A.3 Work, energy and power – IB Physics exam questions with mark scheme

1Standard4 marks

A motor raises a lift of total mass 800 kg through a vertical height of 12 m at constant speed in 15 s. The overall efficiency of the motor and lift system is 75%. Use $g=9.81\ \mathrm{m\,s^{-2}}$.

(a) Calculate the gain in gravitational potential energy of the lift. [1]

(b) Calculate the useful output power. [1]

$\Delta E_p = mg\Delta h = 800\times9.81\times12 = 9.42\times10^4$ J ($9.4\times10^4$ J)(1)
$P_{out} = 94176/15 = 6278\ \mathrm{W}$ ($6.3\times10^3$ W)(1)
Efficiency = useful power output / total power input, so $P_{in}=P_{out}/0.75$(1)
$P_{in} = 8371\ \mathrm{W}$ ($8.4\times10^3$ W)(1)
Answer: (a) $9.4\times10^4$ J (b) $6.3\times10^3$ W (c) $8.4\times10^3$ W
Examiner tip: Divide, not multiply, by the efficiency when finding input power.

2Core5 marks

A child pulls a sledge of mass 25 kg a horizontal distance of 40 m across level snow at a constant speed. The rope is at 35° to the horizontal and the tension in the rope is 80 N, as shown in the diagram.

(a) Calculate the work done on the sledge by the tension in the rope. [2]

(b) Deduce the work done on the sledge by the friction force from the snow. [2]

$W = Fs\cos\theta = 80\times40\times\cos35^\circ$; do not accept $\sin35^\circ$(1)
$= 2.6\times10^{3}\ \mathrm{J}$ (allow 2621 J / 2.62 kJ)(1)
speed is constant so the change in kinetic energy is zero, so the total work done by the resultant force is zero (OWTTE)(1)
work done by friction $= -2.6\times10^{3}\ \mathrm{J}$; sign required for this mark; allow ECF from (a); allow “$2.6\ \mathrm{kJ}$ of work done against friction”(1)
zero, because the weight is perpendicular to the displacement / there is no displacement in the direction of the weight ($\cos90^\circ = 0$)(1)
Answer: (a) $2.6\times10^{3}\ \mathrm{J}$ (b) $-2.6\times10^{3}\ \mathrm{J}$ (c) zero: weight is perpendicular to the displacement
Examiner tip: Only the component of force along the displacement does work: use $\cos\theta$ with $\theta$ measured between the force and the displacement, and remember friction does negative work.

3Core5 marks

The Sankey diagram shows the energy transfers in a gas-fired power station operating at a steady rate. The widths of the arrows are drawn to scale.

(a) Determine the electrical output power of the power station. [1]

(b) Calculate the efficiency of the power station. [1]

(d) Outline how the Sankey diagram is consistent with the principle of conservation of energy. [1]

$1500 - 150 - 760 - 40 = 550\ \mathrm{MW}$(1)
$\eta = 550/1500 = 0.37$ / 37%; allow ECF from (a)(1)
wasted power $= 950\ \mathrm{MW}$ / $1500 - 550$(1)
$E = 950\times10^{6}\times 24\times3600 = 8.2\times10^{13}\ \mathrm{J}$; allow ECF; award [1 max] for an answer in MJ or kW h that is not converted correctly(1)
the width of the input arrow equals the total width of all the output arrows / total energy (power) out equals energy (power) in, so no energy is created or destroyed (OWTTE)(1)
Answer: (a) 550 MW (b) 0.37 (37%) (c) $8.2\times10^{13}\ \mathrm{J}$ (d) input width = sum of output widths
Examiner tip: Convert the day to seconds and MW to W before multiplying: energy in joules needs power in watts and time in seconds.

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4Stretch7 marks

A bullet of mass 12 g travelling horizontally at $380\ \mathrm{m\,s^{-1}}$ embeds itself in a wooden block of mass 2.4 kg. The block hangs at rest from long, light strings, so it is free to swing upwards after the impact.

(a) Show that the kinetic energy of the bullet before the impact is about 870 J. [1]

(b) Calculate the kinetic energy of the block and bullet immediately after the impact. [2]

(c) A bullet of mass $m$ embeds in a block of mass $M$. Using $E_k = \dfrac{p^2}{2m}$, show that the fraction of the bullet's initial kinetic energy that remains as kinetic energy just after the impact is $\dfrac{m}{m+M}$. [2]

(d) Determine the maximum vertical height through which the block rises. [1]

(e) State what happens to the kinetic energy that is not accounted for in (b). [1]

$\tfrac12\times0.012\times380^2 = 866\ \mathrm{J}$; answer to at least 3 s.f. or full substitution needed(1)
momentum conserved: $p = 0.012\times380 = 4.56\ \mathrm{N\,s}$ / $v = 4.56/2.412 = 1.89\ \mathrm{m\,s^{-1}}$(1)
$E_k = \dfrac{4.56^2}{2\times2.412} = 4.3\ \mathrm{J}$ (allow 4.31 J)(1)
momentum $p$ is the same before and after the impact (conserved), so $E_{k,\text{before}} = \dfrac{p^2}{2m}$ and $E_{k,\text{after}} = \dfrac{p^2}{2(m+M)}$(1)
ratio $= \dfrac{p^2/2(m+M)}{p^2/2m} = \dfrac{m}{m+M}$; both steps needed for full credit(1)
$h = \dfrac{E_k}{(m+M)g} = \dfrac{4.31}{2.412\times9.81} = 0.18\ \mathrm{m}$ (allow ECF from (b))(1)
transferred to internal (thermal) energy of the block and bullet / work done deforming the wood (and sound); do not accept “lost” alone(1)
Answer: (a) 866 J (b) 4.3 J (c) $E_{k}$ ratio $= m/(m+M)$ (d) 0.18 m (e) internal/thermal energy of block and bullet (deformation, sound)
Examiner tip: Use conservation of momentum, not energy, for the impact; only after the impact is mechanical energy conserved during the swing.

5Standard6 marks

A toy launcher uses a spring of spring constant $450\ \mathrm{N\,m^{-1}}$. The spring is compressed by 6.0 cm and then released, firing a ball of mass 25 g vertically upwards. Ignore the change in height of the ball while the spring extends and the mass of the spring.

(a) Calculate the elastic potential energy stored in the compressed spring. [1]

(b) Calculate the speed with which the ball leaves the launcher. [2]

(d) When the experiment is carried out, the ball rises only 2.8 m. Determine the average air resistance force on the ball as it rises. [2]

$E_H = \tfrac12\times450\times0.060^2 = 0.81\ \mathrm{J}$; do not accept use of 6.0 (cm)(1)
$\tfrac12\times0.025\times v^2 = 0.81$(1)
$v = 8.0\ \mathrm{m\,s^{-1}}$ (allow 8.05; allow ECF from (a))(1)
$h = \dfrac{0.81}{0.025\times9.81} = 3.3\ \mathrm{m}$ (allow ECF)(1)
energy transferred by air resistance $= 0.81 - 0.025\times9.81\times2.8 = 0.12\ \mathrm{J}$(1)
$F = \dfrac{0.123}{2.8} = 0.044\ \mathrm{N}$ (allow 0.043–0.045 N; allow ECF)(1)
Answer: (a) 0.81 J (b) $8.0\ \mathrm{m\,s^{-1}}$ (c) 3.3 m (d) 0.044 N
Examiner tip: Convert the compression to metres before squaring: 6.0 cm is 0.060 m, and squaring a value in cm gives an answer 10 000 times too big.

6Standard5 marks

An archer draws back the string of a bow. The graph shows how the force $F$ needed to hold the string varies with the draw distance $x$.

(a) State what is represented by the area under the graph. [1]

(b) Determine the energy stored in the bow when the draw distance is 0.50 m. [2]

(c) An arrow of mass 30 g is fired from the bow drawn to 0.50 m. 72% of the stored energy is transferred to the kinetic energy of the arrow. Calculate the speed of the arrow as it leaves the bow. [2]

work done (on the bow / by the archer) / energy transferred to the bow (elastic potential energy stored); do not accept “force” or “power”(1)
area split correctly into a triangle and a trapezium: $\tfrac12\times0.20\times200\ (=20\ \mathrm{J})$ and $0.30\times\dfrac{200+260}{2}\ (=69\ \mathrm{J})$(1)
$89\ \mathrm{J}$ (allow 88–90 J)(1)
$E_k = 0.72\times89 = 64\ \mathrm{J}$(1)
$v = \sqrt{\dfrac{2\times64}{0.030}} = 65\ \mathrm{m\,s^{-1}}$ (allow ECF from (b))(1)
Answer: (a) work done / energy stored (b) 89 J (c) $65\ \mathrm{m\,s^{-1}}$
Examiner tip: For a force that varies, work is the area under the force–displacement graph; $W = Fs$ with the final force overestimates it.

7Hard7 marks

A car of mass 1200 kg experiences a constant rolling resistance of 300 N and an air resistance force $F_{\text{air}} = 0.40v^2$, where $F_{\text{air}}$ is in N and $v$ is the speed in $\mathrm{m\,s^{-1}}$.

(a) Calculate the output power of the engine when the car travels at a constant speed of $30\ \mathrm{m\,s^{-1}}$ on a level road. [2]

(b) The car now climbs a straight hill at a constant speed of $25\ \mathrm{m\,s^{-1}}$. The road rises 1.0 m vertically for every 20 m travelled along the road. Determine the output power of the engine. [3]

(c) The maximum output power of the engine is 60 kW. Show that the maximum speed of the car up the same hill is about $40\ \mathrm{m\,s^{-1}}$. [2]

driving force $= 300 + 0.40\times30^2 = 660\ \mathrm{N}$(1)
$P = Fv = 660\times30 = 2.0\times10^{4}\ \mathrm{W}$ (allow 19.8 kW)(1)
component of weight along the slope $= mg\sin\theta = 1200\times9.81\times\dfrac{1}{20} = 589\ \mathrm{N}$(1)
driving force $= 589 + 300 + 0.40\times25^2 = 1139\ \mathrm{N}$ (allow ECF)(1)
$P = 1139\times25 = 2.8\times10^{4}\ \mathrm{W}$ (allow 28.5 kW; allow ECF)(1)
at maximum speed: $60\,000 = (889 + 0.40v^2)\,v$; allow a check of $P$ at $40\ \mathrm{m\,s^{-1}}$ giving $(889+640)\times40 = 61\ \mathrm{kW} \approx 60\ \mathrm{kW}$(1)
$v = 39.6\ \mathrm{m\,s^{-1}}$ (solved numerically / by trial), so about $40\ \mathrm{m\,s^{-1}}$; answer to more s.f. than given, or a correct check, needed(1)
Answer: (a) $2.0\times10^{4}\ \mathrm{W}$ (b) $2.8\times10^{4}\ \mathrm{W}$ (c) $v = 39.6\ \mathrm{m\,s^{-1}}$
Examiner tip: At constant speed the driving force equals the sum of all resistive forces, including $mg\sin\theta$ on a slope; use that force in $P = Fv$.

8Standard6 marks

The table gives data for three ways of storing energy in a car. Specific energy is the energy available per unit mass; energy density is the energy available per unit volume.

energy store	specific energy / $\mathrm{MJ\,kg^{-1}}$	density / $\mathrm{kg\,m^{-3}}$	efficiency of conversion to useful mechanical energy
petrol	46	740	0.25
hydrogen gas at 700 bar	142	42	0.50
lithium-ion battery pack	0.70	2500	0.85

(a) Calculate the energy density of the compressed hydrogen. [1]

A journey requires 180 MJ of useful mechanical energy.

(b) Determine the mass of petrol that must be burned for this journey. [2]

(d) Hydrogen has the greatest specific energy. Suggest why a hydrogen car still needs a large fuel tank. [1]

$142\times10^{6}\times42 = 6.0\times10^{9}\ \mathrm{J\,m^{-3}}$ (allow $6.0\ \mathrm{GJ\,m^{-3}}$ / $6.0\ \mathrm{MJ\,L^{-1}}$)(1)
energy needed from the fuel $= \dfrac{180}{0.25} = 720\ \mathrm{MJ}$(1)
mass $= \dfrac{720}{46} = 16\ \mathrm{kg}$ (allow 15.7 kg)(1)
energy needed from battery $= \dfrac{180}{0.85} = 212\ \mathrm{MJ}$ / energy density of battery $= 0.70\times10^{6}\times2500 = 1.75\times10^{9}\ \mathrm{J\,m^{-3}}$(1)
volume $= \dfrac{212\times10^{6}}{1.75\times10^{9}} = 0.12\ \mathrm{m^3}$ (allow 121 L; allow mass route: 303 kg / 2500)(1)
hydrogen has a (very) low density (even when compressed), so its energy density / energy per unit volume is low (about one-sixth that of petrol) (OWTTE)(1)
Answer: (a) $6.0\times10^{9}\ \mathrm{J\,m^{-3}}$ (b) 16 kg (c) $0.12\ \mathrm{m^3}$ (d) low density so low energy density
Examiner tip: Divide the useful energy by the efficiency to find the energy the store must supply; multiplying by the efficiency is the most common slip.

9Standard6 marks

A bungee jumper steps off a high platform. The elastic cord becomes taut after the jumper has fallen 15 m, and the jumper first comes momentarily to rest about 40 m below the platform. The jumper then bounces up and down several times before hanging at rest.

Explain the energy transfers that take place from the moment the jumper steps off the platform until they hang at rest. Your answer should refer to the work done by the forces acting on the jumper and to the principle of conservation of energy. [6]

free fall (first 15 m): gravitational potential energy decreases and kinetic energy increases; the weight does positive work on the jumper(1)
air resistance does negative work, so the gain in $E_k$ is slightly less than the loss in $E_p$; the difference becomes internal/thermal energy of the air (and jumper)(1)
once the cord is taut, the cord does negative work on the jumper; elastic potential energy $\tfrac12 k(\Delta x)^2$ is stored in the cord(1)
$E_k$ continues to increase until the tension equals the weight (resultant force zero), then decreases(1)
work done by the resultant force equals the change in kinetic energy (zero at the start and at the lowest point)(1)
at the lowest point $E_k = 0$: the loss in $E_p$ (over 40 m) equals the elastic energy stored plus the energy dissipated(1)
on each bounce energy is transferred back between $E_H$, $E_k$ and $E_p$, but mechanical energy decreases because of air resistance and internal friction/heating in the cord(1)
the oscillations die away; finally the jumper hangs at rest where tension = weight, with the remaining energy stored as $E_H$ and the rest transferred as thermal energy to the surroundings; total energy is conserved throughout(1)

Level	Marks	What it looks like
Level 3	5–6	Coherent account of the whole sequence (free fall, stretching, bouncing, final rest) naming the correct energy stores at each stage. Work done by weight, cord tension and resistive forces is linked to changes in energy, with the distinction between mechanical energy (decreasing) and total energy (conserved) made explicit. Terminology is precise.
Level 2	3–4	Most stages described with correct energy stores, but the role of work done or of resistive forces is only partly explained, or one stage (e.g. why the bouncing stops) is missing or vague. Some imprecise language, e.g. energy “lost”.
Level 1	1–2	Isolated correct statements, e.g. potential energy changes to kinetic energy, or energy is stored in the cord. Little or no reference to work done or to conservation of energy; may contain errors.
Level 0	0	No relevant physics.

Answer: Levels-marked: $E_p \to E_k$ in free fall; $E_p$ and $E_k \to E_H$ once the cord is taut; resistive forces do negative work so mechanical energy decreases and becomes thermal energy; total energy conserved; final rest at equilibrium.
Examiner tip: Top answers say mechanical energy is not conserved because of resistive forces, while total energy is: never write that energy is “lost” without saying where it goes.

10Hard7 marks

A student investigates the efficiency of a small electric motor. The motor winds a thread onto an axle and lifts a mass of $0.200 \pm 0.001\ \mathrm{kg}$ at constant speed through a height of $0.800 \pm 0.005\ \mathrm{m}$. The student records:

potential difference across the motor $= 6.0 \pm 0.1\ \mathrm{V}$
current in the motor $= 0.45 \pm 0.01\ \mathrm{A}$
time to lift the mass, measured with a hand-held stopwatch $= 2.6 \pm 0.2\ \mathrm{s}$

(a) Calculate the gain in gravitational potential energy of the mass. [1]

(b) Calculate the electrical energy supplied to the motor. [1]

(d) Identify the largest source of uncertainty and suggest one change to the procedure that would reduce it. [2]

$\Delta E_p = 0.200\times9.81\times0.800 = 1.57\ \mathrm{J}$(1)
$E = VIt = 6.0\times0.45\times2.6 = 7.0\ \mathrm{J}$ (allow 7.02 J)(1)
$\eta = \dfrac{1.57}{7.02} = 0.22$ (22%); allow ECF(1)
fractional uncertainty $= \dfrac{0.001}{0.200}+\dfrac{0.005}{0.800}+\dfrac{0.1}{6.0}+\dfrac{0.01}{0.45}+\dfrac{0.2}{2.6} = 0.13$ (13%); award for adding fractional/percentage uncertainties of all five quantities(1)
$\eta = 0.22 \pm 0.03$ (allow $22 \pm 3\%$); uncertainty to 1 s.f. and consistent decimal places needed(1)
the time measurement (about 8% of the total 13%) / reaction time with the stopwatch(1)
lift the mass through a greater height (so the time is longer) / use light gates or video analysis to time the motion / repeat and average the time; do not accept “use a more accurate stopwatch” alone(1)
Answer: (a) 1.57 J (b) 7.0 J (c) $0.22 \pm 0.03$ (d) timing (reaction time); increase the height / use light gates
Examiner tip: Add fractional (or percentage) uncertainties for products and quotients; g is taken as exact, and the absolute uncertainty is quoted to 1 s.f.

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