A.3 Work, energy and power: IB Physics HL exam-style questions
Work is energy transferred by a force, W = Fs cos θ, and the work done by the resultant force equals the change in kinetic energy. Without dissipative forces, kinetic, gravitational and elastic potential energy add to a constant; with friction or drag, the shortfall is the work done against them.
Power is the rate of energy transfer, including P = Fv for a vehicle at constant speed. Efficiency, Sankey diagrams and the energy density of fuels link the topic to engines, turbines and power stations, and the area under a force–extension graph gives the work done by a variable force.
43 questions
214 marks
Paper 1A: 25
Paper 1B: 8
Paper 2: 10
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18 practice questions on A.3 Work, energy and power
1A-1A-04
Power and rate of energy transfer·A.3 Work, energy and power
Paper 1AMedium1 mark
Multiple choice · 1 mark3 steps to full marksDeduce
A constant resultant force acts on a body that is initially at rest. At a time t after the force starts to act, the power delivered to the body by the force is P.
What is the power delivered to the body by the force at time 2t?
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Step 1A constant resultant force gives a constant acceleration a = F/m, so from rest the speed is v = at: the speed is proportional to the time.
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All 3 steps must be completed — there is no mark for a part-answer.
Step 2The power delivered by a force is P = Fv. F is constant, so P ∝ v ∝ t.
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Step 3Doubling the time doubles the speed and therefore doubles the power: 2P. (Check: Ek = ½m(at)² ∝ t², and the rate of increase of Ek ∝ t.)
✓ 1
Answer C
Answer: C · 3 stages of work, one mark
Every option, and why
AThis treats the power as constant because the force is constant. The power P = Fv also depends on the speed, which keeps increasing.
BThis assumes that the kinetic energy increases at a steady rate (constant power), so that v ∝ √t. Here it is the force, not the power, that is constant.
CCorrect: P = Fv = F²t/m ∝ t, so doubling the time doubles the power.
DThis is the factor by which the kinetic energy increases (Ek ∝ t²). The power is the rate of transfer of energy, which is proportional to t.
Syllabus understandingA.3 — power P as the rate of work done, as given by P = Fv; the kinetic energy of translational motion; A.2 — Newton's second law F = ma; A.1 — uniformly accelerated motion Command term: Deduce
2A-1A-15
Power and rate of energy transfer·A.3 Work, energy and power
Paper 1AMedium1 mark
Multiple choice · 1 mark3 steps to full marksDeduce
A cargo ship moves at a constant speed through calm water. The total resistive force on the ship is modelled as being proportional to the square of its speed.
The speed of the ship is doubled and is again constant. By what factor does the useful power output of the engines change, and by what factor does the useful energy output for each kilometre travelled change?
Factor for the useful power outputFactor for the useful energy per kilometre
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Step 1At constant speed the driving force equals the resistive force: F = kv². Doubling v multiplies the driving force by 4.
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All 3 steps must be completed — there is no mark for a part-answer.
Step 2Power P = Fv = kv³, so the useful power output increases by a factor of 2³ = 8.
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Step 3Energy for each kilometre = work done by the driving force over 1 km = F × 1000 m ∝ v², a factor of 4. (Check: the power is 8 times larger but each kilometre takes half the time: 8 × ½ = 4.)
✓ 1
Answer D
Answer: D · 3 stages of work, one mark
Every option, and why
AThis treats the resistive force as constant, so that P = Fv only doubles and the work per kilometre is unchanged. The force here grows with v².
BThis takes the power to be proportional to the force alone (∝ v²), forgetting the factor v in P = Fv. The energy per kilometre is right.
CThe power factor is right, but the energy per kilometre has been scaled like the power, ignoring that at double speed each kilometre takes only half the time: 8 × ½ = 4.
DCorrect: P ∝ Fv ∝ v³ gives 8; the work per kilometre ∝ F ∝ v² gives 4.
Syllabus understandingA.3 — that power developed P is the rate of work done, or the rate of energy transfer, as given by P = Fv; that work W done on a body by a constant force depends on the component of the force along the line of displacement as given by W = Fs cos θ; A.2 — Newton’s first law (translational equilibrium at constant velocity) Command term: Deduce
3A-1A-20
Work done by a variable force·A.3 Work, energy and power
Paper 1AMedium1 mark
Multiple choice · 1 mark3 steps to full marksDetermine
A spring of spring constant 200 N m−1 is already stretched by 0.10 m. It is then stretched further to an extension of 0.30 m.
What is the work done in the further stretching?
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Step 1The elastic potential energy stored is ½kx², so the work done is the change in this energy.
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All 3 steps must be completed — there is no mark for a part-answer.
Step 3W = 8.0 J — the area under the force–extension graph between 0.10 m and 0.30 m.
✓ 1
Answer B
Answer: B · 3 stages of work, one mark
Every option, and why
AThis is ½k(0.20)², using the change in extension squared, which ignores the force already present.
BCorrect: ½ × 200 × 0.30² − ½ × 200 × 0.10² = 9.0 − 1.0 = 8.0 J.
CThis is the total energy stored at 0.30 m, forgetting the 1.0 J already stored.
DThis is kΔx = 200 × 0.20 = 40 — a force in newtons, not a work; the ½ and a second factor of extension are missing.
Syllabus understandingA.3 — the work done by a variable force as the area under a force–displacement graph; elastic potential energy EH = ½kΔx² Command term: Determine
4A-1A-42
Sankey diagrams·A.3 Work, energy and power
Paper 1AEasy1 mark
Multiple choice · 1 mark3 steps to full marksDetermine
The Sankey diagram, drawn to scale, shows the energy transferred each second in an electric winch. The motor drives the winch through a gearbox. The arrow for the useful output has not been labelled.
What is the efficiency of the gearbox?
Sankey diagram for the winch, drawn to scale (width of each arrow proportional to power).Show mark scheme
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Step 1Energy is conserved at the motor: the power passed to the gearbox is 500 − 100 = 400 W.
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All 3 steps must be completed — there is no mark for a part-answer.
Step 2At the gearbox: useful output = 400 − 40 = 360 W (the output arrow is 72 % of the width of the input arrow, as drawn).
AThis is the fraction of the gearbox input that is wasted (40/400), not the efficiency.
BThis is the overall efficiency of the winch (360/500): it uses the motor's input instead of the gearbox's input.
CThis is the efficiency of the motor (400/500).
DCorrect: 360 W out of 400 W in.
Syllabus understandingA.3 — that energy transfers can be represented on a Sankey diagram; the principle of the conservation of energy; efficiency η in terms of energy transfer or power Command term: Determine
5A-1A-43
Work done by a constant force·A.3 Work, energy and power
Paper 1AMedium1 mark
Multiple choice · 1 mark3 steps to full marksDetermine
A box of mass m is pulled a distance d up a rough slope inclined at θ to the horizontal. The pulling force F is parallel to the slope and the box moves at constant speed.
What is the work done on the box by the frictional force?
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Step 1Constant speed means zero resultant force along the slope: F = mg sin θ + Ff, so the friction force is Ff = F − mg sin θ, directed down the slope.
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All 3 steps must be completed — there is no mark for a part-answer.
Step 2Friction acts opposite to the displacement (cos 180° = −1), so Wf = −(F − mg sin θ)d.
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Step 3Check: the work done by F, by the weight (−mgd sin θ) and by friction add to zero, as the kinetic energy does not change.
✓ 1
Answer A
Answer: A · 3 stages of work, one mark
Every option, and why
ACorrect: friction = F − mg sin θ down the slope, and it does negative work.
BThis assumes friction alone balances the pull, forgetting that the component of the weight down the slope also opposes the motion.
CThis adds the weight component to the pull instead of letting it share the job of balancing it.
DThe magnitude is right but the sign is wrong: friction acts against the displacement, so it does negative work on the box.
Syllabus understandingA.3 — that work W done on a body by a constant force depends on the component of the force along the line of displacement as given by W = Fs cos θ; A.2 — Newton's first law; friction Command term: Determine
6A-1A-44
Work–energy principle·A.3 Work, energy and power
Paper 1AMedium1 mark
Multiple choice · 1 mark3 steps to full marksDetermine
An electric delivery van of mass 2400 kg speeds up from 12 m s−1 to 18 m s−1 while travelling 150 m along a road. In this distance the road rises through a vertical height of 6.0 m.
What is the average resultant force on the van over the 150 m?
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Step 1The work done by the resultant force equals the change in kinetic energy of the van: Fress = ΔEk.
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All 3 steps must be completed — there is no mark for a part-answer.
Step 2ΔEk = ½ × 2400 × (18² − 12²) = 2.16 × 105 J.
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Step 3Fres = 2.16 × 105/150 = 1440 N ≈ 1.4 kN. The rise of 6.0 m does not enter: the weight component along the slope is already part of the resultant.
✓ 1
Answer B
Answer: B · 3 stages of work, one mark
Every option, and why
AThis uses (Δv)² instead of Δ(v²): ½ × 2400 × 6² /150 = 290 N.
BCorrect: ½ × 2400 × (18² − 12²)/150 = 1.4 kN.
CThis adds the gain in gravitational potential energy (2400 × 9.81 × 6.0 = 1.41 × 105 J) to ΔEk and divides by 150 m: (3.57 × 105)/150 = 2.4 kN. That is the average driving force minus resistive forces, not the resultant force.
DThis omits the ½ in the kinetic energy: 2400 × (18² − 12²)/150 = 2.9 kN.
Syllabus understandingA.3 — that work done by the resultant force on a system is equal to the change in the energy of the system; the kinetic energy of translational motion as given by Ek = ½mv² Command term: Determine
7A-1A-45
Elastic potential energy·A.3 Work, energy and power
Paper 1AEasy1 mark
Multiple choice · 1 mark2 steps to full marksCalculate
The plunger of a pinball machine compresses a light spring of spring constant 250 N m−1 by 6.0 cm. When released, the spring launches a ball of mass 80 g along a horizontal, frictionless track.
What is the speed of the ball as it leaves the spring?
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Step 1All the elastic potential energy becomes kinetic energy: ½k(Δx)² = ½mv².
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All 2 steps must be completed — there is no mark for a part-answer.
ACorrect: 0.060 × √(250/0.080) = 3.35 m s−1 ≈ 3.4 m s−1.
BThis uses EH = k(Δx)² without the ½ while keeping ½mv²: 0.060 × √(2 × 250/0.080) = 4.7 m s−1.
CThis stops at v² = (Δx)²k/m = 11.25 and forgets to take the square root.
DThis uses ½kΔx (not squared) for the elastic energy: √(250 × 0.060/0.080) = 14 m s−1. The units do not even give joules.
Syllabus understandingA.3 — the elastic potential energy as given by EH = ½k(Δx)²; that in the absence of frictional, resistive forces, the total mechanical energy of a system is conserved Command term: Calculate
8A-1A-46
Energy density of fuels·A.3 Work, energy and power
Paper 1AMedium1 mark
Multiple choice · 1 mark3 steps to full marksDetermine
A standby diesel generator delivers a constant electrical power of 5.0 kW. The energy density of diesel (energy released per unit mass when it is burned) is 45.6 MJ kg−1, and the generator converts 32 % of this energy into electrical energy.
What mass of diesel does the generator burn in one hour?
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Notes
Step 1Electrical energy delivered in one hour = 5.0 × 103 × 3600 = 1.8 × 107 J.
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All 3 steps must be completed — there is no mark for a part-answer.
Step 2Energy that must be released by the fuel = useful output/η = 1.8 × 107/0.32 = 5.6 × 107 J.
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Step 3Mass = energy released/energy density = 5.6 × 107/45.6 × 106 = 1.2 kg.
✓ 1
Answer D
Answer: D · 3 stages of work, one mark
Every option, and why
AThis multiplies by the efficiency instead of dividing: 1.8 × 107 × 0.32/45.6 × 106 = 0.13 kg — less fuel than would be needed even at 100 % efficiency.
BThis ignores the efficiency: 1.8 × 107/45.6 × 106 = 0.39 kg, which assumes every joule of chemical energy becomes electrical energy.
CThis divides by the wasted fraction (1 − 0.32 = 0.68): 1.8 × 107/(0.68 × 45.6 × 106) = 0.58 kg.
DCorrect: 1.8 × 107/(0.32 × 45.6 × 106) = 1.2 kg.
Syllabus understandingA.3 — energy density of the fuel sources; efficiency η in terms of energy transfer or power; power as the rate of energy transfer Command term: Determine
9A-1A-61
Energy and projectile motion·A.3 Work, energy and power
Paper 1AEasy1 mark
Multiple choice · 1 mark2 steps to full marksCompare
Two identical balls are projected from the same balcony with the same speed. Ball P is projected horizontally; ball Q is projected at 40° above the horizontal. Both land on the same level ground below. Air resistance is negligible.
Which row compares the speeds of the balls on reaching the ground and the times they spend in the air?
Speed on reaching the groundTime in the air
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Step 1Energy (A.3): ½mv² = ½mu² + mgh for both balls, with the same u and the same drop h, so the landing speeds are equal. The direction of projection does not matter.
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All 2 steps must be completed — there is no mark for a part-answer.
Step 2Kinematics (A.1): the vertical motion decides the time. Q starts with an upward velocity component, so it rises first and then has further to fall; it spends longer in the air than P.
✓ 1
Answer C
Answer: C · 2 stages of work, one mark
Every option, and why
AThe landing speed does not depend on direction because the same kinetic energy is gained from the same loss of gravitational potential energy; and P, with zero initial vertical velocity, reaches the ground sooner.
BThe speeds are equal, but the times are not: Q’s upward vertical component delays its landing.
CCorrect: same energy transfer gives the same speed; the upward component makes Q’s flight longer.
DA longer flight does not mean a greater final speed: the total energy gained depends only on the height fallen.
Syllabus understandingA.3 — the principle of the conservation of energy; gravitational potential energy and kinetic energy; A.1 — the behaviour of projectiles in the absence of fluid resistance Command term: Compare
10A-1A-68
Kinetic energy and momentum·A.3 Work, energy and power
Paper 1AEasy1 mark
Multiple choice · 1 mark3 steps to full marksDeduce
Two bodies P and Q have the same kinetic energy. The mass of P is four times the mass of Q.
What is (momentum of P)/(momentum of Q)?
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Notes
Step 1Ek = p²/2m, so p = √(2mEk).
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All 3 steps must be completed — there is no mark for a part-answer.
Step 2With Ek the same, p ∝ √m.
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Step 3Ratio = √4 = 2. (Equivalently, P moves at half the speed of Q: 4 × ½ = 2.)
✓ 1
Answer C
Answer: C · 3 stages of work, one mark
Every option, and why
AThis inverts the relationship, as if p ∝ 1/√m. It is the ratio of the speeds of P and Q, not of their momenta.
BThis assumes that equal kinetic energies mean equal momenta. Ek depends on p² and on m.
CCorrect: p = √(2mEk) ∝ √m, so the ratio is √4 = 2.
DThis uses p ∝ m, which is true only for bodies with the same speed. P is slower than Q.
Syllabus understandingA.3 — the kinetic energy of translational motion as given by Ek = ½mv² = p²/2m; A.2 — linear momentum p = mvCommand term: Deduce
11A-1A-69
Work done by a variable force·A.3 Work, energy and power
Paper 1AMedium1 mark
Multiple choice · 1 mark3 steps to full marksDetermine
A child pulls a sledge along level snow with a rope that is kept at 30° above the horizontal. The graph shows how the tension T in the rope varies with the horizontal displacement s of the sledge.
What is the work done by the tension on the sledge during the first 10 m?
Tension in the rope against horizontal displacement of the sledge (drawn to scale).Show mark scheme
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Step 1The area under the graph is the work that a force T would do if it acted along the displacement: ½ × 4.0 × 80 + 6.0 × 80 = 160 + 480 = 640 J.
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All 3 steps must be completed — there is no mark for a part-answer.
Step 2Only the component of the tension along the displacement does work: W = Fs cos θ, with θ = 30°.
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Step 3W = 640 × cos 30° = 554 J ≈ 550 J.
✓ 1
Answer B
Answer: B · 3 stages of work, one mark
Every option, and why
AThis uses sin 30°, the vertical component, which is perpendicular to the displacement and does no work: 640 × sin 30° = 320 J.
BCorrect: area 640 J multiplied by cos 30°.
CThis is the area under the graph, ignoring the angle of the rope: the full tension does not act along the displacement.
DThis treats the tension as 80 N for the whole 10 m, ignoring the triangle at the start: 80 × 10 × cos 30° = 690 J.
Syllabus understandingA.3 — that work W done on a body by a constant force depends on the component of the force along the line of displacement as given by W = Fs cos θ; the work done by a varying force as the area under a force–displacement graph Command term: Determine
12A-1A-70
Conservation of mechanical energy·A.3 Work, energy and power
Paper 1AEasy1 mark
Multiple choice · 1 mark3 steps to full marksDeduce
A bungee jumper steps off a platform and falls vertically from rest. The elastic cord becomes taut and stretches until the jumper is momentarily at rest at the lowest point. Air resistance and energy dissipation in the cord are negligible.
Which statements about the jumper at the lowest point are correct? I. The kinetic energy of the jumper is zero. II. The elastic potential energy stored in the cord is equal to the gravitational potential energy lost by the jumper since leaving the platform. III. The resultant force on the jumper is zero.
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Step 1I: the jumper is momentarily at rest, so the kinetic energy is zero. I is correct.
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All 3 steps must be completed — there is no mark for a part-answer.
Step 2II: no energy is dissipated, so mechanical energy is conserved. Kinetic energy is zero at both the start and the lowest point, so all the gravitational potential energy lost is stored as elastic potential energy. II is correct.
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Step 3III: at the lowest point the cord is stretched beyond the equilibrium position, so the tension is greater than the weight. The resultant force is upwards, which is why the jumper moves back up. III is wrong.
✓ 1
Answer A
Answer: A · 3 stages of work, one mark
Every option, and why
ACorrect: zero kinetic energy, and all the gravitational potential energy lost is stored in the cord; the resultant force is upwards, not zero.
BThis confuses the lowest point with the equilibrium position. At the lowest point the tension exceeds the weight.
CThis rejects I, perhaps thinking the jumper is still moving at the lowest point, and accepts III, confusing the lowest point with equilibrium.
DIII is wrong: a zero resultant force at the lowest point would leave the jumper at rest there, but the jumper accelerates back upwards.
Syllabus understandingA.3 — that mechanical energy is the sum of kinetic energy, gravitational potential energy and elastic potential energy; that in the absence of frictional, resistive forces, the total mechanical energy of a system is conserved; A.2 — Newton's second law; elastic restoring force Command term: Deduce
13A-1A-99
Sankey diagrams·A.3 Work, energy and power
Paper 1AMedium1 mark
Multiple choice · 1 mark3 steps to full marksDetermine
The Sankey diagram, drawn to scale, shows the energy transfers each second in the engine of a car that is travelling at a constant speed of 26 m s−1 along a straight, level road. The arrow for the useful output has not been labelled with a value.
What is the total resistive force acting on the car?
Sankey diagram for the car engine, drawn to scale (width of each band proportional to power).Show mark scheme
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Step 1Energy is conserved, so the useful output power = 120 − 48 − 33 = 39 kW (the useful band is 39/120 of the width of the input band).
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All 3 steps must be completed — there is no mark for a part-answer.
Step 2At constant velocity the driving force equals the total resistive force, and the useful power is P = Fv.
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Step 3F = 39 × 103/26 = 1.5 × 103 N = 1.5 kN.
✓ 1
Answer A
Answer: A · 3 stages of work, one mark
Every option, and why
ACorrect: (120 − 48 − 33) kW/26 m s−1 = 1.5 kN.
BThis divides the power carried by the widest output band (the exhaust, 48 kW) by the speed: 48 × 103/26 = 1.8 kN. That band is wasted energy, not the useful output.
CThis uses the total wasted power, 48 + 33 = 81 kW: 81 × 103/26 = 3.1 kN. Only the useful output does work against the resistive forces.
DThis uses the input power from the fuel, 120 × 103/26 = 4.6 kN, ignoring the energy transferred to the surroundings in the engine.
Syllabus understandingA.3 — that energy transfers can be represented on a Sankey diagram; the principle of the conservation of energy; that power developed P is the rate of work done, or the rate of energy transfer, as given by P = ΔW/Δt = Fv; A.2 — Newton’s first law (translational equilibrium at constant velocity) Command term: Determine
14A-1A-100
Kinetic energy and momentum·A.3 Work, energy and power
Paper 1AEasy1 mark
Multiple choice · 1 mark3 steps to full marksDeduce
The momentum of a trolley of constant mass increases by 50 %.
By what percentage does the kinetic energy of the trolley increase?
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Step 1Ek = p²/2m; the mass is constant, so Ek ∝ p².
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All 3 steps must be completed — there is no mark for a part-answer.
Step 2The new momentum is 1.5p, so the kinetic energy becomes 1.5² = 2.25 times its original value.
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Step 3Increase = 2.25 − 1 = 1.25 of the original value: 125 %.
✓ 1
Answer C
Answer: C · 3 stages of work, one mark
Every option, and why
AThis takes Ek ∝ √p: √1.5 = 1.22, an increase of 22 %. The momentum is squared, not square-rooted.
BThis takes Ek ∝ p. Kinetic energy depends on the square of the momentum.
CCorrect: 1.5² = 2.25, an increase of 125 %.
DThis quotes the factor 2.25 as the percentage increase. The new kinetic energy is 225 % of the original, which is an increase of 125 %.
Syllabus understandingA.3 — the kinetic energy of translational motion as given by Ek = ½mv² = p²/2m; A.2 — linear momentum p = mvCommand term: Deduce
15A-1A-101
Work done by a constant force·A.3 Work, energy and power
Paper 1AMedium1 mark
Multiple choice · 1 mark3 steps to full marksDeduce
A parcel is carried up an inclined conveyor belt at constant speed. The parcel does not slip on the belt.
Which row gives the work done on the parcel by the frictional force from the belt and the work done on the parcel by the normal force from the belt?
Work done by the frictional forceWork done by the normal force
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Step 1The velocity is constant, so the resultant force is zero. The only force with a component up the slope is the frictional force from the belt, so it acts up the slope, balancing the component of the weight down the slope.
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All 3 steps must be completed — there is no mark for a part-answer.
Step 2The displacement is also up the slope, so the frictional force does positive work: W = Fs cos 0° = +mgs sin θ, equal to the gain in gravitational potential energy.
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Step 3The normal force is perpendicular to the displacement (cos 90° = 0), so it does no work.
✓ 1
Answer A
Answer: A · 3 stages of work, one mark
Every option, and why
ACorrect: friction acts up the slope, along the displacement, and supplies the gain in gravitational potential energy; the normal force is perpendicular to the displacement.
BThis assumes friction always opposes the motion of the body. Friction opposes relative sliding; here it is the force that drags the parcel up the slope.
CThis assumes a static frictional force cannot do work because there is no sliding. The parcel moves through a displacement along the direction of the force, so work is done.
DThis assumes friction opposes the motion and that the normal force lifts the parcel. The normal force has no component along the displacement.
Syllabus understandingA.3 — that work W done on a body by a constant force depends on the component of the force along the line of displacement as given by W = Fs cos θ; that work done by a force is equivalent to a transfer of energy; A.2 — static friction; Newton’s first law Command term: Deduce
16A-1A-102
Work done by non-conservative forces·A.3 Work, energy and power
Paper 1AMedium1 mark
Multiple choice · 1 mark3 steps to full marksDetermine
A cyclist freewheels (does not pedal) down a long straight hill. The graph shows how the gravitational potential energy and the kinetic energy of the cyclist and bicycle vary with the distance s travelled along the road.
What is the average resistive force on the cyclist and bicycle over the 500 m?
Gravitational potential energy (measured from the bottom of the hill) and kinetic energy of the cyclist and bicycle against distance travelled (drawn to scale).Show mark scheme
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Step 1Mechanical energy = Ek + Ep: 5 + 40 = 45 kJ at the start and 15 + 10 = 25 kJ after 500 m.
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All 3 steps must be completed — there is no mark for a part-answer.
Step 2The loss of mechanical energy, 20 kJ, is the work done against the resistive (non-conservative) forces.
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Step 3Average resistive force = 20 × 103/500 = 40 N.
✓ 1
Answer B
Answer: B · 3 stages of work, one mark
Every option, and why
AThis divides the gain in kinetic energy (10 kJ) by 500 m. That is the work done by the resultant force (weight component minus resistance), not by the resistive force.
BCorrect: loss of mechanical energy (45 − 25) kJ = 20 kJ, divided by 500 m.
CThis divides the loss of gravitational potential energy (30 kJ) by 500 m, as if none of it had become kinetic energy.
DThis adds the loss of gravitational potential energy and the gain in kinetic energy: 40 kJ/500 m = 80 N. The gain in kinetic energy must be subtracted from the loss of potential energy.
Syllabus understandingA.3 — that mechanical energy is the sum of kinetic energy, gravitational potential energy and elastic potential energy; guidance: the change in the total mechanical energy of a system interpreted in terms of the work done on the system by any non-conservative force; that work W done on a body by a constant force depends on the component of the force along the line of displacement as given by W = Fs cos θCommand term: Determine
17A-1A-103
Mechanical energy·A.3 Work, energy and power
Paper 1AEasy1 mark
Multiple choice · 1 mark3 steps to full marksDeduce
A block hangs from a light spring and oscillates vertically. Air resistance and other energy losses are negligible. The elastic potential energy of the spring is ½k(Δx)², where Δx is the extension of the spring from its natural length.
Which sums remain constant during the oscillation? I. kinetic energy + gravitational potential energy II. kinetic energy + elastic potential energy III. kinetic energy + gravitational potential energy + elastic potential energy
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Step 1No resistive forces act, so the total mechanical energy, the sum of the kinetic, gravitational potential and elastic potential energies, is conserved: III is constant.
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All 3 steps must be completed — there is no mark for a part-answer.
Step 2At the lowest point and the highest point the kinetic energy is zero in both cases, but the block is higher at the highest point and the spring is less stretched. So Ek + Ep and Ek + EH each change between these points.
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Step 3Only III remains constant.
✓ 1
Answer C
Answer: C · 3 stages of work, one mark
Every option, and why
AThis treats the block like a projectile and ignores the energy stored in the spring, which changes as the extension changes.
BThis treats the oscillation like that of a mass on a horizontal spring and ignores the change in gravitational potential energy as the block rises and falls.
CCorrect: only the total mechanical energy is conserved; energy is transferred among all three forms.
DThis assumes each pair is conserved separately. Energy moves among all three forms, so neither pair alone stays constant.
Syllabus understandingA.3 — that mechanical energy is the sum of kinetic energy, gravitational potential energy and elastic potential energy; that in the absence of frictional, resistive forces, the total mechanical energy of a system is conserved; the elastic potential energy as given by EH = ½k(Δx)² Command term: Deduce
18A-1A-104
Power and rate of energy transfer·A.3 Work, energy and power
Paper 1AEasy1 mark
Multiple choice · 1 mark3 steps to full marksDeduce
Crane X lifts a load of mass 2M vertically through a height h in a time t. Crane Y lifts a load of mass M vertically through a height 3h in a time 2t. Both loads move at constant speed.
What is (useful power output of X)/(useful power output of Y)?
Show mark scheme
Marking point
Mark
Notes
Step 1Useful power = rate of increase of gravitational potential energy = mgΔh/Δt.
—
All 3 steps must be completed — there is no mark for a part-answer.
Step 2X: 2Mgh/t. Y: Mg(3h)/(2t) = 1.5Mgh/t.
—
Step 3Ratio = 2/1.5 = 4/3.
✓ 1
Answer D
Answer: D · 3 stages of work, one mark
Every option, and why
AThis multiplies the ratio of the work done (2/3) by the ratio of the times the wrong way: (2/3) × (1/2) = 1/3. Y takes longer, which lowers its power.
BThis is the ratio of the work done (2Mgh)/(3Mgh), ignoring the different times.
CThis is the inverted ratio, (power of Y)/(power of X).
DCorrect: (2Mgh/t)/(3Mgh/2t) = 4/3.
Syllabus understandingA.3 — that power developed P is the rate of work done, or the rate of energy transfer, as given by P = ΔW/Δt = Fv; the gravitational potential energy, when close to the surface of the Earth, as given by ΔEp = mgΔhCommand term: Deduce
19A-1A-105
Energy density of fuels·A.3 Work, energy and power
Paper 1AEasy1 mark
Multiple choice · 1 mark3 steps to full marksCalculate
The fuel tank of a hydrogen car has an internal volume of 0.14 m3 and holds 5.6 kg of compressed hydrogen. The energy released per kilogram of hydrogen used is 1.2 × 108 J kg−1.
What is the energy density (energy available per unit volume) of the hydrogen stored in the tank?
All 3 steps must be completed — there is no mark for a part-answer.
Step 2Energy density = energy/volume = 6.72 × 108/0.14.
—
Step 3= 4.8 × 109 J m−3.
✓ 1
Answer D
Answer: D · 3 stages of work, one mark
Every option, and why
AThis divides the energy per kilogram by the density of the stored gas (5.6/0.14 = 40 kg m−3) instead of multiplying: 1.2 × 108/40 = 3.0 × 106.
BThis multiplies the total energy by the volume instead of dividing: 6.72 × 108 × 0.14 = 9.4 × 107.
CThis is the total energy stored in the tank, 6.7 × 108 J, not the energy per unit volume.
DCorrect: 6.72 × 108 J/0.14 m3 = 4.8 × 109 J m−3.
Syllabus understandingA.3 — energy density of the fuel sources Command term: Calculate
20A-1A-106
Elastic potential energy·A.3 Work, energy and power
Paper 1AEasy1 mark
Multiple choice · 1 mark3 steps to full marksDeduce
The graph shows how the force needed to stretch each of two springs, P and Q, varies with the extension. Both springs obey Hooke’s law over the range shown. Each spring is stretched by the same force of 6.0 N.
What is (elastic potential energy stored in P)/(elastic potential energy stored in Q)?
Force against extension for springs P and Q (drawn to scale).Show mark scheme
Marking point
Mark
Notes
Step 1From the graph, the spring constant of P is twice that of Q (8.0 N at 0.10 m compared with 8.0 N at 0.20 m).
—
All 3 steps must be completed — there is no mark for a part-answer.
Step 2With the same force, x = F/k, so EH = ½Fx = F²/2k: the stiffer spring stores less energy.
—
Step 3Ratio = kQ/kP = 1/2. (Check: ½ × 6.0 × 0.075 = 0.225 J and ½ × 6.0 × 0.15 = 0.45 J.)
✓ 1
Answer B
Answer: B · 3 stages of work, one mark
Every option, and why
AThis squares the ratio of the extensions (1/2)² but ignores the different spring constants: EH depends on k as well as on x.
BCorrect: EH = F²/2k, so the ratio is kQ/kP = 1/2.
CThis uses EH = ½kx² as if the extensions were equal, giving kP/kQ = 2. Here the forces are equal, so P stretches only half as far.
DThis takes EH ∝ k², as if the extension were proportional to k (Hooke’s law inverted).
Syllabus understandingA.3 — the elastic potential energy as given by EH = ½k(Δx)²; that work done by a force is equivalent to a transfer of energy; A.2 — Hooke’s law; Tools 3 — the area under a force–extension graph as the work done Command term: Deduce
21A-1A-107
Work–energy principle·A.3 Work, energy and power
Paper 1AEasy1 mark
Multiple choice · 1 mark3 steps to full marksDetermine
A trolley of mass 2.0 kg moves in a straight line at 3.0 m s−1. A resultant force then acts on it in its direction of motion. The graph shows how this resultant force varies with the displacement s of the trolley.
What is the speed of the trolley when s = 4.0 m?
Resultant force on the trolley against its displacement (drawn to scale).Show mark scheme
Marking point
Mark
Notes
Step 1Work done by the resultant force = area under the graph = 6.0 × 2.0 + ½ × 6.0 × 2.0 = 18 J.
—
All 3 steps must be completed — there is no mark for a part-answer.
Step 2This equals the change in kinetic energy: Ek = ½ × 2.0 × 3.0² + 18 = 9.0 + 18 = 27 J.
—
Step 3v = √(2 × 27/2.0) = √27 = 5.2 m s−1.
✓ 1
Answer B
Answer: B · 3 stages of work, one mark
Every option, and why
AThis ignores the kinetic energy the trolley already had: √(2 × 18/2.0) = 4.2 m s−1, as if it started from rest.
BCorrect: 9.0 J + 18 J = 27 J of kinetic energy, so v = 5.2 m s−1.
CThis takes the force as 6.0 N over the whole 4.0 m (24 J): √(2 × 33/2.0) = 5.7 m s−1, ignoring the fall of the force after 2.0 m.
DThis adds speeds instead of energies: 3.0 + 4.2 = 7.2 m s−1. Kinetic energies add; speeds do not.
Syllabus understandingA.3 — that work done by the resultant force on a system is equal to the change in the energy of the system; the kinetic energy of translational motion as given by Ek = ½mv² = p²/2m; that work done by a force is equivalent to a transfer of energy; Tools 3 — the area under a force–displacement graph as the work done Command term: Determine
22A-1A-108
Work done by non-conservative forces·A.3 Work, energy and power
Paper 1AHard1 mark
Multiple choice · 1 mark3 steps to full marksDeduce
A block on a rough horizontal floor is attached to a horizontal spring of spring constant k. The other end of the spring is fixed. The block is pulled until the extension of the spring is x0 and is released from rest. It slides and first comes to rest when the spring is compressed by x1. A frictional force of constant magnitude f acts on the moving block.
What is f?
Show mark scheme
Marking point
Mark
Notes
Step 1The block starts and stops at rest, so the loss of elastic potential energy equals the work done against friction: ½kx0² − ½kx1² = fd.
—
All 3 steps must be completed — there is no mark for a part-answer.
Step 2The distance slid is from extension x0 through the natural length to compression x1: d = x0 + x1.
—
Step 3f = ½k(x0² − x1²)/(x0 + x1) = ½k(x0 − x1).
✓ 1
Answer B
Answer: B · 3 stages of work, one mark
Every option, and why
AThis omits the ½ in the elastic potential energy ½k(Δx)².
CThis divides the energy lost by (x0 − x1), the change in the size of the deformation, instead of by the distance slid, (x0 + x1).
DThis takes the distance slid as x0 only (to the natural length), forgetting the further x1 travelled while compressing the spring.
Syllabus understandingA.3 — guidance: the change in the total mechanical energy of a system interpreted in terms of the work done on the system by any non-conservative force; the elastic potential energy as given by EH = ½k(Δx)²; that work W done on a body by a constant force depends on the component of the force along the line of displacement as given by W = Fs cos θCommand term: Deduce
23A-1A-109
Work–energy principle·A.3 Work, energy and power
Paper 1AMedium1 mark
Multiple choice · 1 mark3 steps to full marksDeduce
A small rubber ball is dropped from rest from the top of a tall tower and falls vertically through still air. The graph shows how its kinetic energy Ek varies with the distance d fallen. Buoyancy is negligible.
Which statements are correct? I. The gradient of the graph at any point is equal to the magnitude of the resultant force on the ball. II. Where the graph is horizontal, the air resistance on the ball is equal in magnitude to its weight. III. Where the graph is horizontal, the total mechanical energy of the ball is constant.
Kinetic energy of the ball against distance fallen (drawn to scale).Show mark scheme
Marking point
Mark
Notes
Step 1I: the work done by the resultant force equals the change in kinetic energy, FresΔd = ΔEk, so the gradient ΔEk/Δd is the resultant force. I is correct.
—
All 3 steps must be completed — there is no mark for a part-answer.
Step 2II: a horizontal graph means zero resultant force, so the air resistance balances the weight (terminal speed). II is correct.
—
Step 3III: at terminal speed Ek is constant but the ball is still falling, so its gravitational potential energy keeps decreasing; the mechanical energy falls at the rate mgv, transferred to the air as thermal energy. III is wrong.
✓ 1
Answer A
Answer: A · 3 stages of work, one mark
Every option, and why
ACorrect: the gradient is the resultant force, which is zero at terminal speed; the mechanical energy still decreases.
BThis accepts III, missing that the gravitational potential energy continues to decrease at terminal speed, and rejects II, although a zero gradient means zero resultant force.
CThis rejects I, perhaps taking the gradient to be the power or the weight. ΔEk/Δd has the unit J m−1 = N and equals the resultant force.
DIII is wrong: a constant kinetic energy does not mean a constant mechanical energy while the ball is still losing height.
Syllabus understandingA.3 — that work done by the resultant force on a system is equal to the change in the energy of the system; guidance: the change in the total mechanical energy of a system interpreted in terms of the work done on the system by any non-conservative force; that mechanical energy is the sum of kinetic energy, gravitational potential energy and elastic potential energy; A.1 — the qualitative effect of fluid resistance, including terminal speed Command term: Deduce
24A-1A-110
Mechanical energy·A.3 Work, energy and power
Paper 1AHard1 mark
Multiple choice · 1 mark3 steps to full marksDeduce
A block of mass m is placed on top of an uncompressed light vertical spring of spring constant k and released from rest. The block moves down, compressing the spring. Air resistance is negligible.
Which row gives the maximum speed of the block and the maximum compression of the spring?
Maximum speedMaximum compression
Show mark scheme
Marking point
Mark
Notes
Step 1The speed is greatest where the resultant force is zero: kx = mg, so x = mg/k. There, ½mv² = mgx − ½kx² = m²g²/2k, so vmax = g√(m/k).
—
All 3 steps must be completed — there is no mark for a part-answer.
Step 2At maximum compression the block is momentarily at rest: mgx = ½kx², so xmax = 2mg/k, twice the compression at equilibrium.
—
Step 3Check: the motion is simple harmonic about x = mg/k with amplitude mg/k and ω = √(k/m), so vmax = ω × amplitude = g√(m/k).
✓ 1
Answer D
Answer: D · 3 stages of work, one mark
Every option, and why
ABoth entries are wrong: the elastic potential energy stored at the equilibrium position is ignored, and the lowest point is confused with the equilibrium position, where the resultant force is zero but the block is still moving.
BThe compression is right, but the speed ignores the energy stored in the spring at the equilibrium position, putting all of mg(mg/k) into kinetic energy.
CThe speed is right, but mg/k is the equilibrium compression. The block passes this point at its maximum speed and goes on to twice this compression.
DCorrect: vmax = g√(m/k) at the equilibrium position, and the block comes momentarily to rest at a compression of 2mg/k.
Syllabus understandingA.3 — that mechanical energy is the sum of kinetic energy, gravitational potential energy and elastic potential energy; that in the absence of frictional, resistive forces, the total mechanical energy of a system is conserved; the elastic potential energy as given by EH = ½k(Δx)²; the gravitational potential energy, when close to the surface of the Earth, as given by ΔEp = mgΔh; A.2 — Hooke’s law; C.1 — simple harmonic motion Command term: Deduce
25A-1A-111
Efficiency and energy transfer·A.3 Work, energy and power
Paper 1AMedium1 mark
Multiple choice · 1 mark3 steps to full marksDetermine
An electric pump in a garden fountain takes in water from the surface of a pond, where the water is at rest, and ejects it through a nozzle 3.0 m above the surface of the pond at a speed of 6.0 m s−1. The pump moves 2.0 kg of water each second. The efficiency of the pump is 60 %.
What is the electrical power input to the pump?
Show mark scheme
Marking point
Mark
Notes
Step 1Each second, 2.0 kg of water gains gravitational potential energy 2.0 × 9.81 × 3.0 = 58.9 J and kinetic energy ½ × 2.0 × 6.0² = 36 J.
—
All 3 steps must be completed — there is no mark for a part-answer.
Step 2Useful output power = 58.9 + 36 = 94.9 W.
—
Step 3Input power = 94.9/0.60 = 158 W.
✓ 1
Answer C
Answer: C · 3 stages of work, one mark
Every option, and why
AThis multiplies the useful power by the efficiency instead of dividing: 94.9 × 0.60 = 57 W, less than the useful output.
BThis ignores the kinetic energy given to the water: 58.9/0.60 = 98 W.
CCorrect: (58.9 + 36) W/0.60 = 158 W.
DThis omits the ½ in the kinetic energy (72 J per second instead of 36 J): (58.9 + 72)/0.60 = 218 W.
Syllabus understandingA.3 — efficiency η in terms of energy transfer or power as given by η = Eoutput/Einput = Poutput/Pinput; that power developed P is the rate of work done, or the rate of energy transfer, as given by P = ΔW/Δt = Fv; the gravitational potential energy, when close to the surface of the Earth, as given by ΔEp = mgΔh; the kinetic energy of translational motion as given by Ek = ½mv² = p²/2mCommand term: Determine
26A-1B-05
Energy transfers in collisions·A.3 Work, energy and power
Paper 1BMedium7 marks
Data-based question7 steps to full marksDetermine
A rubber ball of mass 45 g is dropped from rest from a height of 2.00 m above a concrete floor. A video camera with a vertical scale behind the ball records the maximum height hn that the ball reaches after the nth bounce. Each height has an uncertainty of ±0.01 m. Air resistance is negligible.
A student suggests that the ball loses the same fraction of its kinetic energy at every bounce. The graph shows ln(hn/m) against n.
bounce number n
0
1
2
3
4
5
hn / m
2.00
1.26
0.81
0.53
0.34
0.21
ln(hn/m)
0.69
0.23
−0.21
−0.63
−1.08
−1.56
Graph drawn to scale
(a)
(i)
Test the student's suggestion using at least three of the measured heights.
(2)
(b)
(i)
Determine, using the full range of the graph, the fraction of the kinetic energy retained at each bounce.
(2)
(ii)
Predict the number of the first bounce after which the ball fails to reach a height of 0.030 m.
(1)
(c)
(i)
The specific heat capacity of the rubber is 1.7 × 103 J kg−1 K−1. Assume that half of the kinetic energy lost at the first bounce becomes internal energy of the ball. Estimate the temperature rise of the ball caused by this bounce.
(2)
Show mark scheme
Marking point
Mark
Notes
Part (a)(i)
If the same fraction of energy is lost each time, hn+1/hn is constant; calculates at least three ratios, e.g. 0.63, 0.64, 0.65, 0.64
✓ 1
A test using two heights only scores [1 max].
The ratios are constant (≈ 0.64) within the uncertainty of the data (±0.01 m is ±3 % of 0.34 m), so the suggestion is supported
✓ 1
Conclusion must be consistent with the ratios calculated.
Part (b)(i)
Gradient = (−1.56 − 0.69)/5 = −0.45
✓ 1
Accept −0.43 to −0.47. Using only the first bounce scores [1 max].
ln hn = ln h0 + n ln k, so k = e−0.45 = 0.64
✓ 1
Accept 0.63–0.65. Allow ECF from the gradient. Must be consistent with (a)(i).
Part (b)(ii)
2.00 × 0.64n < 0.030 gives n > ln(0.015)/ln 0.64 = 9.4, so the 10th bounce
✓ 1
Allow ECF from (b)(i). Accept 10 for any k between 0.63 and 0.65.
Part (c)(i)
Energy lost = (1 − 0.64) × 0.045 × 9.81 × 2.00 = 0.32 J
✓ 1
Allow ECF from (b)(i). Accept mg(h0 − h1) = 0.33 J.
ΔT = ½ × 0.32/(0.045 × 1.7 × 103) = 2.1 × 10−3 K
✓ 1
Accept 2.0–2.2 × 10−3 K.
Answers: (a)(i) ratio ≈ 0.64; supported · (b)(i) 0.64 · (b)(ii) 10th bounce · (c)(i) 2.1 × 10−3 K (the remaining parts are explanations — see the table above)
Syllabus understandingA.3 — the principle of the conservation of energy; transfers between kinetic and gravitational potential energy; B.1 — specific heat capacity Q = mcΔT; Tools 3 — testing a relationship with at least three data points, linearising an exponential relationship with logarithms, using the full data range Command term: Determine
27A-1B-16
Elastic potential energy·A.3 Work, energy and power
Paper 1BMedium7 marks
Data-based question7 steps to full marksDetermine
A ball of mass 25.0 g rests on a light vertical spring of spring constant 320 N m−1. The spring is compressed by a distance x, measured with a ruler to ±0.1 cm, and released. The maximum height h reached by the ball above its release position is found from a video recording.
If all the elastic potential energy is transferred to gravitational potential energy of the ball, ½kx² = mgh, so h = (k/2mg)x².
x / cm
2.0
2.5
3.0
3.5
4.0
4.5
h / m
0.23
0.37
0.53
0.72
0.93
1.19
Graph drawn to scale. The point for x = 4.5 cm has not been plotted.
(a)
(i)
Calculate x² for x = 4.5 cm, in m², together with its absolute uncertainty.
(2)
(ii)
Determine the gradient of the graph of h against x², using the full range of the data. Give its unit.
(2)
(iii)
Determine the fraction of the stored elastic potential energy that is transferred to gravitational potential energy of the ball.
(2)
(iv)
Suggest one reason why your answer to (a)(iii) is less than 1.
(1)
Show mark scheme
Marking point
Mark
Notes
Part (a)(i)
x² = (0.045)² = 20.3 × 10−4 m²
✓ 1
Accept 20.25 × 10−4.
Fractional uncertainty = 2 × 0.1/4.5 = 4.4 %, so ±0.9 × 10−4 m²
✓ 1
Absolute uncertainty to 1 s.f.
Part (a)(ii)
Line through the origin and (20.3 × 10−4, 1.19): gradient = 1.19/20.3 × 10−4 = 586
Air resistance on the rising ball; OR the spring is not light, so some energy becomes kinetic and gravitational potential energy of the spring; OR energy dissipated as internal energy in the spring (and sound)
✓ 1
Allow ECF from (a)(iii). Any one physically justified reason.
Answers: (a)(i) (20.3 ± 0.9) × 10−4 m² · (a)(ii) 586 m−1 · (a)(iii) 0.90 (the remaining parts are explanations — see the table above)
Syllabus understandingA.3 — the elastic potential energy EH = ½kΔx²; the gravitational potential energy ΔEp = mgΔh; conservation of energy; Tools 3 — propagation of uncertainty for a power, linearising a relationship, gradient with units, comparing an experimental value with a theoretical one Command term: Determine
28A-1B-17
Work–energy principle·A.3 Work, energy and power
Paper 1BHard7 marks
Data-based question7 steps to full marksDetermine
A team tests a new runner coating on a sledge of mass 12.0 kg. The sledge is released from rest on a uniform snow slope inclined at 20.0° to the horizontal, from different heights h above a pair of timing gates at the bottom of the slope, which measure its speed v. A constant frictional force f acts along the slope; air resistance is negligible.
Applying the work–energy principle gives v² = 2h(g − f/(m sin 20.0°)). The graph shows v² against h with the line of best fit. The specific latent heat of fusion of ice is 3.34 × 105 J kg−1.
h / m
v / m s−1
v² / m² s−2
0.50
2.41
5.8
1.00
3.62
13.1
1.50
4.53
20.5
2.00
5.22
27.2
2.50
5.84
34.1
3.00
6.46
41.7
Graph drawn to scale
(a)
(i)
Determine the gradient of the line of best fit. Give its unit.
(2)
(ii)
Hence determine f.
(2)
(iii)
For the run from h = 3.00 m, half of the work done against friction melts snow at 0 °C under the runners. Estimate the mass of snow melted.
(2)
(iv)
The line meets the h axis at about 0.08 m. State and explain whether this affects your answer to (a)(ii).
(1)
Show mark scheme
Marking point
Mark
Notes
Part (a)(i)
Large triangle on the line, e.g. from (0.083, 0) to (3.00, 41.6): 41.6/2.92 = 14.2
✓ 1
Accept 13.8–14.6.
Unit m s−2
✓ 1
Part (a)(ii)
Gradient = 2(g − f/(m sin 20.0°)), so f = m sin 20.0° (g − gradient/2) = 12.0 × 0.342 × (9.81 − 7.12)
✓ 1
Allow ECF from (a)(i).
f = 11 N
✓ 1
Accept 10–12 N.
Part (a)(iii)
Work done against friction = f × h/sin 20.0° = 11 × 3.00/0.342 = 97 J
✓ 1
Allow ECF from (a)(ii).
Mass = ½ × 97/3.34 × 105 = 1.4 × 10−4 kg
✓ 1
Accept 1.3–1.6 × 10−4 kg. Using the vertical height instead of the distance along the slope scores 0 for the first mark.
Part (a)(iv)
No: a constant offset in h (e.g. the gates measure v about 8 cm above the level from which h was measured) shifts the line sideways without changing its gradient, and f depends only on the gradient
✓ 1
Allow ECF from (a)(ii): the answer must refer to the method actually used.
Answers: (a)(i) 14.2 m s−2 · (a)(ii) 11 N · (a)(iii) 1.4 × 10−4 kg (the remaining parts are explanations — see the table above)
Syllabus understandingA.3 — work done by the resultant force equals the change in energy of the system; the change in mechanical energy as the work done by a non-conservative force; W = Fs cos θ; B.1 — specific latent heat Q = mL; Tools 3 — gradient with units, systematic error from an intercept Command term: Determine
29A-1B-22
Efficiency and energy transfer·A.3 Work, energy and power
Paper 1BMedium6 marks
Data-based question6 steps to full marksDetermine
A small electric winch, powered from a 6.00 V supply, lifts a hook carrying a load of mass M through a height of 0.800 m at a steady speed. For each load the current I is read from an ammeter (±0.01 A) and the time t for the lift is measured with a stopwatch (±0.2 s). The electrical energy supplied is E = VIt; the uncertainty in the supply voltage is negligible.
The graph shows E against M.
M / kg
I / A
t / s
E / J
0.100
0.34
2.3
4.69
0.200
0.38
2.8
6.38
0.300
0.42
3.3
8.32
0.400
0.47
3.5
9.87
0.500
0.51
3.8
11.63
0.600
0.55
4.0
Graph drawn to scale. The point for M = 0.600 kg has not been plotted.
(a)
(i)
Calculate E for M = 0.600 kg together with its absolute uncertainty.
(2)
(ii)
The gradient of the graph is the extra electrical energy needed for each extra kilogram lifted. Determine, using the full range of the data, the efficiency with which this extra energy is transferred to gravitational potential energy of the load.
(2)
(iii)
Determine the intercept of the line of best fit on the E axis and suggest what it represents. Hence explain why the overall efficiency of lifting the 0.600 kg load is smaller than your answer to (a)(ii).
Gradient from (0.100, 4.69) to (0.600, 13.2): (13.2 − 4.69)/0.500 = 17 J kg−1
✓ 1
Allow ECF from (a)(i). Accept 16–18 J kg−1.
Efficiency = gh/gradient = 9.81 × 0.800/17 = 0.46
✓ 1
Accept 0.43–0.49.
Part (a)(iii)
Intercept ≈ 3.0 J (accept 2.5–3.5 J): energy supplied even with no load — raising the hook and overcoming friction / heating in the winch
✓ 1
Allow ECF from (a)(ii).
Overall efficiency = 0.600 × 9.81 × 0.800/13.2 = 0.36, smaller than 0.46 because the fixed 3 J is wasted as well
✓ 1
Allow ECF from (a)(i).
Answers: (a)(i) 13.2 ± 0.9 J · (a)(ii) 0.46 · (a)(iii) intercept 3.0 J; overall efficiency 0.36 (the remaining parts are explanations — see the table above)
Syllabus understandingA.3 — gravitational potential energy ΔEp = mgΔh; efficiency = useful energy output/total energy input; B.5 — electrical power P = IV; Tools 3 — propagation of uncertainty for a product, gradient using the full data range, extrapolating to an intercept Command term: Determine
30A-1B-26
Energy density of fuels·A.3 Work, energy and power
Paper 1BMedium7 marks
Data-based question6 steps to full marksDetermine
A spirit burner containing ethanol heats 0.250 kg of water in a thin metal can. The mass Δm of ethanol burned is found by weighing the burner before and after heating on a balance that reads to ±0.01 g, so each value of Δm is ±0.02 g. The temperature rise ΔT of the water is ±0.4 K. The specific heat capacity of water is 4180 J kg−1 K−1.
Only a fraction η of the energy released by the burning ethanol is transferred to the water, so ΔT = (ηEd/mwcw)Δm, where Ed = 29.7 MJ kg−1 is the energy density of ethanol. The graph shows ΔT against Δm.
Δm / g
0.40
0.80
1.20
1.60
2.00
2.40
ΔT / K
3.7
7.3
10.9
14.7
18.1
21.7
Graph drawn to scale: line of best fit (solid), lines of maximum and minimum gradient (dashed).
(a)
(i)
Determine the gradient of the line of best fit. Give its unit.
(2)
(ii)
The lines of maximum and minimum gradient pass through the ends of the error bars of the first and last points. Determine the absolute uncertainty in the gradient.
(2)
(b)
(i)
Determine η, together with its absolute uncertainty.
(2)
(ii)
Suggest one change to the apparatus that would increase the value of η.
(1)
Show mark scheme
Marking point
Mark
Notes
Part (a)(i)
Gradient using the full range, e.g. (21.7 − 3.7)/(2.40 − 0.40) = 9.0 K g−1
✓ 1
Accept 8.8–9.2 K g−1.
Unit K g−1 (or 9.0 × 103 K kg−1)
✓ 1
Accept °C g−1.
Part (a)(ii)
Maximum gradient (22.1 − 3.3)/2.00 = 9.4 K g−1; minimum gradient (21.3 − 4.1)/2.00 = 8.6 K g−1
✓ 1
Accept values read from the dashed lines within ±0.1 K g−1.
Uncertainty = (9.4 − 8.6)/2 = ±0.4 K g−1
✓ 1
Allow ECF from (a)(i) for the comparison. Accept the larger of (max − best) and (best − min).
Part (b)(i)
Energy reaching the water per kilogram of ethanol = gradient × mwcw = 9.0 × 103 × 0.250 × 4180 = 9.40 × 106 J kg−1, so η = 9.40/29.7 = 0.32
✓ 1
Allow ECF from (a)(i). Gradient must be converted to K kg−1; a power-of-ten error scores 0 for this mark. Accept η = 0.31–0.33.
Fractional uncertainty of η = that of the gradient, 0.4/9.0 = 4.4 %, so η = 0.32 ± 0.01
✓ 1
Allow ECF from (a)(ii). Accept 32 % ± 1 % (or ± 1.4 %). The uncertainties in mw, cw and Ed are negligible.
Part (b)(ii)
Reduce the energy transferred to the surroundings: e.g. surround the flame and can with a draught shield; put a lid on the can and insulate its sides; place the can closer to the flame; use a can with a blackened base
✓ 1
Do not accept "use more accurate instruments" or "repeat the experiment": these do not change η.
Answers: (a)(i) 9.0 K g−1 · (a)(ii) ±0.4 K g−1 · (b)(i) η = 0.32 ± 0.01 (the remaining parts are explanations — see the table above)
Syllabus understandingA.3 — energy density of the fuel sources; efficiency η in terms of energy transfer as given by η = Eoutput/Einput; B.1 — specific heat capacity, Q = mcΔT; Tools 3 — gradient with units; lines of maximum and minimum gradient; propagating the uncertainty of a gradient to a derived quantity Command term: Determine
31A-1B-33
Power and rate of energy transfer·A.3 Work, energy and power
Paper 1BEasy7 marks
Data-based question5 steps to full marksDetermine
A student of mass (58.0 ± 0.5) kg runs up a flight of stairs as quickly as possible. The vertical height of the flight, measured with a tape measure, is (3.40 ± 0.02) m. A partner times five runs with a stopwatch. The student is at rest at the bottom before each run.
The useful power developed by the student is taken to be the rate of increase of their gravitational potential energy.
Run
1
2
3
4
5
t / s
3.62
3.48
3.71
3.55
3.44
(a)
(i)
Determine the mean time for a run and its absolute uncertainty.
(2)
(ii)
Determine the useful power developed by the student, together with its absolute uncertainty.
(2)
(b)
(i)
A cereal bar of mass 0.040 kg releases 1.7 × 107 J kg−1 when it is metabolised. The student’s muscles transfer 22 % of this energy to useful work. Determine the number of runs up the stairs that the energy from one bar could provide.
(2)
(ii)
Suggest why the rate at which the student’s muscles do useful work during a run is greater than the power found in (a)(ii).
(1)
Show mark scheme
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Mark
Notes
Part (a)(i)
Mean = 17.80/5 = 3.56 s
✓ 1
Uncertainty = half the range = (3.71 − 3.44)/2 = ±0.14 s
✓ 1
Accept ±0.1 s or ±0.135 s. Do not accept ±0.01 s (the precision of the stopwatch): the spread of repeated readings is far larger.
Allow ECF from (a)(i). Accept ±28 W to ±30 W; absolute uncertainty to 1 or 2 s.f. Accept 543 ± 30 W.
Part (b)(i)
Energy from the bar = 0.040 × 1.7 × 107 = 6.8 × 105 J; useful energy = 0.22 × 6.8 × 105 = 1.50 × 105 J
✓ 1
Number of runs = 1.50 × 105/(58.0 × 9.81 × 3.40) = 1.50 × 105/1934 = 77
✓ 1
Accept 77–78 (or "77 complete runs"). Dividing by 0.22 instead of multiplying (1600 runs) scores [1 max].
Part (b)(ii)
The student also gains kinetic energy (starting from rest), and moves the arms and legs, so more work is done than mgh alone; OR the centre of mass is raised slightly higher than the vertical height of the stairs at the top
✓ 1
Any one valid reason. Do not accept "friction" or "heat losses": these are not useful work.
Answers: (a)(i) 3.56 ± 0.14 s · (a)(ii) (5.4 ± 0.3) × 102 W · (b)(i) 77 (the remaining parts are explanations — see the table above)
Syllabus understandingA.3 — that power developed P is the rate of work done, or the rate of energy transfer, as given by P = ΔW/Δt = Fv; the gravitational potential energy, when close to the surface of the Earth, as given by ΔEp = mgΔh; efficiency η in terms of energy transfer or power as given by η = Eoutput/Einput = Poutput/Pinput; energy density of the fuel sources; Tools 3 — mean and half-range uncertainty of repeated readings; propagation of fractional uncertainties through a product and quotient Command term: Determine
32A-1B-34
Power and rate of energy transfer·A.3 Work, energy and power
Paper 1BMedium7 marks
Data-based question6 steps to full marksDetermine
A battery-powered model car of mass 1.50 kg is driven up a straight ramp at a constant speed of 0.400 m s−1, set by its speed controller. The angle θ of the ramp to the horizontal is changed, and for each angle the electrical power input P to the motor is found from ammeter and voltmeter readings. Each value of P has an uncertainty of ±0.10 W.
A constant resistive force f acts on the car, and the motor and gears transfer a constant fraction η of the electrical power input to useful work done on the car. It can be shown that P = (mgv/η) sin θ + fv/η. The graph shows P against sin θ.
θ / °
sin θ
P / W
3
0.052
1.97
6
0.105
2.76
9
0.156
3.49
12
0.208
4.21
15
0.259
4.98
18
0.309
5.69
Graph drawn to scale, with the line of best fit.
(a)
(i)
Determine the gradient of the line of best fit. Give its unit.
(2)
(ii)
Determine η.
(2)
(b)
(i)
Use the intercept on the P axis to determine f.
(2)
(ii)
The speed controller is later found to be faulty: the true speed of the car was 5 % less than 0.400 m s−1 in every run. Deduce whether your answer to (b)(i) is affected.
(1)
Show mark scheme
Marking point
Mark
Notes
Part (a)(i)
Large triangle on the line, e.g. from (0, 1.23) to (0.300, 5.56): gradient = 4.33/0.300 = 14.4
✓ 1
Accept 13.8–15.0.
Unit W
✓ 1
sin θ has no unit.
Part (a)(ii)
Gradient = mgv/η, so η = 1.50 × 9.81 × 0.400/14.4
✓ 1
Allow ECF from (a)(i).
η = 0.41
✓ 1
Accept 0.39–0.43.
Part (b)(i)
Intercept = 1.2 W
✓ 1
Accept 1.1–1.35 W.
Intercept = fv/η, so f = 1.2 × 0.41/0.400 = 1.2 N
✓ 1
Allow ECF from (a)(ii). Accept 1.1–1.4 N. Omitting η (f = 1.2/0.400 = 3.0 N) scores [1 max].
Part (b)(ii)
Not affected: f = intercept × η/v = intercept × mg/gradient, in which v cancels (the value of η is 5 % too large, but f is unchanged)
✓ 1
Allow ECF from (a)(ii) and (b)(i). The conclusion must be supported by the cancellation of v; "not affected" alone scores [0].
Answers: (a)(i) 14.4 W · (a)(ii) 0.41 · (b)(i) 1.2 N (the remaining parts are explanations — see the table above)
Syllabus understandingA.3 — that power developed P is the rate of work done, or the rate of energy transfer, as given by P = ΔW/Δt = Fv; efficiency η in terms of energy transfer or power as given by η = Eoutput/Einput = Poutput/Pinput; that work W done on a body by a constant force depends on the component of the force along the line of displacement as given by W = Fs cos θ; A.2 — translational equilibrium on a slope; Tools 3 — linearising a relationship, gradient and intercept of a straight-line graph with units, the effect of a systematic error on a derived quantity Command term: Determine
33A-1B-35
Elastic potential energy·A.3 Work, energy and power
Paper 1BMedium7 marks
Data-based question6 steps to full marksDetermine
A student investigates the energy transfer in a toy catapult. The cord of the catapult is pulled back a distance x with a newton-meter (±0.5 N), and the graph of force F against x is plotted.
The catapult is clamped to the edge of a bench so that a ball of mass 0.0250 kg leaves it horizontally from a height of (1.20 ± 0.01) m above the floor. The cord is pulled back 0.150 m and released. The horizontal distance R from the launch point to where the ball lands is measured in five trials. Air resistance is negligible.
x / m
0.025
0.050
0.075
0.100
0.125
0.150
F / N
3.9
9.7
16.6
26.3
36.5
49.8
Trial
1
2
3
4
5
R / m
6.21
6.40
6.29
6.44
6.31
Graph drawn to scale, with a smooth curve of best fit through the origin.
(a)
(i)
Estimate the elastic potential energy stored in the cord when x = 0.150 m.
(2)
(ii)
State and explain whether a trapezium estimate of the area overestimates or underestimates the energy stored.
(1)
(b)
(i)
Determine the speed with which the ball leaves the catapult, together with its absolute uncertainty.
(3)
(ii)
Determine the fraction of the stored elastic potential energy that is transferred to kinetic energy of the ball.
(1)
Show mark scheme
Marking point
Mark
Notes
Part (a)(i)
Area under the curve, by counting squares or by trapezia, e.g. 0.025 × [½(0 + 49.8) + 3.9 + 9.7 + 16.6 + 26.3 + 36.5]
✓ 1
A single triangle ½ × 0.150 × 49.8 = 3.7 J scores [0]: the graph is not linear.
= 2.9 J
✓ 1
Accept 2.8–3.1 J.
Part (a)(ii)
Overestimates: the curve becomes steeper (it is concave upwards), so each straight chord lies above the curve and the trapezia include area that is not under it
✓ 1
The reason must refer to the shape of the curve.
Part (b)(i)
Mean R = 6.33 m and time of fall t = √(2 × 1.20/9.81) = 0.495 s
✓ 1
A.1: the vertical motion is independent of the horizontal motion.
v = 6.33/0.495 = 12.8 m s−1
✓ 1
Allow ECF from the mean range and the time.
Fractional uncertainty = 0.115/6.33 + ½ × 0.01/1.20 = 0.018 + 0.004 = 0.022, so v = 12.8 ± 0.3 m s−1
✓ 1
Allow ECF from the value of v. The factor ½ for t ∝ √h is required; ignoring the height uncertainty (±0.2 m s−1) is accepted.
Part (b)(ii)
Ek = ½ × 0.0250 × 12.8² = 2.05 J; fraction = 2.05/2.9 = 0.71
✓ 1
Allow ECF from (a)(i) and (b)(i). Accept 0.66–0.73.
Answers: (a)(i) 2.9 J · (b)(i) 12.8 ± 0.3 m s−1 · (b)(ii) 0.71 (the remaining parts are explanations — see the table above)
Syllabus understandingA.3 — the elastic potential energy as given by EH = ½k(Δx)²; the kinetic energy of translational motion as given by Ek = ½mv² = p²/2m; efficiency η in terms of energy transfer or power as given by η = Eoutput/Einput = Poutput/Pinput; that work done by a force is equivalent to a transfer of energy; A.1 — projectile motion in the absence of fluid resistance; Tools 3 — area under a curve, half-range uncertainty, uncertainty in a square root Command term: Determine
34A-2-04
Power and rate of energy transfer·A.3 Work, energy and power
Paper 2Hard13 marks
Short answer & extended response9 steps to full marksDetermine
A cyclist and bicycle have a combined mass of 80 kg. The resistive force on them is modelled as F = kv², where v is the speed and k is a constant. On a horizontal road the cyclist reaches a steady speed of 12.0 m s−1 while delivering a power of 300 W to the bicycle.
(a)
(i)
Show that k is about 0.17 in SI units.
(2)
(ii)
Determine the fundamental SI unit of k.
(1)
(b)
(i)
The cyclist stops pedalling and freewheels down a straight road inclined at 5.0° to the horizontal. Determine the terminal speed.
(2)
(ii)
The cyclist climbs the same road at a steady speed, again delivering 300 W. Determine this speed.
(3)
(c)
(i)
Calculate the mechanical power needed to ride up this road at 12.0 m s−1.
(1)
(ii)
An electrically assisted bicycle lets the cyclist ride up this road at 12.0 m s−1 while still delivering 300 W. The rest is supplied by a motor of efficiency 80 % powered by a 36 V battery. Calculate the current from the battery.
(2)
(iii)
The battery stores 0.50 kW h of energy. Determine the distance the cyclist can climb at 12.0 m s−1 before the battery is empty.
(2)
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Notes
Part (a)(i)
At steady speed the driving force equals the resistive force: F = P/v = 300/12.0 = 25 N
✓ 1
k = 25/12.0² = 0.174
✓ 1
Must see 0.174 or full substitution.
Part (a)(ii)
k = F/v²: kg m s−2/(m² s−2) = kg m−1
✓ 1
Answer in fundamental units required; "N s² m−2" alone scores 0.
Part (b)(i)
At terminal speed mg sin 5.0° = kv²: 80 × 9.81 × sin 5.0° = 68.4 N
✓ 1
v = √(68.4/0.174) = 19.8 m s−1
✓ 1
Allow ECF from (a)(i). Accept 19.6–20.2 m s−1.
Part (b)(ii)
Driving force = mg sin 5.0° + kv², so 300 = v(68.4 + 0.174v²)
✓ 1
Allow ECF from (a)(i) and (b)(i).
First estimate ignoring drag: v ≈ 300/68.4 = 4.39 m s−1; drag at this speed is only about 3.3 N
✓ 1
Or any valid numerical/iterative solution.
v = 4.2 m s−1
✓ 1
Accept 4.1–4.3 m s−1. 4.4 m s−1 (drag ignored) scores [2 max].
Part (c)(i)
P = (68.4 + 25) × 12.0 = 1121 W
✓ 1
Allow ECF from (a)(i) and (b)(i). Accept 1.1 kW.
Part (c)(ii)
Motor output = 1121 − 300 = 821 W; electrical input = 821/0.80 = 1026 W
✓ 1
Allow ECF from (c)(i).
I = P/V = 1026/36 = 28.5 A
✓ 1
Accept 28–29 A. 23 A (efficiency ignored) scores [1 max].
Part (c)(iii)
0.50 kW h = 0.50 × 3.6 × 106 = 1.8 × 106 J; time = 1.8 × 106/1026 = 1754 s
✓ 1
Allow ECF from (c)(ii).
Distance = 12.0 × 1754 = 21 km
✓ 1
Accept 20–22 km.
Answers: (a)(ii) kg m−1 · (b)(i) 19.8 m s−1 · (b)(ii) 4.2 m s−1 · (c)(i) 1121 W · (c)(ii) 28.5 A · (c)(iii) 21 km (the remaining parts are explanations — see the table above)
Syllabus understandingA.3 — that power developed P is the rate of work done, or the rate of energy transfer, as given by P = Fv; efficiency η = Poutput/Pinput; A.1 — the qualitative effect of fluid resistance, including terminal speed (the model F = kv² is given); A.2 — free-body diagrams and translational equilibrium; B.5 — electrical power P = IV; Tool 3 — work with fundamental units Command term: Determine
35A-2-24
Efficiency and energy transfer·A.3 Work, energy and power
Paper 2Medium13 marks
Short answer & extended response8 steps to full marksDetermine
A disused mine shaft is converted into a gravity energy store. When electricity is plentiful, electric winches raise heavy blocks of total mass 5.0 × 105 kg through a height of 250 m. When electricity is needed, the blocks are lowered at a constant speed of 1.2 m s−1 and the winches act as generators. The winch motors have an efficiency of 80 % and the generators an efficiency of 90 %.
(a)
(i)
Show that the energy stored when all the blocks are raised through 250 m is about 1.2 × 109 J.
(1)
(ii)
A lithium-ion battery stores 0.60 MJ kg−1. Compare the energy stored per kilogram of blocks with this value.
(2)
(b)
(i)
Calculate the electrical power output while the blocks descend.
(2)
(ii)
Calculate the time for which the store can deliver this power.
(1)
(c)
(i)
Determine the electrical energy needed to raise all the blocks and the overall efficiency of one complete store-and-return cycle.
(2)
(ii)
The generators deliver their output at 11 kV through cables of total resistance 0.80 Ω. Determine the fraction of the output power that is dissipated in the cables.
(3)
(iii)
Outline a Sankey diagram for one complete cycle that starts with 100 MJ of electrical energy supplied to the motors.
(2)
Show mark scheme
Marking point
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Notes
Part (a)(i)
ΔEp = 5.0 × 105 × 9.81 × 250 = 1.23 × 109 J
✓ 1
Must see 1.23 × 109 or full substitution.
Part (a)(ii)
Energy per kilogram = gΔh = 9.81 × 250 = 2452 J kg−1
✓ 1
Allow ECF from (a)(i).
The battery stores about 0.60 × 106/2452 ≈ 240 times more per kilogram
Power dissipated = I²R = 482² × 0.80 = 1.9 × 105 W
✓ 1
Fraction = 1.9 × 105/5.3 × 106 = 0.035 (3.5 %)
✓ 1
Accept 3–4 %.
Part (c)(iii)
100 MJ splits into 20 MJ dissipated in the motors and 80 MJ stored as gravitational potential energy
✓ 1
Allow ECF from (c)(i) efficiencies.
80 MJ splits into 8 MJ dissipated in the generators and 72 MJ electrical output; arrow widths proportional to the energies
✓ 1
Accept the cable losses shown as a further branch of about 2.5 MJ.
Answers: (a)(ii) 2452 J kg−1 · (b)(i) 5.3 × 106 W · (b)(ii) 208 s · (c)(i) 1.5 × 109 J; 72 % · (c)(ii) 3.5 % (the remaining parts are explanations — see the table above)
Syllabus understandingA.3 — gravitational potential energy ΔEp = mgΔh; energy density; power P = Fv; efficiency; Sankey diagrams; B.5 — electrical power and the heating effect of a current, P = I²RCommand term: Determine
36A-2-25
Work–energy principle·A.3 Work, energy and power
Paper 2Hard12 marks
Short answer & extended response8 steps to full marksDetermine
A lorry of mass 3.2 × 104 kg descends a long hill, keeping a constant speed by using its brakes. Its brakes then fail, and the driver steers into an escape lane. The lane slopes upwards at 8.0° to the horizontal and is filled with deep, loose gravel. The lorry enters the lane at 28 m s−1 and stops after travelling 110 m along it. Assume that the resistive force of the gravel is constant.
(a)
(i)
Before the brakes failed, the lorry descended a vertical height of 150 m at constant speed. 40 % of the energy dissipated by braking was absorbed by the brake drums, of total mass 120 kg and specific heat capacity 450 J kg−1 K−1. Estimate the rise in temperature of the brake drums.
(2)
(ii)
State one assumption made in your estimate, and suggest why it explains the brake failure.
(1)
(b)
(i)
Calculate the kinetic energy of the lorry as it enters the lane and the gravitational potential energy it gains before it stops.
(2)
(ii)
Determine the resistive force exerted by the gravel.
(2)
(iii)
Show that the stopping distance on an upward slope of angle θ is u²/[2(g sin θ + F/m)], and hence explain why a fully loaded lorry stops in a longer distance than an empty one, if F is the same for both.
(2)
(c)
(i)
Calculate the rate at which the gravel dissipates energy as the lorry enters the lane.
(1)
(ii)
The same gravel bed is laid on level ground. Determine the stopping distance from 28 m s−1, and explain why escape lanes are built sloping upwards.
(2)
Show mark scheme
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Mark
Notes
Part (a)(i)
Energy to drums = 0.40 × 3.2 × 104 × 9.81 × 150 = 1.9 × 107 J
✓ 1
At constant speed all the gravitational potential energy lost is dissipated.
ΔT = 1.9 × 107/(120 × 450) = 349 K
✓ 1
Accept 340–360 K.
Part (a)(ii)
No energy is lost from the drums to the air during the descent (or the 40 % is spread uniformly); a rise of several hundred kelvin so overheats the brakes that friction falls (brake fade)
✓ 1
Allow ECF from (a)(i). Any valid assumption with a link to failure.
Work done by the gravel = loss of mechanical energy = 1.25 × 107 − 4.8 × 106 = 7.7 × 106 J
✓ 1
Allow ECF from (b)(i).
F = 7.7 × 106/110 = 7.0 × 104 N
✓ 1
Accept 6.9–7.1 × 104 N.
Part (b)(iii)
Work–energy: ½mu² = (mg sin θ + F)s, so s = u²/[2(g sin θ + F/m)]
✓ 1
Symbolic working required.
Larger m makes F/m smaller, so the denominator is smaller and s is larger (the slope term g sin θ is the same for both)
✓ 1
Part (c)(i)
P = Fv = 7.0 × 104 × 28 = 2.0 × 106 W
✓ 1
Allow ECF from (b)(ii).
Part (c)(ii)
All the kinetic energy is dissipated by the gravel: 1.25 × 107/7.0 × 104 = 178 m
✓ 1
Allow ECF from (b)(ii). Accept 175–180 m.
On the slope about 40 % of the energy becomes gravitational potential energy, so a shorter lane is needed; the gravel stops the stored energy from returning the lorry down the slope
✓ 1
Answers: (a)(i) 349 K · (b)(i) 1.25 × 107 J; 4.8 × 106 J · (b)(ii) 7.0 × 104 N · (c)(i) 2.0 × 106 W · (c)(ii) 178 m (the remaining parts are explanations — see the table above)
Syllabus understandingA.3 — work done W = Fs cos θ; the change in mechanical energy equals the work done by non-conservative forces; power P = Fv; B.1 — specific heat capacity Command term: Determine
37A-2-26
Elastic potential energy·A.3 Work, energy and power
Paper 2Easy12 marks
Short answer & extended response7 steps to full marksDetermine
A child bounces on a pogo stick. The combined mass of the child and the stick is 32 kg. The graph shows how the force needed to compress the spring of the stick varies with its compression. At the lowest point of each bounce the spring is compressed by 0.12 m and the child and stick are momentarily at rest.
Force F needed to compress the spring against the compression (drawn to scale).
(a)
(i)
Use the graph to determine the spring constant of the spring.
(1)
(ii)
Determine the elastic potential energy stored at the lowest point.
(2)
(iii)
Show that, if no energy were dissipated, the centre of mass of the child and stick would rise about 0.28 m from the lowest point to the highest point.
(2)
(b)
(i)
The centre of mass actually rises 0.22 m. Calculate the efficiency of the transfer from elastic to gravitational potential energy.
(2)
(ii)
The child makes 2.1 bounces per second at a steady height. Calculate the average power the child must supply.
(2)
(c)
(i)
The child stands still on the stick. Calculate the compression of the spring.
(1)
(ii)
The child then bounces gently without the stick leaving the ground. Calculate the frequency of the resulting simple harmonic motion.
(2)
Show mark scheme
Marking point
Mark
Notes
Part (a)(i)
k = gradient = 1440/0.12 = 1.2 × 104 N m−1
✓ 1
Accept 1.1–1.3 × 104 N m−1.
Part (a)(ii)
Area under the graph (triangle) = ½ × 0.12 × 1440
✓ 1
= 86 J
✓ 1
Allow ECF from (a)(i) if ½kx² is used. Accept 86.4 J.
Energy dissipated per bounce = 86.4 − 69.1 = 17.3 J
✓ 1
Allow ECF from (b)(i).
P = 17.3 × 2.1 = 36 W
✓ 1
Accept 35–37 W.
Part (c)(i)
x = mg/k = 32 × 9.81/1.2 × 104 = 0.026 m
✓ 1
Allow ECF from (a)(i).
Part (c)(ii)
f = (1/2π)√(k/m)
✓ 1
The oscillation is about the equilibrium position of (c)(i).
f = (1/2π)√(1.2 × 104/32) = 3.1 Hz
✓ 1
Allow ECF from (a)(i).
Answers: (a)(i) 1.2 × 104 N m−1 · (a)(ii) 86 J · (b)(i) 0.80 · (b)(ii) 36 W · (c)(i) 0.026 m · (c)(ii) 3.1 Hz (the remaining parts are explanations — see the table above)
Syllabus understandingA.3 — elastic potential energy EH = ½kΔx² and the work done as the area under a force–extension graph; efficiency; power; A.2 — Hooke's law; C.1 — the period of a mass–spring system T = 2π√(m/k) Command term: Determine
38A-2-50
Conservation of energy·A.3 Work, energy and power
Paper 2Easy11 marks
Short answer & extended response8 steps to full marksDetermine
In a drop-tower ride, a gondola carrying riders, of total mass 4.8 × 103 kg, is raised at a constant speed of 2.0 m s−1 by an electric motor to a height of 56.0 m above its lowest point. It is then released from rest and falls freely through 40.0 m. Magnetic brakes then bring it to rest over the final 16.0 m. Air resistance and friction are negligible.
(a)
(i)
Calculate the useful power output of the motor while the gondola is being raised.
(1)
(ii)
The efficiency of the motor is 75 %. Determine the electrical energy supplied to the motor to raise the gondola through 56.0 m.
(2)
(b)
(i)
Show that the speed of the gondola at the end of the free fall is about 28 m s−1.
(1)
(ii)
Determine the average braking force exerted on the gondola by the magnetic brakes.
(2)
(iii)
Calculate the magnitude of the average acceleration of the gondola during braking, as a multiple of g.
(1)
(c)
(i)
The brakes consist of copper fins on the gondola that pass between strong permanent magnets. Explain why a force opposing the motion acts on the gondola.
(2)
(ii)
The copper fins have a total mass of 120 kg. The specific heat capacity of copper is 385 J kg−1 K−1. Assume that all the energy transferred by the brakes becomes internal energy of the fins. Calculate the rise in temperature of the fins in one ride.
(2)
Show mark scheme
Marking point
Mark
Notes
Part (a)(i)
P = Fv = mgv = 4.8 × 103 × 9.81 × 2.0 = 9.4 × 104 W
✓ 1
At constant speed the lifting force equals the weight.
Part (a)(ii)
ΔEp = 4.8 × 103 × 9.81 × 56.0 = 2.64 × 106 J
✓ 1
Energy supplied = 2.64 × 106/0.75 = 3.5 × 106 J
✓ 1
Multiplying by 0.75 (2.0 × 106 J) scores [1 max].
Part (b)(i)
½mv² = mgh, so v = √(2 × 9.81 × 40.0) = 28.0 m s−1
✓ 1
Must see 28.0 or the full substitution.
Part (b)(ii)
Work done by the brakes = kinetic energy at the start of braking + gravitational potential energy lost during braking: F × 16.0 = ½ × 4.8 × 103 × 28.0² + 4.8 × 103 × 9.81 × 16.0 (= mg × 56.0)
✓ 1
Allow ECF from (b)(i). The 16.0 m of descent during braking is the hidden step.
F = 2.64 × 106/16.0 = 1.6 × 105 N
✓ 1
Accept 1.65 × 105 N. Ignoring the potential energy lost during braking (1.2 × 105 N) scores [1 max].
Allow ECF from (b)(ii). ALT: a = v²/2s = 28.0²/32.0 = 24.5 m s−2. Using F/m (3.5g) scores [0].
Part (c)(i)
The magnetic flux through regions of the moving copper fins changes, so emfs are induced and eddy currents flow in the fins (Faraday’s law)
✓ 1
D.4. Do not accept "the magnets attract the copper": copper is not magnetic.
By Lenz’s law the induced currents are in a direction that opposes the change producing them, so the force of the magnetic field on the currents opposes the relative motion of fins and magnets
✓ 1
Accept: force on a current-carrying conductor in a magnetic field opposes the motion.
Part (c)(ii)
Energy transferred by the brakes = loss of gravitational potential energy over the whole 56.0 m = 2.64 × 106 J
✓ 1
Allow ECF from (a)(ii) or (b)(ii) (F × 16.0 m).
ΔT = 2.64 × 106/(120 × 385) = 57 K
✓ 1
Accept 56–58 K.
Answers: (a)(i) 9.4 × 104 W · (a)(ii) 3.5 × 106 J · (b)(ii) 1.6 × 105 N · (b)(iii) 2.5g · (c)(ii) 57 K (the remaining parts are explanations — see the table above)
Syllabus understandingA.3 — the principle of the conservation of energy; the gravitational potential energy, when close to the surface of the Earth, as given by ΔEp = mgΔh; the kinetic energy of translational motion as given by Ek = ½mv² = p²/2m; that power developed P is the rate of work done, or the rate of energy transfer, as given by P = ΔW/Δt = Fv; efficiency η in terms of energy transfer or power as given by η = Eoutput/Einput = Poutput/Pinput; guidance: the change in the total mechanical energy of a system interpreted in terms of the work done on the system by any non-conservative force; B.1 — specific heat capacity Q = mcΔT; D.4 — Faraday’s and Lenz’s laws Command term: Determine
39A-2-51
Kinetic energy and momentum·A.3 Work, energy and power
Paper 2Medium13 marks
Short answer & extended response9 steps to full marksDetermine
A pile driver is used to drive a steel pile vertically into the ground. A hammer of mass 1500 kg is raised 2.40 m above the top of the pile, which has a mass of 600 kg, and is released from rest. The hammer falls freely, strikes the pile and stays in contact with it, and the hammer and pile then move down together until they stop. In each blow the pile moves 0.090 m into the ground.
(a)
(i)
Calculate the speed of the hammer just before it strikes the pile.
(1)
(ii)
Calculate the speed of the hammer and pile immediately after the impact.
(2)
(iii)
A hammer of mass M strikes a pile of mass m and they move off together. Show, using Ek = p²/2m, that the fraction of the hammer’s kinetic energy transferred to other forms in the impact is m/(M + m), and hence that this fraction is about 0.3 for this pile driver.
(2)
(b)
(i)
Determine the average resistive force exerted by the ground on the pile while the hammer and pile move into the ground.
(3)
(ii)
A heavier hammer is dropped from a smaller height so that it has the same kinetic energy just before the impact. Explain why it drives the pile further into the ground.
(2)
(c)
(i)
The hammer is raised through 2.40 m in 4.0 s at a constant speed by a winch. Calculate the useful power output of the winch.
(1)
(ii)
The winch is driven by a diesel engine of efficiency 30 %. The energy density of diesel fuel is 3.6 × 1010 J m−3. Determine the volume of diesel used to raise the hammer once.
(2)
Show mark scheme
Marking point
Mark
Notes
Part (a)(i)
v = √(2 × 9.81 × 2.40) = 6.86 m s−1
✓ 1
Accept 6.9 m s−1.
Part (a)(ii)
Momentum is conserved in the impact: 1500 × 6.86 = (1500 + 600)v
✓ 1
Allow ECF from (a)(i). The impact is so short that the impulse of the external forces (weight, ground) is negligible.
v = 4.90 m s−1
✓ 1
Accept 4.9 m s−1.
Part (a)(iii)
The momentum p is the same before and after, so Ekafter/Ekbefore = [p²/2(M + m)]/[p²/2M] = M/(M + m)
✓ 1
Symbolic working required. Allow ECF from (a)(ii) only for the numerical check.
Fraction transferred = 1 − M/(M + m) = m/(M + m) = 600/2100 = 0.286
✓ 1
Must see 0.286 or 600/2100. Numerical check from the speeds: 1 − (½ × 2100 × 4.90²)/(½ × 1500 × 6.86²) = 0.286 is accepted for this mark only.
Part (b)(i)
Ek after impact = ½ × 2100 × 4.90² = 2.52 × 104 J
✓ 1
Allow ECF from (a)(ii).
Gravitational potential energy lost in the 0.090 m = 2100 × 9.81 × 0.090 = 1.85 × 103 J
✓ 1
The hidden step: the hammer and pile are still falling while they stop.
F × 0.090 = 2.52 × 104 + 1.85 × 103, so F = 3.0 × 105 N
✓ 1
Accept 3.0–3.1 × 105 N. 2.8 × 105 N (potential energy ignored) scores [2 max].
Part (b)(ii)
By (a)(iii) the fraction of the kinetic energy kept after the impact is M/(M + m), which is larger for a larger hammer mass M
✓ 1
Allow ECF from (a)(iii).
So more energy is available to do work against the (same) resistive force of the ground, and the distance moved, ≈ energy/force, is larger
✓ 1
Do not accept "a heavier hammer has more momentum" without the energy argument.
Part (c)(i)
P = 1500 × 9.81 × 2.40/4.0 = 8.8 × 103 W
✓ 1
Accept 8.83 kW.
Part (c)(ii)
Energy from the fuel = 1500 × 9.81 × 2.40/0.30 = 1.18 × 105 J
Answers: (a)(i) 6.86 m s−1 · (a)(ii) 4.90 m s−1 · (b)(i) 3.0 × 105 N · (c)(i) 8.8 × 103 W · (c)(ii) 3.3 × 10−6 m3(the remaining parts are explanations — see the table above)
Syllabus understandingA.3 — the kinetic energy of translational motion as given by Ek = ½mv² = p²/2m; that work done by the resultant force on a system is equal to the change in the energy of the system; guidance: the change in the total mechanical energy of a system interpreted in terms of the work done on the system by any non-conservative force; that power developed P is the rate of work done, or the rate of energy transfer, as given by P = ΔW/Δt = Fv; efficiency η in terms of energy transfer or power as given by η = Eoutput/Einput = Poutput/Pinput; energy density of the fuel sources; A.2 — conservation of linear momentum; energy considerations in inelastic collisions Command term: Determine
40A-2-52
Power and rate of energy transfer·A.3 Work, energy and power
Paper 2Medium12 marks
Short answer & extended response9 steps to full marksDetermine
A lift car and its passengers have a total mass of 1400 kg. The car is connected by a cable that passes over a pulley at the top of the lift shaft to a counterweight of mass 1200 kg. A motor turns the pulley and exerts a driving force on the cable system. The cable and pulley have negligible mass, and friction is negligible. The graph shows how the upward velocity of the car varies with time during one trip.
Upward velocity of the lift car against time for one trip (drawn to scale).
(a)
(i)
Determine the height risen by the car during the trip.
(1)
(b)
(i)
Show that the increase in gravitational potential energy of the system of car and counterweight during the trip is about 5.5 × 104 J.
(2)
(ii)
Calculate the power output of the motor while the car moves at constant speed.
(2)
(iii)
Determine the maximum power output of the motor during the trip.
(3)
(iv)
Deduce the force that the motor must exert on the cable system during the final 2.0 s of the trip, and comment on the direction of the energy transfer between the motor and the moving masses.
(2)
(c)
(i)
The motor has an efficiency of 85 % and is supplied from a 400 V source. Calculate the current in the motor while the car moves at constant speed.
(2)
Show mark scheme
Marking point
Mark
Notes
Part (a)(i)
Area under the graph = ½ × 2.0 × 2.0 + 12.0 × 2.0 + ½ × 2.0 × 2.0 = 28 m
✓ 1
A.1.
Part (b)(i)
The car gains 1400 × 9.81 × 28 J and the counterweight loses 1200 × 9.81 × 28 J
During the first 2.0 s the acceleration is 1.0 m s−2 and both the car and the counterweight accelerate: F − (1400 − 1200)g = (1400 + 1200) × 1.0
✓ 1
The hidden step: the total mass 2600 kg is accelerated. Using 1400 kg only gives 6.7 × 103 W and scores [2 max].
F = 1962 + 2600 = 4.56 × 103 N
✓ 1
Allow ECF from (b)(ii) for (1400 − 1200)g.
The power is greatest at the end of the acceleration, where v = 2.0 m s−1: P = 4.56 × 103 × 2.0 = 9.1 × 103 W
✓ 1
Accept 9.12 kW.
Part (b)(iv)
(1400 − 1200) × 9.81 − 2600 × 1.0 = −6.4 × 102 N: a force of 640 N opposite to the direction of motion of the car
✓ 1
Allow ECF from (b)(iii) method. Accept 638 N.
The motor does negative work, so energy is transferred from the moving car and counterweight to the motor (the kinetic energy is not all needed to lift the extra 200 kg)
✓ 1
Accept: the motor acts as a brake or could act as a generator. The comment must be consistent with the sign found.
Part (c)(i)
Electrical power input = 3.92 × 103/0.85 = 4.6 × 103 W
✓ 1
Allow ECF from (b)(ii).
I = P/V = 4.62 × 103/400 = 12 A
✓ 1
Accept 11.5 A. 9.8 A (efficiency ignored) scores [1 max].
Answers: (a)(i) 28 m · (b)(ii) 3.9 × 103 W · (b)(iii) 9.1 × 103 W · (b)(iv) 6.4 × 102 N opposing the motion · (c)(i) 12 A (the remaining parts are explanations — see the table above)
Syllabus understandingA.3 — that power developed P is the rate of work done, or the rate of energy transfer, as given by P = ΔW/Δt = Fv; the gravitational potential energy, when close to the surface of the Earth, as given by ΔEp = mgΔh; efficiency η in terms of energy transfer or power as given by η = Eoutput/Einput = Poutput/Pinput; that work done by the resultant force on a system is equal to the change in the energy of the system; that work done by a force is equivalent to a transfer of energy; A.1 — displacement as the area under a velocity–time graph; A.2 — Newton’s second law; B.5 — electrical power P = IVCommand term: Determine
41A-2-53
Mechanical energy·A.3 Work, energy and power
Paper 2Hard14 marks
Short answer & extended response10 steps to full marksDetermine
A bungee jumper of mass 72 kg steps off a high bridge and falls vertically from rest. She is attached to the bridge by an elastic cord of unstretched length 20.0 m that obeys Hooke’s law with spring constant 150 N m−1. Treat the jumper as a point mass, and ignore air resistance, the mass of the cord and energy dissipation in the cord unless told otherwise.
(a)
(i)
Calculate the speed of the jumper when the cord first becomes taut.
(1)
(ii)
Show that the extension of the cord when the jumper is at her lowest point is about 19 m.
(2)
(iii)
Determine the maximum speed of the jumper.
(3)
(iv)
Determine the magnitude and direction of the acceleration of the jumper at her lowest point.
(2)
(b)
(i)
In reality energy is dissipated in the cord and the air, and after several oscillations the jumper hangs at rest. Determine the total energy dissipated from the moment she steps off the bridge.
(2)
(ii)
While she hangs at rest, the jumper is pulled down a short distance and released, so that she oscillates with the cord always taut. Explain why the oscillations are simple harmonic and calculate their period.
(2)
(c)
(i)
For cords made from the same material, the product of spring constant and unstretched length, kL, is constant. The jumper uses a cord of the same material with twice the unstretched length. Deduce, without calculating new values, that the maximum tension in the cord is unchanged.
(2)
Show mark scheme
Marking point
Mark
Notes
Part (a)(i)
v = √(2 × 9.81 × 20.0) = 19.8 m s−1
✓ 1
Part (a)(ii)
At the lowest point Ek = 0, so the gravitational potential energy lost over the whole fall equals the elastic potential energy: 72 × 9.81 × (20.0 + x) = ½ × 150 × x²
✓ 1
The fall is (20.0 + x), not 20.0 m: using 20.0 m only (x = 13.7 m) scores [0].
75x² − 706x − 1.41 × 104 = 0, so x = 19.2 m
✓ 1
Must see 19.2 m or the quadratic with its positive root.
Part (a)(iii)
The speed is greatest where the resultant force is zero: kx = mg, so x = 72 × 9.81/150 = 4.71 m
✓ 1
Taking the maximum speed as the speed at which the cord becomes taut (19.8 m s−1) scores [0] for the question.
½mv² = mg(20.0 + 4.71) − ½ × 150 × 4.71²
✓ 1
Allow ECF from (a)(i) if done in two stages: ½mv² = ½m(19.8)² + mg(4.71) − ½k(4.71)².
Accept 30.2 m s−2; direction required. Using 19 m gives 29.8 m s−2 and is accepted.
Part (b)(i)
At rest the extension is 4.71 m, so the jumper has fallen 24.7 m: gravitational potential energy lost = 72 × 9.81 × 24.7 = 1.75 × 104 J; elastic potential energy stored = ½ × 150 × 4.71² = 1.66 × 103 J
Accept 1.58 × 104 J. The change in total mechanical energy is the work done by the non-conservative forces.
Part (b)(ii)
The weight is constant and the tension changes by k × (displacement from the equilibrium position), so the resultant force is −k × displacement: the acceleration is proportional to the displacement and directed towards the equilibrium position
✓ 1
C.1. Reference to displacement from the equilibrium position is required.
T = 2π√(m/k) = 2π√(72/150) = 4.4 s
✓ 1
Accept 4.35 s.
Part (c)(i)
Write k = C/L and x = λL: the energy equation mg(L + x) = ½kx² becomes mg(1 + λ) = ½Cλ², which does not contain L, so the extension is the same fraction λ of the unstretched length
✓ 1
Allow ECF from (a)(ii) method. Symbolic argument required; numerical substitution of new values scores [1 max].
Maximum tension = kx = (C/L)(λL) = Cλ, independent of L (so the maximum acceleration is also unchanged)
✓ 1
Answers: (a)(i) 19.8 m s−1 · (a)(iii) 20.9 m s−1 · (a)(iv) 30 m s−2 upwards · (b)(i) 1.6 × 104 J · (b)(ii) 4.4 s (the remaining parts are explanations — see the table above)
Syllabus understandingA.3 — that mechanical energy is the sum of kinetic energy, gravitational potential energy and elastic potential energy; the principle of the conservation of energy; the elastic potential energy as given by EH = ½k(Δx)²; the gravitational potential energy, when close to the surface of the Earth, as given by ΔEp = mgΔh; guidance: the change in the total mechanical energy of a system interpreted in terms of the work done on the system by any non-conservative force; A.2 — Hooke’s law; Newton’s second law; C.1 — conditions for simple harmonic motion; the period of a mass–spring system T = 2π√(m/k) Command term: Determine
42A-2-54
Conservation of energy·A.3 Work, energy and power
Paper 2Hard14 marks
Short answer & extended response11 steps to full marksDetermine
A pole vaulter of mass 68 kg runs up and plants her pole with a horizontal speed of 9.20 m s−1; at this instant her centre of mass is 1.10 m above the ground. The pole bends, storing elastic potential energy, and then straightens, lifting her. Her centre of mass passes over the bar 5.60 m above the ground with a horizontal speed of 0.80 m s−1. Treat the vaulter as a point mass at her centre of mass and ignore air resistance.
Force exerted along the pole against the decrease Δ in the distance between its ends as it bends (drawn to scale).
(a)
(i)
Calculate the kinetic energy of the vaulter as she plants the pole.
(1)
(ii)
Show that, if all of this kinetic energy were transferred to gravitational potential energy, her centre of mass would rise to about 5.4 m above the ground.
(2)
(iii)
Assume that the pole returns all the energy stored in it. Determine the work done by the vaulter’s muscles during the vault, from planting the pole to passing over the bar.
(3)
(b)
(i)
The graph shows how the force exerted along the pole varies with the decrease Δ in the distance between its ends. At maximum bend Δ = 1.50 m. Estimate the elastic potential energy stored in the pole at maximum bend.
(2)
(ii)
Calculate the fraction of the vaulter’s kinetic energy at take-off that is stored in the pole at maximum bend.
(1)
(c)
(i)
Show that the rise of the centre of mass calculated in (a)(ii) does not depend on the mass of the vaulter, and deduce the percentage change in this rise if the take-off speed is increased by 2.0 %.
(2)
(ii)
After clearing the bar the vaulter falls onto a landing mat. Her centre of mass is 0.80 m above the ground when she first touches the mat. Determine the speed at which she reaches the mat.
(1)
(iii)
The mat reduces her vertical velocity to zero in 0.40 s. Determine the average vertical force exerted on her by the mat.
(2)
Show mark scheme
Marking point
Mark
Notes
Part (a)(i)
Ek = ½ × 68 × 9.20² = 2.88 × 103 J
✓ 1
Part (a)(ii)
Δh = v²/2g = 9.20²/(2 × 9.81) = 4.31 m
✓ 1
Allow ECF from (a)(i) (Δh = Ek/mg).
Height = 1.10 + 4.31 = 5.41 m
✓ 1
Must see 5.41 m or the full substitution. The initial height of 1.10 m must be added.
Part (a)(iii)
Mechanical energy at the bar = 68 × 9.81 × 5.60 + ½ × 68 × 0.80² = 3.736 × 103 + 22 = 3.757 × 103 J
✓ 1
Kinetic energy at the bar must be included.
Mechanical energy at take-off = 2.878 × 103 + 68 × 9.81 × 1.10 = 3.612 × 103 J
✓ 1
Allow ECF from (a)(i).
Work done by the muscles = increase in mechanical energy = 1.5 × 102 J
✓ 1
Accept 140–150 J. Heights measured from the take-off position give the same result.
Part (b)(i)
Area under the graph from 0 to 1.50 m, e.g. by counting squares (each 0.1 m × 200 N square = 20 J) or by trapezia
✓ 1
≈ 2.1 × 103 J
✓ 1
Accept 1.9–2.3 × 103 J. Treating the area as a rectangle (1.50 × 1800 = 2.7 × 103 J) or a triangle (1.3 × 103 J) scores [1 max].
Part (b)(ii)
2.08 × 103/2.88 × 103 = 0.72
✓ 1
Allow ECF from (a)(i) and (b)(i). Accept 0.66–0.80.
Part (c)(i)
½mv² = mgΔh, so Δh = v²/2g: m cancels
✓ 1
Allow ECF from (a)(ii).
Δh ∝ v², so the rise increases by 1.02² − 1 = 0.040, i.e. 4.0 % (about 0.17 m)
✓ 1
Accept 4 %. "2 %" scores [0] for this mark.
Part (c)(ii)
½mv² = ½m(0.80)² + mg(5.60 − 0.80), so v = √(0.64 + 2 × 9.81 × 4.80) = 9.7 m s−1
✓ 1
Accept 9.74 m s−1. Ignoring the 0.80 m s−1 gives 9.70 m s−1 and is accepted.
Part (c)(iii)
Vertical component of velocity at contact = √(2 × 9.81 × 4.80) = 9.70 m s−1; rate of change of momentum = 68 × 9.70/0.40 = 1.65 × 103 N
✓ 1
Allow ECF from (c)(ii) (using 9.7 m s−1 is accepted).
Force from the mat − weight = 1.65 × 103 N, so the force = 1.65 × 103 + 68 × 9.81 = 2.3 × 103 N
✓ 1
Accept 2.3–2.4 × 103 N. Omitting the weight (1.6 × 103 N) scores [1 max].
Answers: (a)(i) 2.88 × 103 J · (a)(iii) 1.5 × 102 J · (b)(i) 2.1 × 103 J · (b)(ii) 0.72 · (c)(ii) 9.7 m s−1 · (c)(iii) 2.3 × 103 N (the remaining parts are explanations — see the table above)
Syllabus understandingA.3 — the principle of the conservation of energy; that mechanical energy is the sum of kinetic energy, gravitational potential energy and elastic potential energy; the kinetic energy of translational motion as given by Ek = ½mv² = p²/2m; the gravitational potential energy, when close to the surface of the Earth, as given by ΔEp = mgΔh; that work done by a force is equivalent to a transfer of energy; that work done by the resultant force on a system is equal to the change in the energy of the system; A.2 — Newton’s second law in terms of the rate of change of momentum; impulse; Tools 3 — the area under a force–displacement graph as the work done Command term: Determine
43A-2-65
Energy transfer in a wind turbine·A.3 Work, energy and power
Paper 2Hard20 marks
Short answer & extended response14 steps to full marksDetermine
A horizontal-axis wind turbine has three blades that sweep a circle of diameter 90 m. Its hub is 80 m above the ground. The graph shows the electrical power output of the turbine for different wind speeds. The density of air is 1.2 kg m−3.
Electrical output power P of the turbine against wind speed (drawn to scale).
(a)
(i)
Show that the kinetic energy carried per second by wind of speed v through an area A perpendicular to the wind is ½ρAv³.
(2)
(ii)
Show that ½ρAv³ has the fundamental SI units of power.
(1)
(iii)
Determine the overall efficiency of the turbine when the wind speed is 10 m s−1.
(2)
(b)
(i)
At a wind speed of 10.0 m s−1 the air passes through the blades at an average speed of 7.0 m s−1 and leaves well behind the turbine at 4.0 m s−1. Determine the force exerted by the air on the rotor.
(3)
(ii)
Calculate the torque of this force about the base of the tower.
(1)
(iii)
Show that, in this model, the power removed from the wind is about 2.2 MW, and suggest why it is greater than the electrical output read from the graph.
(2)
(c)
(i)
At a wind speed of 10 m s−1 the rotor turns at 16 revolutions per minute. Calculate the speed of the tips of the blades.
(1)
(ii)
Each blade has a mass of 6.5 × 103 kg and its centre of mass is 15 m from the axis of rotation. Determine the magnitude of the resultant force that the hub exerts on one blade. Ignore the weight of the blade.
(2)
(iii)
A damaged blade is repaired so that the centre of mass of the rotor no longer lies on its axis of rotation. Explain why the tower then vibrates when the rotor turns.
(2)
(d)
(i)
Assume that the overall efficiency of the turbine stays at the value found in (a)(iii). Determine the wind speed at which the turbine would reach its rated output of 2.0 MW, and comment on your answer with reference to the graph.
(2)
(ii)
Explain why no turbine can transfer all of the kinetic energy of the wind passing through it.
(2)
Show mark scheme
Marking point
Mark
Notes
Part (a)(i)
Mass of air passing per second = ρ × volume per second = ρAv
✓ 1
Symbolic working required.
Kinetic energy per second = ½(ρAv)v² = ½ρAv³
✓ 1
Part (a)(ii)
kg m−3 × m² × m³ s−3 = kg m² s−3 = (kg m s−2)(m s−1) = N m s−1 = W
✓ 1
Fundamental units kg m² s−3 must be seen.
Part (a)(iii)
Power in the wind = ½ × 1.2 × π × 45² × 10³ = 3.82 × 106 W
✓ 1
Allow ECF from (a)(i). Using 90 m as the radius scores 0 for this mark.
Output read from the graph ≈ 1.14 MW; efficiency = 1.14/3.82 = 0.30
✓ 1
Accept 1.1–1.2 MW and 0.28–0.32.
Part (b)(i)
Mass flow = ρA × 7.0 = 1.2 × 6362 × 7.0 = 5.34 × 104 kg s−1
✓ 1
Allow ECF from (a)(iii) area.
Rate of change of momentum of the air = 5.34 × 104 × (10.0 − 4.0)
✓ 1
The speed 7.0 m s−1 sets the mass flow; the change of speed is 6.0 m s−1.
F = 3.2 × 105 N
✓ 1
Accept 3.1–3.3 × 105 N.
Part (b)(ii)
τ = 3.2 × 105 × 80 = 2.6 × 107 N m
✓ 1
Allow ECF from (b)(i).
Part (b)(iii)
½ × 5.34 × 104 × (10.0² − 4.0²) = 2.24 × 106 W
✓ 1
Allow ECF from (b)(i) mass flow.
Energy is dissipated between the air and the grid: drag and turbulence at the blades, friction in the bearings and gearbox, and heating in the generator
✓ 1
Any one valid loss.
Part (c)(i)
ω = 16 × 2π/60 = 1.68 rad s−1; v = ωr = 1.676 × 45 = 75 m s−1
✓ 1
Using 90 m as the radius (151 m s−1) scores 0.
Part (c)(ii)
The force is the centripetal force on the blade, whose mass may be treated as concentrated at its centre of mass: a = ω²r = 1.676² × 15 = 42.1 m s−2
✓ 1
Allow ECF from (c)(i) for ω. Using 45 m scores 0 for this mark.
F = 6.5 × 103 × 42.1 = 2.7 × 105 N (about four times the weight of the blade)
✓ 1
Accept 2.7–2.8 × 105 N.
Part (c)(iii)
The centre of mass of the rotor now moves in a circle about the axis, so a resultant (centripetal) force must act on the rotor; it is provided by the shaft and bearings
✓ 1
Do not accept "the rotor is unbalanced" without reference to a force.
The direction of this force turns with the rotor, so (Newton’s third law) the rotor exerts on the tower a force whose direction changes once every revolution: a periodic force that makes the tower vibrate
✓ 1
Accept a reference to a periodic driving force at the rotation frequency (possible resonance).
Part (d)(i)
Power in the wind needed = 2.0 × 106/0.30 = 6.7 × 106 W
✓ 1
Allow ECF from (a)(iii).
v = (2P/ρA)1/3 = (2 × 6.7 × 106/(1.2 × 6362))1/3 = 12 m s−1, the speed at which the graph reaches 2.0 MW
✓ 1
Accept 11.8–12.3 m s−1. The comment must refer to the graph.
Part (d)(ii)
To remove all the kinetic energy the air would have to leave the blades at zero speed
✓ 1
Allow ECF from (a)(i) and (b)(iii).
Air that has stopped would block the air arriving behind it, so the mass flow through the rotor would fall to zero; some kinetic energy must remain to carry the air away
✓ 1
Do not accept "because of friction".
Answers: (a)(iii) 0.30 · (b)(i) 3.2 × 105 N · (b)(ii) 2.6 × 107 N m · (c)(i) 75 m s−1 · (c)(ii) 2.7 × 105 N · (d)(i) 12 m s−1(the remaining parts are explanations — see the table above)
Syllabus understandingA.3 — kinetic energy; energy density and power as the rate of transfer of energy; efficiency; A.2 — Newton's second law in the form F = Δp/Δt; Newton's third law; centripetal acceleration a = ω²r and v = ωr; that circular motion is caused by a centripetal force; A.4 — the torque τ of a force about an axis; Tools — fundamental SI units Command term: Determine
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