting the kinetic energy. Each parcel starts at rest, so it gains ½mv2 as well as mgh.

19IALM-M2-4-FB-09 M2 · A2Medium9 marks

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.

trailer400 kgcar1200 kgresistance 100 Nresistance 300 Nmotionαroad inclined at α to the horizontal, sin α = 1/14 (true angle)Figure 1
(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)
Show mark scheme
AnswerMarkNotes
Part (a)
total weight component = 1600 × 9.8 × 1/14 = 1120 NB1
30 000 ÷ v = 300 + 100 + 1120 (= 1520)M1
v = 19.7 m s−1A1
Part (b)
driving force = 30 000 ÷ 10 = 3000 NM1
3000 − 1520 = 1600aM1
a = 0.925 m s−2A1
Part (c)
weight component on the trailer = 400 × 9.8 × 1/14 = 280 NB1
trailer: T − 100 − 280 = 400 × 0.925M1
T = 750 NA1

Exam tip: Use the whole system for the acceleration (the tension is internal), then one vehicle on its own for the tension.

20IALM-M2-4-FB-10 M2 · A2Medium9 marks

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.

0123450102030405060kinetic energy / Jdistance moved down the plane / mFigure 1
(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
AnswerMarkNotes
Part (a)
½ × 2.5 × u2 = 10M1
u = 2.83 m s−1 (√8)A1
Part (b)
PE lost = 2.5 × 9.8 × 5 × 0.6 = 73.5 JB1
KE gained = 50 − 10 = 40 J (from the graph)B1
work against friction = 73.5 − 40 = 33.5 JA1
Part (c)
F = 33.5 ÷ 5 = 6.7 NM1
cos α = 0.8, R = 2.5 × 9.8 × 0.8 = 19.6 NB1
μ = 6.7 ÷ 19.6 = 0.342A1
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 movedB1

Exam tip: The gradient of a kinetic energy–distance graph is the resultant force in the direction of motion (here 8 N).

21IALM-M2-4-B-11 M2 · A2Hard12 marks

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
AnswerMarkNotes
Part (a)
mg sin α = 9000 × 9.8 ÷ 50 = 1764 NB1
120 000 ÷ v = 3600 + 1764M1
v = 22.4 m s−1 (22.37)A1
Part (b)
driving force = 120 000 ÷ 12 = 10 000 NM1
10 000 − 3600 − 1764 = 9000aM1
a = 0.515 m s−2A1
Part (c)
KE gained = ½ × 9000 × (182 − 122) = 810 000 JB1
PE gained = 9000 × 9.8 × 600 ÷ 50 = 1 058 400 JB1
work against resistance = 3600 × 600 = 2 160 000 JB1
work by engine = 810 000 + 1 058 400 + 2 160 000 = 4 028 400 J, then ÷ 40M1
average rate = 101 kW (100 710 W)A1
Part (d)
let the resistance increase with speed / vary with the road surface or windB1

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.

22IALM-M2-4-B-12 M2 · A2Hard12 marks

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.

20°2 kg30 N15°
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)
Show mark scheme
AnswerMarkNotes
Part (a)
R + 30 sin 15° = 2 × 9.8 cos 20°M1
R = 10.7 N (10.65)A1
Part (b)
30 cos 15° × 4M1
= 116 J (115.9)A1
Part (c)
F = 0.3 × 10.65 = 3.196 N; work = 3.196 × 4M1
= 12.8 JA1
Part (d)
2 × 9.8 × 4 sin 20°M1
= 26.8 JA1
Part (e)
work by pulling force = KE gained + PE gained + work against frictionM1
115.9 = ½ × 2 × v2 + 26.8 + 12.8 (ft)A1
v = 8.74 m s−1A1
Part (f)
it acts perpendicular to the plane, so perpendicular to the motion, and does no workB1

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.

23IALM-M2-4-B-13 M2 · A2Hard11 marks

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
AnswerMarkNotes
Part (a)
600 000 ÷ v = 15 000M1
v = 40 m s−1A1
Part (b)
driving force = 600 000 ÷ 24 = 25 000 NM1
25 000 − 15 000 = 80 000aM1
a = 0.125 m s−2A1
Part (c)
mg sin α = 80 000 × 9.8 ÷ 100 = 7840 NB1
600 000 ÷ v = 15 000 + 7840M1
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 000dM1 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.

24IALM-M2-4-B-14 M2 · A2Hard13 marks

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.

ABC18 m30 m (rough)smoothsmooth
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
AnswerMarkNotes
Part (a)
mechanical energy conserved on the smooth sectionM1
½ × 120 × v2 = 120 × 9.8 × 18A1
v = 18.8 m s−1A1
Part (b)
KE at B = 21 168 J, KE at C = ½ × 120 × 122 = 8640 JB1
30R = 21 168 − 8640M1
R = 418 N (417.6)A1
Part (c)
120 × 9.8 × h = 8640M1
h = 7.35 mA1
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 BB1
417.6d = 8640M1
d = 20.7 m from CA1
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 CB1

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.

25IALM-M2-4-FB-11 M2 · A2Hard12 marks

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.

roughsmoothA3 kgB4 kg1.5 m30°60°Figure 1
(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)
Show mark scheme
AnswerMarkNotes
Part (a)
component of the weight of B down its face = 4 × 9.8 sin 60° = 33.95 NB1
for A: 3 × 9.8 sin 30° = 14.7 N and limiting friction 0.25 × 3 × 9.8 cos 30° = 6.365 NB1
33.95 > 14.7 + 6.365 = 21.07, so B moves down and A moves upA1
Part (b)
PE lost by B = 4 × 9.8 × 1.5 sin 60° = 50.92 JB1
PE gained by A = 3 × 9.8 × 1.5 sin 30° = 22.05 JB1
work against friction = 6.365 × 1.5 = 9.548 JB1
½ × (3 + 4)v2 = 50.92 − 22.05 − 9.548 (= 19.32)M1
v = 2.35 m s−1A1
Part (c)
KE of A = ½ × 3 × 2.3502 = 8.282 JB1
8.282 = (14.7 + 6.365)dM1
d = 0.393 mA1
Part (d)
weight component 14.7 N > limiting friction 6.365 N, so A cannot stay at restB1

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.

26IALM-M2-4-FB-12 M2 · A2Hard13 marks

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.

0.000.020.040.060.080.100.000.010.020.030.040.050.060.071 / v (s m⁻¹)sin α(0, 0.020)(0.05, 0.040)(0.10, 0.060)Figure 1
(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)
Show mark scheme
AnswerMarkNotes
Part (a)
at maximum speed the acceleration is zero, so driving force = resistance + weight componentM1
P/v = R + 2000g sin αM1
dividing by P gives the resultA1
Part (b)
gradient = (0.060 − 0.020) ÷ 0.10 = 0.4M1
2000 × 9.8 ÷ P = 0.4, P = 49 000A1
intercept R/P = 0.020M1
R = 0.020 × 49 000 = 980A1
Part (c)
1 ÷ 0.020 = 50 m s−1B1
Part (d)
driving force = 49 000 ÷ 15 = 3266.7 NM1
3266.7 − 980 − 2000 × 9.8 × 0.05 = 2000aM1
a = 0.653 m s−2A1
Part (e)
component of the weight down the road = 2000 × 9.8 × 0.05 = 980 NB1
this equals the resistance of 980 N, so the resultant force is zero and the speed is constantB1

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.

27IALM-M2-4-FB-13 M2 · A2Hard14 marks

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.

010020030040050060070080090010001100energy / kJ1000work doneby the engine294gain inpotential energy320work done againstresistance?not showngain inkinetic energyFigure 1
(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
AnswerMarkNotes
Part (a)
1250 × 9.8 × h = 294 000M1
h = 24 mA1
sin α = 24 ÷ 400 = 0.06A1
Part (b)
400R = 320 000M1
R = 800 NA1
Part (c)
gain in KE = 1000 − 294 − 320M1
= 386 kJA1
½ × 1250v2 = ½ × 1250 × 102 + 386 000M1
v = 26.8 m s−1A1
Part (d)
1 000 000 ÷ 20M1
= 50 kWA1
Part (e)
driving force = 60 000 ÷ 26.79 = 2240 NM1
2240 − 800 − 1250 × 9.8 × 0.06 (= 735) = 1250aM1
a = 0.564 m s−2A1

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.

28IALM-M2-4-FB-14 M2 · A2Hard11 marks

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.

024681012141618200.00.51.01.52.02.5v / m s⁻¹time t / sFigure 1
(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
AnswerMarkNotes
Part (a)
½ × 4 × 2 + 10 × 2 + ½ × 5 × 2M1
= 29 mA1
Part (b)
the lift starts and ends at rest, so there is no overall change in KEB1
work = 800 × 9.8 × 29 + 300 × 29M1
= 236 060 J = 236 kJA1
Part (c)
T = 800 × 9.8 + 300 = 8140 NM1
P = 8140 × 2 = 16 280 W = 16.3 kWA1
Part (d)
first 4 s: a = 0.5 m s−2, T − 7840 − 300 = 800 × 0.5, T = 8540 NM1
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 stageA1
it occurs at t = 4 s, just before the lift stops acceleratingB1
Part (e)
a = −0.4 m s−2: T = 7840 + 300 − 800 × 0.4 = 7820 NB1

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.