A.3 Work, energy and power: IB Physics SL exam-style questions
Work is a transfer of energy, W = Fs cos θ, and the work done by the resultant force equals the change in kinetic energy. Without friction, mechanical energy (kinetic, gravitational and elastic) is conserved; with friction, the missing energy is the work done by the non-conservative force.
Power is the rate of transfer of energy, including P = Fv for a vehicle at constant speed. Efficiency, Sankey diagrams and the energy density of fuels connect this topic to real engineering and energy problems.
27 questions
200 marks
Paper 1A: 10
Paper 1B: 6
Paper 2: 11
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27 practice questions on A.3 Work, energy and power
1A-1A-09
Work done by a force·A.3 Work, energy and power
Paper 1AEasy1 mark
Multiple choice · 1 mark2 steps to full marksCalculate
A box is dragged 4.0 m along horizontal ground by a force of 25 N applied at 60° to the horizontal.
What is the work done on the box by this force?
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Marking point
Mark
Notes
Step 1Only the component of the force along the direction of motion does work, so W = Fs cos θ.
—
All 2 steps must be completed — there is no mark for a part-answer.
Step 2W = 25 × 4.0 × cos 60° = 25 × 4.0 × 0.500 = 50 J.
✓ 1
Answer C
Answer: C · 2 stages of work, one mark
Every option, and why
AThis is Fs, ignoring the angle. Only the horizontal component of the force does work here.
BThis uses sin 60°. The angle in W = Fs cos θ is measured between the force and the displacement.
CCorrect: W = Fs cos θ = 25 × 4.0 × cos 60° = 50 J.
DThe work would be zero only if the force were perpendicular to the displacement, that is at 90°.
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 θCommand term: Calculate
2A-1A-10
Power & efficiency·A.3 Work, energy and power
Paper 1AMedium1 mark
Multiple choice · 1 mark3 steps to full marksDetermine
The Sankey diagram represents the energy transfers in an electric motor for every 500 J of electrical energy supplied.
What is the efficiency of the motor?
Diagram NOT accurately drawnShow mark scheme
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Notes
Step 1A Sankey diagram conserves energy: the branch widths must add up to the input width.
—
All 3 steps must be completed — there is no mark for a part-answer.
Step 2The two loss branches carry 120 J and 80 J, so the useful output is 500 − 120 − 80 = 300 J.
—
Step 3η = Eoutput / Einput = 300 / 500 = 0.60, that is 60 %.
✓ 1
Answer A
Answer: A · 3 stages of work, one mark
Every option, and why
ACorrect: the useful output is 500 − 120 − 80 = 300 J, so η = 300/500 = 0.60.
BThis is the fraction of the input that is wasted, (120 + 80)/500. Efficiency is the useful fraction, not the wasted one.
CThis takes the thermal branch as the useful output. In a motor, thermal energy is a loss.
DThis takes the sound branch alone as the useful output.
Syllabus understandingA.3 — that energy transfers can be represented on a Sankey diagram; efficiency η in terms of energy transfer or power Command term: Determine
3A-1A-11
Conservation of energy·A.3 Work, energy and power
Paper 1AMedium1 mark
Multiple choice · 1 mark1 step to full marksDeduce
A crate is pulled up a rough ramp by a rope parallel to the ramp, and its speed increases. Over a certain distance the rope does work W on the crate, the crate gains kinetic energy ΔEk and gravitational potential energy ΔEp, and the work done against friction is Wf.
Which equation is correct?
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Notes
Step 1Energy is conserved: the energy transferred to the crate by the rope becomes kinetic energy, gravitational potential energy and, through the work done against friction, internal energy of the crate and ramp, so W = ΔEk + ΔEp + Wf.
✓ 1
Answer D
Answer: D · 1 stage of work, one mark
Every option, and why
AThis treats friction as a source of energy. Work done against friction is energy the rope must supply in addition, dissipated as internal energy.
BThis treats the gain in height as releasing energy. Moving up the ramp increases the gravitational potential energy, which the rope must supply.
CThis treats the rope's work as the work done by the resultant force. The resultant force also includes the weight component and friction, so the rope does more work than the gain in kinetic energy.
DCorrect: all the work done by the rope is accounted for as kinetic energy, gravitational potential energy and the energy dissipated against friction.
Syllabus understandingA.3 — the principle of the conservation of energy; that work done by the resultant force on a system is equal to the change in the energy of the system; the change in the total mechanical energy of a system is interpreted in terms of the work done by any non-conservative force Command term: Deduce
4A-1A-12
Conservation of energy·A.3 Work, energy and power
Paper 1AMedium1 mark
Multiple choice · 1 mark2 steps to full marksIdentify
A ball is released from rest and falls vertically. Air resistance is negligible.
Which graph shows how the kinetic energy Ek of the ball varies with time t after release?
Sketch graphs, not to scaleShow mark scheme
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Notes
Step 1The ball accelerates uniformly from rest, so v = gt.
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All 2 steps must be completed — there is no mark for a part-answer.
Step 2Ek = ½mv² = ½mg²t², so Ek ∝ t²: a curve through the origin whose gradient increases.
✓ 1
Answer B
Answer: B · 2 stages of work, one mark
Every option, and why
AA straight line is the shape of v against t, or of Ek against the distance fallen (Ek = mgd). Against time, Ek ∝ v² ∝ t².
BCorrect: Ek = ½mg²t², a parabola through the origin that gets steeper as the ball speeds up.
CA curve that levels off describes a ball approaching terminal speed. With negligible air resistance the ball keeps accelerating.
DIt is the total mechanical energy, Ek + Ep, that stays constant. The kinetic energy starts at zero and increases.
Syllabus understandingA.3 — that in the absence of frictional, resistive forces, the total mechanical energy of a system is conserved; the kinetic energy of translational motion as given by Ek = ½mv² Command term: Identify
5A-1A-13
Elastic potential energy·A.3 Work, energy and power
Paper 1AMedium1 mark
Multiple choice · 1 mark2 steps to full marksDetermine
A block is held against a spring on a frictionless horizontal surface, compressing the spring by x. When the block is released, the spring pushes it away. The spring obeys Hooke's law.
The spring is now compressed by 2x instead. What happens to the elastic potential energy stored before release and to the speed of the block after it leaves the spring?
Elastic potential energy storedSpeed of the block
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Notes
Step 1EH = ½k(Δx)², so doubling the compression stores 2² = 4 times as much energy.
—
All 2 steps must be completed — there is no mark for a part-answer.
Step 2With no friction all of it becomes kinetic energy, ½mv², so v ∝ √EH and the speed doubles.
✓ 1
Answer D
Answer: D · 2 stages of work, one mark
Every option, and why
AThis takes the stored energy to be proportional to the compression, as the force is. The energy is the area under the force–compression graph, ½k(Δx)².
BThis takes the stored energy to be proportional to the compression and then correctly takes v ∝ √EH.
CThe energy is right, but the speed has been scaled like the energy. Kinetic energy ∝ v², so four times the energy gives twice the speed.
DCorrect: ½k(Δx)² gives × 4 for the energy, and ½mv² = EH gives × √4 = × 2 for the speed.
Syllabus understandingA.3 — the elastic potential energy as given by EH = ½k(Δx)²; that if mechanical energy is conserved, work is the amount of energy transformed between different forms of mechanical energy in a system Command term: Determine
6A-1B-05
Elastic potential energy·A.3 Work, energy and power
Paper 1BMedium11 marks
Data-based question11 steps to full marksDetermine
A trolley of mass 0.800 ± 0.004 kg on a horizontal track is pushed against a spring fixed to the end of the track, compressing it by a distance x. When the trolley is released the spring pushes it away, and a light gate placed just beyond the point where the trolley leaves the spring measures its speed v.
The uncertainty in x is ±0.1 cm and the uncertainty in v is ±0.02 m s−1.
x / cm
2.0
3.0
4.0
5.0
6.0
7.0
v / m s−1
0.34
0.48
0.66
0.82
1.00
1.15
Graph drawn to scale
(a)
The spring constant is k. Show that, if no energy is dissipated, v = x√(k/m).
(2)
(b)
Draw the line of best fit on the graph.
(1)
(c)
Determine the gradient of the line in s−1.
(2)
(d)
Determine k.
(2)
(e)
The uncertainty in the gradient is ±3 %. Determine the absolute uncertainty in k.
(2)
(f)
Loading the spring with known weights gives k = 250 ± 5 N m−1. Compare this with your answer, and suggest a reason for any difference.
(2)
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Part (a)
elastic potential energy is transferred to kinetic energy: ½kx² = ½mv²
✓ 1
v² = kx²/m so v = x√(k/m)
✓ 1
Part (b)
single straight line through all error bars, passing through «or very close to» the origin
✓ 1
Part (c)
gradient from a large triangle with x converted to m
The gradient is squared, so its fractional uncertainty is doubled.
Δk = 0.065 × 218 ≈ 14 N m−1
✓ 1
Accept 13–16 N m−1.
Part (f)
range 204–232 N m−1 lies below 245–255 N m−1, so the launch value is significantly smaller
✓ 1
Allow ECF.
some elastic energy is not transferred to the trolley's kinetic energy «friction, kinetic energy of the spring itself, vibration, sound», so v is systematically low
✓ 1
OWTTE.
Answers: (c) 16.5 s−1 · (d) 218 N m−1 · (e) ±14 N m−1(the remaining parts are explanations — see the table above)
Syllabus understandingA.3 — that if mechanical energy is conserved, work is the amount of energy transformed between different forms of mechanical energy in a system, such as the elastic potential energy as given by EH = ½k(Δx)² Command term: Determine
7A-1B-06
Power & efficiency·A.3 Work, energy and power
Paper 1BEasy8 marks
Data-based question8 steps to full marksCalculate
A small electric motor winds up a thread to lift a load of mass m vertically through 0.800 ± 0.005 m at a steady speed. The potential difference V across the motor and the current I in it are read from meters, and the time t for the lift is measured with a hand-held stopwatch that reads to 0.01 s. The uncertainty in t is estimated as ±0.2 s.
m / kg
V / V
I / A
t / s
Efficiency / %
0.100
6.02
0.21
3.10
20.0
0.200
5.98
0.30
3.42
25.6
0.300
5.95
0.39
3.81
0.400
5.91
0.49
4.35
24.9
0.500
5.86
0.60
5.12
21.8
(a)
Explain why the uncertainty in t is much larger than the resolution of the stopwatch.
(1)
(b)
(i)
For the 0.300 kg load, calculate the useful output power of the motor.
(1)
(ii)
Hence complete the table.
(2)
(c)
Determine the percentage uncertainty in the output power for the 0.300 kg load. The uncertainty in m is negligible.
(2)
(d)
The student claims that the efficiency of the motor increases as the load increases. Comment on this claim.
(1)
(e)
Suggest one change to the method that would reduce the percentage uncertainty in t.
(1)
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Notes
Part (a)
the stopwatch is started and stopped by hand, so the human reaction time «≈ 0.2 s» dominates
✓ 1
OWTTE.
Part (b)(i)
P = mgh/t = 0.300 × 9.81 × 0.800 / 3.81 = 0.618 W
✓ 1
Accept 0.62 W.
Part (b)(ii)
input power = VI = 5.95 × 0.39 = 2.32 W
✓ 1
efficiency = 0.618/2.32 = 26.6 %
✓ 1
Accept 27 %. Allow ECF from (b)(i).
Part (c)
0.005/0.800 + 0.2/3.81 «= 0.0062 + 0.0525»
✓ 1
= 5.9 % «≈ 6 %»
✓ 1
Accept 5–6 %. Award [1] max if the percentage uncertainties are not added.
Part (d)
the efficiency rises to a maximum of about 27 % near 0.3 kg and then falls «to 22 % at 0.500 kg», by more than the uncertainty, so the claim is not supported
✓ 1
Allow ECF from (b)(ii).
Part (e)
use light gates / a motion sensor to time the lift OR increase the height of the lift so t is longer
✓ 1
Do not accept 'repeat the readings' alone.
Answers: (b)(i) 0.618 W · (b)(ii) 26.6 % · (c) 5.9 % (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; efficiency η in terms of energy transfer or power as given by η = Eoutput/Einput = Poutput/PinputCommand term: Calculate
8A-2-09
Conservation of energy·A.3 Work, energy and power
Paper 2Medium12 marks
Short answer & extended response12 steps to full marksDetermine
A lorry of mass 1.2 × 104 kg travels down a long, straight hill at a constant speed of 15 m s−1. The road down the hill is 3.0 km long and the lorry descends a vertical height of 150 m. The average total resistive force from air resistance and rolling friction is 1.8 kN. The rest of the energy is transferred to the brakes.
The brake discs have a total mass of 60 kg and are made of steel of specific heat capacity 450 J kg−1 K−1.
Figure 1 — Sankey diagram for (c), to be completed
(a)
(i)
State the change in the kinetic energy of the lorry during the descent.
(1)
(ii)
Calculate the decrease in the gravitational potential energy of the lorry.
(1)
(b)
Show that the energy transferred to the brakes is about 12 MJ.
(2)
(c)
Complete, on Figure 1, the Sankey diagram for the energy transfers during the descent.
(2)
(d)
The brake discs are at 20 °C at the top of the hill and their maximum safe temperature is 350 °C. Assume that all the energy transferred to the brakes remains in the discs. Determine whether the discs exceed their maximum safe temperature.
(3)
(e)
Outline why, in practice, the discs reach a lower temperature than the one you found in (d).
(1)
(f)
Calculate the average rate at which energy is transferred to the brakes.
(2)
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Notes
Part (a)(i)
zero «the speed is constant»
✓ 1
Part (a)(ii)
ΔEp = 1.2 × 104 × 9.81 × 150 = 1.77 × 107 J
✓ 1
Accept 17.7 MJ or 18 MJ.
Part (b)
work done against the resistive forces = 1800 × 3000 = 5.4 × 106 J
✓ 1
energy to the brakes = 1.77 × 107 − 5.4 × 106 «= 1.23 × 107 J»
✓ 1
Must see full substitution OR answer to 3 s.f.
Part (c)
two output bands labelled: thermal energy in the brakes «12.3 MJ» AND thermal energy from work against air resistance/rolling friction «5.4 MJ»
✓ 1
band widths in proportion: brakes ≈ 69 % and resistive forces ≈ 31 % of the input width «brakes band about twice as wide»
✓ 1
Allow ECF from (b). The two output widths must add up to the input width.
Part (d)
ΔT = Q/mc = 1.23 × 107 / (60 × 450) = 454 K
✓ 1
Accept 444 K from 12 MJ.
final temperature = 20 + 454 = 474 °C
✓ 1
Allow ECF from (b).
474 °C > 350 °C so the discs exceed their maximum safe temperature
✓ 1
The decision must be consistent with the candidate’s temperature.
Part (e)
energy is transferred from the hot discs to the surrounding air «by convection and radiation» during the descent
✓ 1
Do not accept “heat is lost” without a destination or mechanism.
Part (f)
time = 3000 / 15 = 200 s
✓ 1
P = 1.23 × 107 / 200 = 6.1 × 104 W
✓ 1
Allow ECF from (b). Accept 60 kW from 12 MJ.
Answers: (a)(ii) 1.77 × 107 J · (b) 1.23 × 107 J · (d) 474 °C; exceeds · (f) 6.1 × 104 W (the remaining parts are explanations — see the table above)
Syllabus understandingA.3 — the principle of the conservation of energy; that energy transfers can be represented on a Sankey diagram; that work done by a force is equivalent to a transfer of energy; that power developed P is the rate of work done, or the rate of energy transfer; B.1 — quantitative analysis of thermal energy transfers Q with the use of specific heat capacity c as given by Q = mcΔTCommand term: Determine
9A-2-10
Elastic potential energy·A.3 Work, energy and power
Paper 2Medium12 marks
Short answer & extended response12 steps to full marksDetermine
A railway wagon of mass 1.8 × 104 kg rolls at 1.5 m s−1 along a level track into a buffer stop. The buffer stop has two identical spring buffers side by side, and both are compressed by the same amount. Figure 1 shows how the force F needed to compress one buffer varies with its compression x. Friction is negligible.
Figure 1 — graph drawn to scale
(a)
Determine the spring constant of one buffer.
(1)
(b)
Show that the kinetic energy of the wagon is about 20 kJ.
(1)
(c)
Each buffer can be compressed by at most 0.30 m. Determine whether the buffers can stop the wagon.
(3)
(d)
Calculate the maximum deceleration of the wagon.
(2)
(e)
Suggest why the buffers are also fitted with dampers that transfer energy to internal energy.
(2)
(f)
The buffers are replaced by stiffer ones that also stop the wagon. Explain the effect on the maximum deceleration of the wagon.
(3)
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Notes
Part (a)
gradient = 90 × 103 / 0.30 = 3.0 × 105 N m−1
✓ 1
Accept any correct pair of values read from the line.
Part (b)
½ × 1.8 × 104 × 1.5² «= 2.03 × 104 J»
✓ 1
Must see full substitution OR answer to 3 s.f.
Part (c)
energy stored in the two buffers: 2 × ½kx² = 2.03 × 104 J
✓ 1
x = √(2.03 × 104 / 3.0 × 105) = 0.26 m
✓ 1
Allow ECF from (a) and (b).
0.26 m < 0.30 m so the buffers stop the wagon
✓ 1
Award [1 max] for using one buffer only «x = 0.37 m, “cannot stop it”».
Part (d)
maximum force = 2kx = 2 × 3.0 × 105 × 0.260 = 1.56 × 105 N
✓ 1
Allow ECF from (c).
a = 1.56 × 105 / 1.8 × 104 = 8.7 m s−2
✓ 1
Award [2] for CNA.
Part (e)
without dampers the elastic potential energy stored is all returned to the wagon as kinetic energy
✓ 1
so the wagon would rebound at 1.5 m s−1; dampers dissipate the energy so the wagon comes to rest
✓ 1
OWTTE.
Part (f)
the same kinetic energy is stored, ½kx² = constant, so the maximum compression is smaller
✓ 1
the maximum force kx is larger «kx = √(2kE), which increases with k»
✓ 1
so the maximum deceleration is greater
✓ 1
MP3 only scores if MP2 scores.
Answers: (a) 3.0 × 105 N m−1 · (b) 2.03 × 104 J · (c) 0.26 m; stops the wagon · (d) 8.7 m s−2(the remaining parts are explanations — see the table above)
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; A.2 — elastic restoring force FH following Hooke’s law as given by FH = −kx where k is the spring constant Command term: Determine
10A-1A-32
Conservation of energy·A.3 Work, energy and power
Paper 1AEasy1 mark
Multiple choice · 1 mark1 step to full marksDetermine
A ball is released from rest at a height of 5.0 m above the ground. Air resistance is negligible. Take g = 9.81 m s−2.
What is its speed just before it lands?
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Notes
Step 1With no resistive forces the mechanical energy is conserved, so the loss of gravitational potential energy equals the gain in kinetic energy: mgh = ½mv². The mass cancels, giving v = √(2 × 9.81 × 5.0) = 9.9 m s⁻¹.
✓ 1
Answer B
Answer: B · 1 stage of work, one mark
Every option, and why
AThis is √(gh). The factor of 2 has been lost — v² = 2gh, not gh.
BCorrect: mgh = ½mv², so v = √(2 × 9.81 × 5.0) = 9.9 m s⁻¹.
CThis is gh. Check the algebra — you need the square root of 2gh.
DThis is v² = 2gh = 98, left unrooted. The units give it away: m² s⁻², not m s⁻¹.
Syllabus understandingA.3 — that in the absence of frictional, resistive forces, the total mechanical energy of a system is conserved Command term: Determine
11A-1A-33
Kinetic energy & momentum·A.3 Work, energy and power
Paper 1AMedium1 mark
Multiple choice · 1 mark3 steps to full marksDetermine
Two bodies, of mass 2.0 kg and 8.0 kg, have the same momentum of magnitude 12 kg m s−1.
What is the kinetic energy of the 2.0 kg body, and how does it compare with that of the 8.0 kg body?
Ek of the 2.0 kg body / JRatio Ek(2.0 kg) : Ek(8.0 kg)
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Notes
Step 1The momenta are given, not the speeds, so use the form of the kinetic energy written in terms of momentum: Ek = p²/2m.
—
All 3 steps must be completed — there is no mark for a part-answer.
Step 2For the 2.0 kg body: Ek = 12² / (2 × 2.0) = 144/4 = 36 J.
—
Step 3For the 8.0 kg body: Ek = 144 / 16 = 9.0 J, so the ratio is 36 : 9.0 = 4 : 1. At equal momentum, energy is inversely proportional to mass.
✓ 1
Answer A
Answer: A · 3 stages of work, one mark
Every option, and why
ACorrect: E_k = p²/2m = 144/4 = 36 J and 144/16 = 9.0 J, a ratio of 4 : 1.
BThese are the values for the 8.0 kg body, and the ratio is inverted. For a given momentum the lighter body has the more kinetic energy.
CThe energy is right but the ratio is upside down. E_k = p²/2m, so a smaller mass means a larger energy.
D72 J is pv, not ½pv. The factor of ½ in the kinetic energy has been dropped.
Syllabus understandingA.3 — the kinetic energy of translational motion as given by Ek = ½mv² = p²/2mCommand term: Determine
12A-1A-34
Energy density of fuels·A.3 Work, energy and power
Paper 1AMedium1 mark
Multiple choice · 1 mark2 steps to full marksCalculate
A diesel generator is 40 % efficient. The energy density of its fuel is 45 MJ kg−1.
What mass of fuel is used when the generator supplies 9.0 MJ of electrical energy?
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Notes
Step 1Energy that must be released by the fuel: Einput = Eoutput/η = 9.0/0.40 = 22.5 MJ.
—
All 2 steps must be completed — there is no mark for a part-answer.
Step 2Mass of fuel = 22.5 MJ ÷ 45 MJ kg−1 = 0.50 kg.
✓ 1
Answer D
Answer: D · 2 stages of work, one mark
Every option, and why
AThis multiplies by the efficiency instead of dividing: 9.0 × 0.40/45. Only 40 % of the fuel energy is useful, so more fuel is needed than 9.0 MJ alone would suggest, not less.
BThis ignores the efficiency, assuming all the fuel's energy becomes electrical energy: 9.0/45.
CThis divides by the wasted fraction, 0.60, instead of the useful fraction, 0.40: 9.0/0.60/45.
DCorrect: the fuel must release 9.0/0.40 = 22.5 MJ, which needs 22.5/45 = 0.50 kg.
Syllabus understandingA.3 — energy density of the fuel sources; efficiency η in terms of energy transfer or power as given by η = Eoutput/EinputCommand term: Calculate
13A-1A-35
Work done by a force·A.3 Work, energy and power
Paper 1AMedium1 mark
Multiple choice · 1 mark1 step to full marksDeduce
A puck on a smooth horizontal table is attached by a string to a fixed pin. The puck moves in a horizontal circle around the pin at constant speed.
Which statement about the work done on the puck by the tension in the string is correct?
Show mark scheme
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Notes
Step 1In W = Fs cos θ, the tension points to the centre and every small displacement is along the tangent, so θ = 90° and no work is done at any stage of the motion — consistent with the constant speed (no change in kinetic energy).
✓ 1
Answer A
Answer: A · 1 stage of work, one mark
Every option, and why
ACorrect: the tension is perpendicular to every small displacement, so cos 90° = 0 and no energy is transferred; this is why the speed stays constant.
BA force does work only if it has a component along the displacement. A non-zero force acting on a moving body can still do no work.
CNo work is done over any part of a revolution, not only over a whole one. The work depends on the angle between the force and each small displacement, not on the overall displacement.
DIf the tension did work, the kinetic energy of the puck would change. The speed is constant, so no energy is transferred.
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 — that circular motion is caused by a centripetal force acting perpendicular to the velocity Command term: Deduce
14A-1B-14
Work done by a force·A.3 Work, energy and power
Paper 1BHard11 marks
Data-based question11 steps to full marksDetermine
The manufacturer of a sports slingshot publishes the force F needed to hold its rubber band at each draw distance x. The published data are shown in the table and plotted on the graph.
A student draws the slingshot to x = 0.40 m and launches a projectile of mass 15.0 ± 0.1 g. A chronograph records the launch speed as 44.0 ± 0.5 m s−1.
x / m
0.00
0.05
0.10
0.15
0.20
0.25
0.30
0.35
0.40
F / N
0
13
25
39
52
67
83
101
119
Graph drawn to scale
(a)
Draw the curve of best fit on the graph.
(1)
(b)
Explain why the energy stored in the band at full draw is equal to the area under the graph.
(1)
(c)
Estimate the energy stored in the band when x = 0.40 m.
(2)
(d)
Show that the kinetic energy of the projectile is about 15 J.
(1)
(e)
Determine the absolute uncertainty in the kinetic energy of the projectile.
(2)
(f)
The uncertainty in your estimate in (c) is about ±5 %. Determine the efficiency of the slingshot together with its absolute uncertainty.
(2)
(g)
The manufacturer claims an efficiency of 75 %. Comment on this claim.
(1)
(h)
Suggest where the energy that does not reach the projectile goes.
(1)
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Notes
Part (a)
smooth curve through all points, rising slightly more steeply at large x
✓ 1
Do not accept point-to-point joining or a single straight line.
Part (b)
the work done by the drawing force is ∫F dx «force × distance for each small step», and this work is stored as elastic energy
✓ 1
OWTTE. Do not accept ½Fx or ½kx² — the graph is not linear.
Part (c)
area by counting squares OR trapezium rule on the tabulated values
✓ 1
Trapezium rule gives 22.0 J.
22 J
✓ 1
Accept 21–23 J.
Part (d)
½ × 0.0150 × 44.0² = 14.5 J
✓ 1
Must see conversion of g to kg and full substitution OR answer to 3 s.f.
Speed is squared, so its fractional uncertainty is doubled.
ΔEk = 0.029 × 14.5 ≈ 0.4 J
✓ 1
Accept 0.4–0.5 J.
Part (f)
η = 14.5/22 = 0.66
✓ 1
Allow ECF from (c).
fractional uncertainty = 0.029 + 0.05 = 0.08, so η = 0.66 ± 0.05 «66 ± 5 %»
✓ 1
Accept ±0.05 to ±0.06.
Part (g)
the largest value consistent with the data, 71 %, is below 75 %, so the claim is not supported «for this projectile»
✓ 1
Allow ECF from (f).
Part (h)
kinetic energy of the band and pouch OR internal energy «heating» of the rubber OR sound
✓ 1
Any one.
Answers: (c) 22 J · (d) 14.5 J · (e) ±0.4 J · (f) 0.66 ± 0.05 (the remaining parts are explanations — see the table above)
Syllabus understandingA.3 — that work done by a force is equivalent to a transfer of energy; the elastic potential energy stored; efficiency η in terms of energy transfer or power as given by η = Eoutput/EinputCommand term: Determine
15A-1B-15
Conservation of energy·A.3 Work, energy and power
Paper 1BHard11 marks
Data-based question11 steps to full marksDetermine
A glider is released from rest on an inclined air track. A card of width 20.0 ± 0.2 mm on the glider passes through a light gate lower down the track, and the timer records the time t for which the beam is blocked. The vertical height h through which the glider falls before reaching the gate is varied. The student calculates the speed v = 0.0200/t and tests whether mechanical energy is conserved.
The error bars show an uncertainty of ±2 % in each value of v².
h / m
t / ms
v / m s−1
v² / m² s−2
0.050
22.54
0.887
0.787
0.100
15.31
1.31
1.71
0.150
12.27
1.63
2.66
0.200
10.60
0.250
9.39
2.13
4.54
0.300
8.59
2.33
5.42
Graph drawn to scale
(a)
Complete the table.
(2)
(b)
Show that, if mechanical energy is conserved, the graph of v² against h is a straight line of gradient 2g.
(1)
(c)
Outline why the uncertainty in v² is ±2 %.
(1)
(d)
Draw the line of best fit on the graph.
(1)
(e)
Determine the gradient of the line and its absolute uncertainty, using lines of maximum and minimum gradient.
(3)
(f)
Discuss whether the data show that mechanical energy is conserved.
(1)
(g)
The line of best fit meets the h axis at a positive value. Suggest a systematic error that would cause this, and state whether it affects your conclusion in (f).
(2)
Show mark scheme
Marking point
Mark
Notes
Part (a)
v = 0.0200/0.01060 = 1.89 «m s−1»
✓ 1
v² = 3.56 «m² s−2»
✓ 1
Accept 3.54–3.58. Allow ECF from their v.
Part (b)
mgh = ½mv², so v² = 2gh: proportional to h with gradient 2g
✓ 1
Part (c)
the fractional uncertainty in the card width is 0.2/20.0 = 1 % «timing uncertainty negligible», and v is squared, so 2 × 1 % = 2 %
✓ 1
Part (d)
single straight line through all error bars, not forced through the origin
✓ 1
Part (e)
gradient of best line = 18.6 m s−2
✓ 1
Accept 18.3–19.0 m s−2. Unit required at least once.
maximum and minimum gradient lines drawn through all error bars
✓ 1
uncertainty = ½(max − min) ≈ 0.4 m s−2, so gradient = 18.6 ± 0.4 m s−2
✓ 1
Accept 0.3–0.7 m s−2. MP3 is for matching precision.
Part (f)
2g = 19.6 m s−2 lies above the range of the gradient, so some mechanical energy is dissipated «air resistance, residual friction»; energy is not conserved
✓ 1
Allow ECF from (e).
Part (g)
h systematically measured too large by about 8 mm «eg measured from the wrong point on the glider or gate»
✓ 1
Accept a zero error of the ruler or a mis-placed reference mark.
no: a constant error in h shifts the line sideways but does not change its gradient
✓ 1
Answers: (a) 1.89 m s−1, 3.56 m² s−2 · (e) 18.6 ± 0.4 m s−2(the remaining parts are explanations — see the table above)
Syllabus understandingA.3 — that in the absence of frictional, resistive forces, the total mechanical energy of a system is conserved; the gravitational potential energy, when close to the surface of the Earth as given by ΔEp = mgΔhCommand term: Determine
16A-1B-16
Energy density of fuels·A.3 Work, energy and power
Paper 1BEasy9 marks
Data-based question9 steps to full marksDetermine
A candle stands on a top-pan balance that reads to 0.01 g, under a metal can containing 0.150 kg of water. Every 2 minutes the mass of the candle and the temperature of the water are recorded. Δm is the mass of wax burnt and ΔT is the rise in temperature of the water since the start. The specific heat capacity of water is 4180 J kg−1 K−1.
Time / min
Mass of candle / g
Temperature / °C
Δm / g
ΔT / K
0
24.63
19.4
0.00
0.0
2
24.44
23.8
0.19
4.4
4
24.22
28.2
0.41
8.8
6
24.02
32.9
0.61
13.5
8
23.81
37.2
0.82
17.8
10
23.62
41.7
1.01
22.3
Graph drawn to scale
(a)
State the absolute uncertainty in a value of Δm.
(1)
(b)
Draw the line of best fit on the graph.
(1)
(c)
Determine the gradient of the line.
(2)
(d)
Determine the energy transferred to the water per kilogram of wax burnt.
(2)
(e)
The accepted energy density of candle wax is 43 MJ kg−1. Calculate the percentage of the energy released by the wax that reaches the water.
(1)
(f)
Suggest two improvements to the apparatus that would make the result closer to the accepted value.
(2)
Show mark scheme
Marking point
Mark
Notes
Part (a)
±0.02 g «two readings, each ±0.01 g, are subtracted»
✓ 1
Accept ±0.01 g only with the reasoning that the balance reads half a division.
Part (b)
single straight line through the points «and the origin»
✓ 1
Part (c)
gradient from a large triangle on the line
✓ 1
22 K g−1
✓ 1
Accept 21–23 K g−1.
Part (d)
energy per gram = mc × gradient = 0.150 × 4180 × 21.9 = 1.38 × 104 J g−1
✓ 1
Allow ECF from (c).
= 14 MJ kg−1 «1.38 × 107 J kg−1»
✓ 1
Accept 13–15 MJ kg−1.
Part (e)
13.8/43 × 100 = 32 %
✓ 1
Allow ECF from (d).
Part (f)
any one: a draught shield round the flame; a lid on the can / lagging round the can; place the can closer to the flame «so more of the hot gas reaches it»
✓ 1
any second, different improvement, eg a can with a blackened base or a copper can; stir the water before each reading
✓ 1
Do not accept 'repeat the experiment'.
Answers: (c) 22 K g−1 · (d) 14 MJ kg−1 · (e) 32 % (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 or power Command term: Determine
17A-2-24
Work done by a force·A.3 Work, energy and power
Paper 2Easy12 marks
Short answer & extended response12 steps to full marksExplain
A weightlifter raises a barbell of mass 120 kg from rest on the floor to rest above his head, a vertical rise of 1.9 m, in 2.4 s. He holds it stationary above his head and then lets go, so that it falls 1.9 m back to the floor. Air resistance is negligible.
(a)
The lift.
(i)
Calculate the work done by the weightlifter on the barbell.
(2)
(ii)
Calculate the average power developed by the weightlifter in raising the barbell.
(1)
(iii)
State the work done on the barbell by the gravitational force during the lift.
(1)
(iv)
Explain, by considering the work done by the resultant force, why your answers to (a)(i) and (a)(iii) are equal in magnitude.
(2)
(b)
State why the weightlifter does no work on the barbell while he holds it stationary.
(1)
(c)
(i)
Calculate the speed of the barbell just before it reaches the floor.
(2)
(ii)
A rubber mat on the floor is compressed by 4.0 cm as it brings the barbell to rest. Determine the average resultant force on the barbell while it is being stopped.
(3)
Show mark scheme
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Notes
Part (a)(i)
the barbell starts and ends at rest, so the work done equals the gain in gravitational potential energy mgΔh
✓ 1
120 × 9.81 × 1.9 = 2.24 × 103 J
✓ 1
Award [2] for CNA.
Part (a)(ii)
P = 2.24 × 103/2.4 = 930 W
✓ 1
Accept 0.93 kW. Allow ECF from (a)(i).
Part (a)(iii)
−2.24 × 103 J
✓ 1
The negative sign is required.
Part (a)(iv)
work done by the resultant force = change in kinetic energy of the barbell
✓ 1
the barbell is at rest at the start and the end, so the change in kinetic energy is zero and the two works add to zero
✓ 1
OWTTE.
Part (b)
the displacement of the barbell is zero, so W = Fs cos θ = 0
✓ 1
Accept “the barbell does not move”.
Part (c)(i)
½mv² = mgh so v = √(2 × 9.81 × 1.9)
✓ 1
= 6.1 m s−1
✓ 1
Accept 6.11 m s−1.
Part (c)(ii)
kinetic energy at impact = 2.24 × 103 J «= mgh»
✓ 1
Allow ECF from (c)(i).
work done by the resultant force over the stopping distance = loss of kinetic energy: F × 0.040 = 2.24 × 103
✓ 1
F = 5.6 × 104 N
✓ 1
Award [2] max for 5.7 × 104 N «force exerted by the mat, not the resultant force».
Answers: (a)(i) 2.24 × 103 J · (a)(ii) 930 W · (a)(iii) −2.24 × 103 J · (c)(i) 6.1 m s−1 · (c)(ii) 5.6 × 104 N (the remaining parts are explanations — see the table above)
Syllabus understandingA.3 — 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; 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 power developed P is the rate of work done Command term: Explain
18A-2-25
Sankey diagrams·A.3 Work, energy and power
Paper 2Medium13 marks
Short answer & extended response13 steps to full marksDetermine
An electric bicycle and its rider have a total mass of 90 kg. The rider does not pedal. The bicycle climbs a straight hill at a constant speed of 4.0 m s−1; the road rises 1.0 m vertically for every 20 m travelled along it. The battery supplies a current of 8.0 A to the motor at a potential difference of 36 V. The motor and drive transfer 75 % of the electrical energy they receive into work done on the bicycle; the rest becomes thermal energy.
When fully charged, the battery can deliver a total charge of 5.0 × 104 C at 36 V.
The diagram is an incomplete Sankey diagram for the energy transfers each second during the climb.
Sankey diagram, band widths drawn to scale
(a)
(i)
Calculate the electrical power supplied by the battery.
(1)
(ii)
Calculate the time for which a fully charged battery can supply this current.
(2)
(b)
(i)
Show that the rate at which the bicycle and rider gain gravitational potential energy is about 180 W.
(2)
(ii)
Determine the total resistive force acting on the bicycle and rider.
(3)
(c)
Complete the Sankey diagram, labelling each branch you add with the energy transfer it represents and its value in watts.
(2)
(d)
Calculate the overall efficiency of the bicycle, taking the gain in gravitational potential energy as the useful output.
(1)
(e)
The battery has a mass of 3.6 kg. Calculate the energy density of the battery, and suggest one reason why a battery is used in the bicycle rather than a small petrol engine, even though petrol has an energy density of about 46 MJ kg−1.
(2)
Show mark scheme
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Notes
Part (a)(i)
P = IV = 8.0 × 36 = 288 W
✓ 1
Part (a)(ii)
t = Q/I = 5.0 × 104/8.0
✓ 1
= 6.3 × 103 s
✓ 1
Accept 6250 s or 1.7 h.
Part (b)(i)
vertical speed = 4.0 × (1.0/20) = 0.20 m s−1 OR component of weight along the road = 90 × 9.81/20 = 44.1 N
✓ 1
P = 90 × 9.81 × 0.20 = 177 «W»
✓ 1
Must see full substitution OR answer to 3 s.f.
Part (b)(ii)
power transferred to the bicycle = 0.75 × 288 = 216 W
✓ 1
power used against resistive forces = 216 − 177 = 39 W
✓ 1
F = P/v = 39/4.0 = 9.9 N
✓ 1
Accept 9.0–10 N «9.0 N follows from using 180 W». Allow ECF.
Part (c)
the 216 W band divided into two labelled branches: work done against resistive forces «≈ 39 W» AND gain in gravitational potential energy «≈ 177 W»
✓ 1
Allow ECF from (b). Accept “thermal energy / air and road” for the resistive branch.
branch widths in proportion: the resistive branch about one-fifth of the width of the 216 W band, the two new branches together equal to the 216 W band
✓ 1
By eye.
Part (d)
η = 177/288 = 0.61 «= 61 %»
✓ 1
Accept 0.63 «from 180 W». Allow ECF.
Part (e)
energy stored = QV = 5.0 × 104 × 36 = 1.8 × 106 J, so energy density = 0.50 MJ kg−1
✓ 1
any one: the electric motor transfers a much larger fraction of its input to useful work than a small petrol engine OR no exhaust gases are emitted where the bicycle is used OR it is quieter OR the battery can be recharged from renewable sources
✓ 1
Answers: (a)(i) 288 W · (a)(ii) 6.3 × 103 s · (b)(i) 177 W · (b)(ii) 9.9 N · (d) 61 % · (e) 0.50 MJ kg−1(the remaining parts are explanations — see the table above)
Syllabus understandingA.3 — that energy transfers can be represented on a Sankey diagram; 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; energy density of the fuel sources; B.5 — direct current as a flow of charge carriers as given by I = Δq/Δt; electrical power as given by P = IVCommand term: Determine
19A-2-26
Conservation of energy·A.3 Work, energy and power
Paper 2Medium13 marks
Short answer & extended response13 steps to full marksDetermine
A roller-coaster car of total mass 480 kg passes point A, at the top of the first drop, at 2.0 m s−1. Graph 1 shows how the height of the track above the lowest point B varies with the distance along the track. The track from A to B is 78 m long and the track from B to the next crest C is 52 m long.
Graph 1, drawn to scaleGraph 2, axes for (b)
(a)
Assume first that no energy is dissipated.
(i)
Calculate the loss of gravitational potential energy of the car between A and B.
(1)
(ii)
Show that the speed of the car at B would be about 30 m s−1.
(1)
(b)
Sketch, on graph 2, how the kinetic energy of the car would vary with the distance along the track from A to C if no energy were dissipated.
(2)
(c)
In practice the car passes B at 27 m s−1.
(i)
Determine the energy dissipated between A and B.
(3)
(ii)
Calculate the average resistive force on the car between A and B.
(1)
(iii)
The average resistive force between B and C is the same as between A and B. Determine whether the car reaches C.
(3)
(d)
Suggest two reasons why the resistive force on the car is not constant along the track.
(2)
Show mark scheme
Marking point
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Notes
Part (a)(i)
ΔEp = 480 × 9.81 × 45 = 2.12 × 105 J
✓ 1
Part (a)(ii)
½ × 480 × v² = ½ × 480 × 2.0² + 2.12 × 105 so v = 29.8 «m s−1»
✓ 1
Must see full substitution OR answer to 3 s.f. Do not award if the kinetic energy at A is omitted «29.7 m s−1».
Part (b)
shape: kinetic energy rises from a small value at A to a maximum at B «78 m», then falls to a lower value at C, mirroring the height profile «smooth curve»
✓ 1
values: about 2.1 × 105 J at B AND about 8.1 × 104 J at C
✓ 1
Allow ± 0.1 × 105 J. Allow ECF from (a).
Part (c)(i)
total energy at A = 960 + 2.12 × 105 = 2.129 × 105 J «relative to B»
✓ 1
Ek at B = ½ × 480 × 27² = 1.75 × 105 J
✓ 1
energy dissipated = 3.8 × 104 J
✓ 1
Award [2] max for 3.7 × 104 J «kinetic energy at A omitted».
Part (c)(ii)
F = 3.79 × 104/78 = 490 N
✓ 1
Accept 470–490 N. Allow ECF.
Part (c)(iii)
energy needed to rise to C = 480 × 9.81 × 28 = 1.32 × 105 J
✓ 1
work done against the resistive force = 486 × 52 = 2.5 × 104 J
✓ 1
1.57 × 105 J < 1.75 × 105 J «the kinetic energy at B», so the car reaches C
✓ 1
MP3 requires a comparison. The car arrives with about 1.8 × 104 J. Allow ECF.
Part (d)
air resistance increases with speed «so it is largest near B»
✓ 1
the friction in the wheels depends on the normal force, which changes with the slope and curvature of the track «largest in the dip at B»
✓ 1
Accept any other valid physical reason.
Answers: (a)(i) 2.12 × 105 J · (a)(ii) 29.8 m s−1 · (c)(i) 3.8 × 104 J · (c)(ii) 490 N · (c)(iii) yes (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; that in the absence of frictional, resistive forces, the total mechanical energy of a system is conserved; that work done by the resultant force on a system is equal to the change in the energy of the system Command term: Determine
20A-2-27
Energy density of fuels·A.3 Work, energy and power
Paper 2Hard12 marks
Short answer & extended response12 steps to full marksDetermine
A delivery lorry travels 400 km each day. The average resistive force on it is 2.2 kN. Two ways of powering the lorry are compared.
energy density of fuel / MJ kg−1
overall efficiency of drive system
diesel engine
45
35 %
hydrogen fuel cell and electric motor
120
50 %
(a)
State what is meant by the energy density of a fuel.
(1)
(b)
(i)
Show that the work done against the resistive force each day is about 9 × 108 J.
(1)
(ii)
Calculate the mass of diesel and the mass of hydrogen that would be needed each day.
(3)
(c)
The hydrogen is stored as a gas at a pressure of 35 MPa and a temperature of 290 K. The molar mass of hydrogen is 2.0 g mol−1.
(i)
The lorry has a hydrogen tank of internal volume 0.55 m³. Using the ideal gas model, determine whether the tank can hold enough hydrogen for one day.
(3)
(ii)
Suggest why the ideal gas model may not be reliable for this calculation, and outline how this affects your conclusion in (c)(i).
(2)
(d)
Outline one advantage and one disadvantage of hydrogen compared with diesel as the energy source for the lorry.
(2)
Show mark scheme
Marking point
Mark
Notes
Part (a)
the energy released per unit mass of the fuel «when it is used/burned»
✓ 1
Accept J kg−1 as part of the answer.
Part (b)(i)
W = Fs = 2.2 × 103 × 4.0 × 105 = 8.8 × 108 «J»
✓ 1
Part (b)(ii)
energy from diesel = 8.8 × 108/0.35 = 2.5 × 109 J
✓ 1
mass of diesel = 2.51 × 109/4.5 × 107 = 56 kg
✓ 1
mass of hydrogen = 8.8 × 108/(0.50 × 1.2 × 108) = 15 kg
✓ 1
Accept 14.7 kg. Award [1] max if both efficiencies are ignored «20 kg, 7.3 kg».
0.50 m³ < 0.55 m³, so «according to the model» the tank is large enough
✓ 1
MP3 requires a comparison.
Part (c)(ii)
at such a high pressure the gas is very dense: the volume of the molecules «and the forces between them» can no longer be neglected
✓ 1
the margin «0.50 m³ against 0.55 m³» is small, so the conclusion is not reliable: the real gas may need a larger volume and the tank may be too small
✓ 1
OWTTE.
Part (d)
advantage: higher energy density so a smaller mass of fuel is needed OR no carbon dioxide is emitted where the lorry is used «only water»
✓ 1
disadvantage: must be stored at very high pressure in a heavy, bulky tank OR low energy per unit volume OR much hydrogen is produced from fossil fuels or with large energy input
✓ 1
Answers: (b)(i) 8.8 × 108 J · (b)(ii) diesel 56 kg, hydrogen 15 kg · (c)(i) 0.50 m³; large enough by the ideal-gas model (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 or power; that work W done on a body by a constant force depends on the component of the force along the line of displacement; B.3 — the equations governing the behaviour of ideal gases as given by PV = nRT; the temperature, pressure and density conditions under which an ideal gas is a good approximation of a real gas Command term: Determine
21A-2-28
Power & efficiency·A.3 Work, energy and power
Paper 2Easy11 marks
Short answer & extended response11 steps to full marksCalculate
A chairlift carries skiers from the bottom station to the top station of a ski slope, which is 320 m higher. The chairs are fixed to a continuous cable loop: chairs move up full on one side and come down empty on the other. When the lift is busy it carries 1200 skiers per hour. The average mass of a skier with equipment is 85 kg.
(a)
(i)
Calculate the gain in gravitational potential energy of one skier carried from the bottom to the top.
(1)
(ii)
Show that the useful power needed to raise the skiers is about 90 kW.
(1)
(b)
Explain why the weight of the chairs does not need to be included in (a)(ii).
(2)
(c)
When the lift is busy, the input power to its electric motor is 130 kW. The cable moves at 2.5 m s−1.
(i)
Calculate the efficiency of the lift.
(1)
(ii)
Calculate the force that the drive wheel must exert on the cable to provide the useful power.
(2)
(d)
The operator wants to carry 1500 skiers per hour. The maximum input power of the motor is 150 kW. Determine whether this is possible at the same efficiency.
(3)
(e)
Suggest why the motor still needs an input of power when the lift runs with no skiers.
(1)
Show mark scheme
Marking point
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Notes
Part (a)(i)
85 × 9.81 × 320 = 2.67 × 105 J
✓ 1
Part (a)(ii)
2.67 × 105 × 1200/3600 = 8.89 × 104 «W»
✓ 1
Must see full substitution OR answer to 3 s.f.
Part (b)
for every chair moving up there is a chair «of the same mass» moving down
✓ 1
the gravitational potential energy gained by the rising chairs equals that lost by the falling chairs, so no net energy is needed for the chairs
✓ 1
OWTTE.
Part (c)(i)
η = 8.89 × 104/1.3 × 105 = 0.68 «68 %»
✓ 1
Allow ECF.
Part (c)(ii)
P = Fv so F = P/v
✓ 1
F = 8.89 × 104/2.5 = 3.6 × 104 N
✓ 1
Award [2] for CNA.
Part (d)
useful power needed = 8.89 × 104 × 1500/1200 = 1.11 × 105 W
✓ 1
input power needed = 1.11 × 105/0.684 = 163 kW
✓ 1
163 kW > 150 kW, so it is not possible
✓ 1
ALTERNATIVE: greatest useful power = 0.684 × 150 = 103 kW ✓ < 111 kW needed ✓ so not possible ✓. MP3 requires a comparison. Allow ECF.
Part (e)
energy is still dissipated by friction in the wheels, bearings and cable «and by air resistance and heating in the motor»
✓ 1
Do not accept “to lift the chairs”.
Answers: (a)(i) 2.67 × 105 J · (a)(ii) 8.89 × 104 W · (c)(i) 68 % · (c)(ii) 3.6 × 104 N · (d) 163 kW needed; not possible (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 = Poutput/Pinput; the gravitational potential energy, when close to the surface of the Earth as given by ΔEp = mgΔhCommand term: Calculate
22A-1A-43
Power & efficiency·A.3 Work, energy and power
Paper 1AHard1 mark
Multiple choice · 1 mark2 steps to full marksDetermine
A car travels at a constant speed along a level road. The total resistive force on the car is proportional to the square of its speed. The useful output power of the engine is P1 at speed v and P2 at speed 2v.
What is P2/P1?
Show mark scheme
Marking point
Mark
Notes
Step 1At constant speed the resultant force is zero, so the driving force equals the resistive force, which increases by 2² = 4.
—
All 2 steps must be completed — there is no mark for a part-answer.
Step 2P = Fv: the force is 4 times larger and the speed 2 times larger, so P2/P1 = 4 × 2 = 8.
✓ 1
Answer C
Answer: C · 2 stages of work, one mark
Every option, and why
AThis doubles the speed in P = Fv but keeps the driving force the same. The resistive force, and therefore the driving force, also increases.
BThis multiplies the force by 4 but leaves the speed unchanged in P = Fv.
CCorrect: F ∝ v² and P = Fv, so P ∝ v³ and doubling the speed gives 2³ = 8.
DThis squares both factors, 2² × 2². The force goes as v², but the speed enters P = Fv only once.
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 = FvCommand term: Determine
23A-1B-20
Power & efficiency·A.3 Work, energy and power
Paper 1BMedium10 marks
Data-based question10 steps to full marksDetermine
A cycling-science database lists the steady power P that one rider must deliver, measured with a power meter, to ride at constant speed v on a level road in still air.
A student tests the hypothesis that the total resistive force on the rider is proportional to v², so that P = kvn with n = 3.
v / km h−1
v / m s−1
P / W
lg(v / m s−1)
lg(P / W)
25
6.94
78.8
0.842
1.897
30
8.33
124
0.921
2.093
35
9.72
192
0.988
2.283
40
11.1
272
45
12.5
387
1.097
2.588
50
13.9
514
1.143
2.711
Graph drawn to scale
(a)
Explain why n = 3 if the resistive force is proportional to v².
(2)
(b)
Complete the table.
(1)
(c)
Draw the line of best fit on the graph.
(1)
(d)
Determine n.
(2)
(e)
Discuss whether the data support the hypothesis.
(2)
(f)
Determine the total resistive force on the rider at 40 km h−1.
(2)
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Notes
Part (a)
at constant speed the driving force equals the resistive force, and P = Fv
✓ 1
if F ∝ v² then P ∝ v³ «lg P = 3 lg v + lg k, gradient 3»
✓ 1
Part (b)
1.046 and 2.435
✓ 1
Both needed. Accept 2 or 3 d.p.
Part (c)
single straight line with points evenly scattered on either side
✓ 1
Part (d)
gradient from a large triangle on the line
✓ 1
n = 2.7
✓ 1
Accept 2.6–2.85.
Part (e)
n is clearly less than 3 «the difference is larger than the scatter allows», so the hypothesis is not fully supported
✓ 1
Allow ECF from (d).
part of the resistive force «rolling resistance, friction in the drive» does not increase as v², so P rises less steeply than v³
✓ 1
OWTTE.
Part (f)
v = 40/3.6 = 11.1 m s−1
✓ 1
F = P/v = 272/11.1 = 24.5 N
✓ 1
Accept 24–26 N.
Answers: (b) 1.046, 2.435 · (d) n = 2.7 · (f) 24 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; Tool 3 — construct and interpret graphs using logarithmic scales Command term: Determine
24A-2-32
Friction & inclined planes·A.3 Work, energy and power
Paper 2Easy10 marks
Short answer & extended response10 steps to full marksCalculate
In a warehouse, parcels slide down a straight chute inclined at 25° to the horizontal. A parcel of mass 1.6 kg is released from rest at the top and slides 3.5 m to the bottom of the chute, where its speed is 3.6 m s−1.
Diagram NOT accurately drawn
(a)
(i)
Calculate the vertical height through which the parcel falls.
(1)
(ii)
Calculate the loss of gravitational potential energy of the parcel.
(1)
(b)
Determine the average frictional force on the parcel.
(3)
(c)
(i)
Show that the normal force on the parcel is about 14 N.
(1)
(ii)
Calculate the coefficient of dynamic friction between the parcel and the chute.
(1)
(d)
Calculate the time taken by the parcel to slide down the chute.
(2)
(e)
Parcels sometimes arrive at the bottom too quickly. Suggest one change to the chute that would reduce their speed at the bottom.
(1)
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Notes
Part (a)(i)
h = 3.5 sin 25° = 1.48 m
✓ 1
Part (a)(ii)
ΔEp = 1.6 × 9.81 × 1.48 = 23.2 J
✓ 1
Allow ECF from (a)(i).
Part (b)
Ek at the bottom = ½ × 1.6 × 3.6² = 10.4 J
✓ 1
energy dissipated = 23.2 − 10.4 = 12.8 J
✓ 1
Ff = 12.8/3.5 = 3.7 N
✓ 1
Award [3] for CNA. Allow ECF.
Part (c)(i)
FN = mg cos 25° = 1.6 × 9.81 × cos 25° = 14.2 «N»
✓ 1
Must see full substitution OR answer to 3 s.f.
Part (c)(ii)
μd = 3.67/14.2 = 0.26
✓ 1
Allow ECF from (b).
Part (d)
the forces on the parcel are constant, so its acceleration is uniform and s = (u + v)t/2: 3.5 = ½ × (0 + 3.6) × t
✓ 1
t = 1.9 s
✓ 1
Accept 1.94 s. Award [2] for CNA. ALTERNATIVE: a = v²/2s = 1.85 m s−2 ✓ and t = v/a = 1.9 s ✓.
Part (e)
reduce the angle of the chute OR give the chute a rougher surface «larger coefficient of friction» OR make the chute shorter from a lower starting point
✓ 1
Accept any one valid change.
Answers: (a)(i) 1.48 m · (a)(ii) 23.2 J · (b) 3.7 N · (c)(i) 14.2 N · (c)(ii) 0.26 · (d) 1.9 s (the remaining parts are explanations — see the table above)
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 gravitational potential energy as given by ΔEp = mgΔh; A.2 — surface frictional force on a body in motion as given by Ff = μdFN; normal force FN is the component of the contact force acting perpendicular to the surface; A.1 — the equations of motion for solving problems with uniformly accelerated motion Command term: Calculate
25A-2-33
Power & efficiency·A.3 Work, energy and power
Paper 2Medium12 marks
Short answer & extended response12 steps to full marksDraw
A car of mass 1300 kg travels along a straight, level road. Its engine delivers a constant useful output power of 45 kW. The graph shows how the total resistive force on the car varies with its speed.
Graph drawn to scale
(a)
Determine the acceleration of the car at a speed of 15 m s−1.
(3)
(b)
(i)
Draw, on the graph, a curve showing how the driving force varies with speed between 15 m s−1 and 50 m s−1.
(2)
(ii)
Hence determine the top speed of the car.
(2)
(c)
Explain why the acceleration of the car decreases as its speed increases, even though the output power is constant.
(2)
(d)
The car climbs a straight hill on which the road rises 1.0 m for every 10 m travelled along it. The output power is still 45 kW and the resistive force varies with speed as shown on the graph. Determine the top speed of the car on the hill.
(3)
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Notes
Part (a)
driving force = P/v = 4.5 × 104/15 = 3000 N
✓ 1
resistive force from the graph = 480 N
✓ 1
Accept 460–500 N.
a = (3000 − 480)/1300 = 1.9 m s−2
✓ 1
Allow ECF.
Part (b)(i)
at least three correct points, eg 3000 N at 15 m s−1, 1500 N at 30 m s−1, 900 N at 50 m s−1
✓ 1
Allow ± 100 N.
smooth curve with a negative gradient whose magnitude decreases as the speed increases «not a straight line», crossing the resistive-force curve
✓ 1
Part (b)(ii)
the top speed is where the driving force equals the resistive force «the curves cross: resultant force zero»
✓ 1
35 m s−1
✓ 1
Accept 34–36 m s−1. Allow ECF from the candidate’s curve.
Part (c)
the driving force F = P/v decreases as the speed increases
✓ 1
while the resistive force increases, so the resultant force, and hence a = F/m, decreases
✓ 1
Part (d)
component of the weight along the road = 1300 × 9.81 × 0.10 = 1280 N
✓ 1
at top speed the driving force P/v equals the resistive force + 1280 N «eg the resistive-force curve raised by 1280 N crosses the driving-force curve»
✓ 1
Accept a trial: at 22 m s−1 P/v = 2050 N and the opposing force is 1960 N.
v = 23 m s−1
✓ 1
Accept 21–24 m s−1. Allow ECF from the candidate’s curve in (b)(i).
Answers: (a) 1.9 m s−2 · (b)(ii) 35 m s−1 · (d) 23 m s−1(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; A.2 — Newton’s three laws of motion; that free-body diagrams can be analysed to find the resultant force on a system Command term: Draw
26A-2-36
Work done by a force·A.3 Work, energy and power
Paper 2Easy10 marks
Short answer & extended response10 steps to full marksCalculate
A body of mass 1.5 kg is initially at rest on a horizontal, frictionless surface. A horizontal force is applied to it. The graph shows how the force varies with the displacement of the body.
Graph drawn to scale
(a)
State what is represented by the area under a force–displacement graph.
(1)
(b)
(i)
Calculate the work done by the force over the first 2.0 m.
(1)
(ii)
Show that the total work done by the force over 5.0 m is 105 J.
(1)
(c)
Calculate the speed of the body after it has moved 5.0 m.
(2)
(d)
Draw, on the graph, a line representing the constant force that would do the same work on the body over the 5.0 m.
(2)
(e)
Outline why the acceleration of the body decreases between 2.0 m and 5.0 m.
(1)
(f)
The same force–displacement graph applies to a body of mass 3.0 kg. Deduce the speed of this body after it has moved 5.0 m.
(2)
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Notes
Part (a)
the work done by the force «the energy transferred»
✓ 1
Part (b)(i)
30 × 2.0 = 60 J
✓ 1
Part (b)(ii)
60 + ½ × 3.0 × 30 = 60 + 45 = 105 «J»
✓ 1
Must see the area of the triangle added.
Part (c)
½ × 1.5 × v² = 105
✓ 1
v = 11.8 m s−1
✓ 1
Award [2] for CNA.
Part (d)
horizontal line at 21 N
✓ 1
Allow ± 1 N.
drawn from 0 to 5.0 m «so that the area under it is 105 J»
✓ 1
Part (e)
the force «the resultant force» decreases, and a = F/m with m constant
✓ 1
Part (f)
the same work is done, so the body gains the same kinetic energy «105 J»
✓ 1
v = √(2 × 105/3.0) = 8.4 m s−1
✓ 1
Accept 11.8/√2. Allow ECF from (c).
Answers: (b)(i) 60 J · (b)(ii) 105 J · (c) 11.8 m s−1 · (d) 21 N · (f) 8.4 m s−1(the remaining parts are explanations — see the table above)
Syllabus understandingA.3 — 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; the kinetic energy of translational motion as given by Ek = ½mv² Command term: Calculate
27A-2-38
Work done by a force·A.3 Work, energy and power
Paper 2Hard13 marks
Short answer & extended response13 steps to full marksDetermine
A sledge of mass 24 kg is pulled from rest for 18 m across level snow. The rope is held at 32° above the horizontal and the tension in it is a constant 95 N. The coefficient of dynamic friction between the sledge and the snow is 0.22. Air resistance is negligible.
Diagram NOT accurately drawn
(a)
(i)
Calculate the work done on the sledge by the tension over the 18 m.
(1)
(ii)
State why the vertical component of the tension does no work on the sledge.
(1)
(b)
(i)
Show that the normal force on the sledge is about 185 N.
(2)
(ii)
Determine the speed of the sledge after it has travelled 18 m.
(3)
(c)
The rope is now held at 45° above the horizontal, with the same tension. Determine whether the sledge gains more or less kinetic energy over the 18 m than before.
(3)
(d)
Explain why there is an angle between 0° and 45° at which the kinetic energy gained by the sledge is greatest.
(3)
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Notes
Part (a)(i)
W = Fs cos θ = 95 × 18 × cos 32° = 1.45 × 103 J
✓ 1
Part (a)(ii)
it is perpendicular to the displacement «cos 90° = 0»
Must see full substitution OR answer to 4 s.f. «185.1 N».
Part (b)(ii)
Ff = 0.22 × 185 = 40.7 N
✓ 1
work done by the resultant force = 1.45 × 103 − 40.7 × 18 = 717 J = kinetic energy gained
✓ 1
Accept (80.6 − 40.7) × 18.
v = √(2 × 717/24) = 7.7 m s−1
✓ 1
Award [3] for CNA. Award [2] max for 6.6 m s−1 «friction found from mg».
Part (c)
FN = 235.4 − 95 sin 45° = 168 N, so Ff = 37.0 N
✓ 1
resultant force = 95 cos 45° − 37.0 = 30.2 N, so kinetic energy = 543 J
✓ 1
543 J < 717 J, so the sledge gains less kinetic energy
✓ 1
MP3 requires a comparison. Allow ECF from (b)(ii).
Part (d)
increasing the angle reduces the horizontal component of the tension «T cos θ», reducing the forward force
✓ 1
but it increases the vertical component, which reduces FN and therefore the frictional force
✓ 1
at small angles the reduction in friction is the larger effect «T cos θ hardly changes», at large angles the loss of horizontal component is, so the kinetic energy is greatest at an angle in between
✓ 1
Do not expect the value «about 12°». MP3 requires both effects to be compared.
Answers: (a)(i) 1.45 × 103 J · (b)(i) 185 N · (b)(ii) 7.7 m s−1 · (c) 543 J < 717 J; less (the remaining parts are explanations — see the table above)
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 the resultant force on a system is equal to the change in the energy of the system; A.2 — surface frictional force on a body in motion as given by Ff = μdFNCommand term: Determine
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