IB Physics SL A.3 Work, energy and power Questions

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Practise IB Physics A.3 by interpreting force–displacement data, tracing energy transfers and calculating power, efficiency and fuel requirements.

141 questions in this syllabus scope, with question parts counted together.

Exam points & syllabus

Exam points

Account for energy with named intermediate stores and physical destinations; distinguish conserved total energy from mechanical energy or useful output.
Separate no-loss motion from impacts or other dissipative stages; a stated retained-energy fraction determines the next ideal rise or speed.
Trace useful energy through the actual devices: include the turbine or generator, and name the moving or internal store required by the scheme.
Explain a loss by naming its location or destination; bare friction, energy destruction or spread out alone may not earn credit.
Match work to the force component along displacement, and distinguish a real path from zero net displacement around a closed route.
Apply net work to the change in kinetic energy, retaining initial motion and avoiding extra forces already included in a stated resultant.
Read force–distance areas over the requested interval; if the graph shows applied force, subtract resisting work before finding kinetic energy.
Distinguish an instantaneous force or acceleration from accumulated work: decreasing positive force can still increase kinetic energy.
On a slope, use vertical height for gravitational energy and the full travel path for friction; include both braking and the weight component.
Track separate driving and stopping intervals, and compare available energy with resisting work along the actual route before stating a conclusion.
Conserve the appropriate stores over an ideal flight or swing, using the actual vertical height change and any initial kinetic energy.
In a spring or rope system, associate each store with its body and distinguish total fall distance from extension beyond natural length.
Use elastic energy’s quadratic extension dependence; released work is initial minus final stored energy, not the energy left in the spring.
Equal spring energy need not give equal launch speed; parallel springs under the same load require summing each spring’s changed stored energy.
A temporary fall in total kinetic energy can be elastic storage during a collision; read all relevant bodies at the requested instant.
Use squared-speed and momentum relationships carefully: same-mass energy ratios scale as v², while at fixed momentum energy scales as 1/m.
Keep graph variables distinct: constant acceleration gives linear speed–time, quadratic kinetic energy–time and linear kinetic energy–distance.
Use centre-of-mass or average height for distributed water and vertical height for point masses; supplied weight already includes g.
Average power uses energy change over the matching interval; for uniform acceleration do not replace it with final-speed Fv.
At steady speed, driving power balances resistance through Fv; uphill motion or multiple power sources require the correct total force or power.
Separate energy-gain rate, dissipation rate and useful driving power; input may exceed the kinetic-energy increase even in steady operation.
For flow or repeated events, multiply energy per unit mass or per event by the matching rate, then apply efficiency once.
Under a declared uniform incoming-air model, derive available power from kinetic energy per time and mass flow; treat it as an ideal input bound.
For efficiency, define the useful-output boundary and compare like energy or power quantities; given wasted energy is not total input.
For solar systems, use incident intensity × area, external useful output and the averaging interval stated; count panels and conversion stages.
Read fuel-energy units before calculating: convert between energy per mass and per volume using mass density, and include efficiency in consumption.

Covered syllabus points

A.3.1—Conservation of energy A.3.2—Work as energy transfer A.3.3—Sankey diagrams A.3.4—Work by constant force A.3.5—Work-energy change A.3.6—Mechanical energy A.3.7—Mechanical energy conservation A.3.8—Mechanical energy transformations A.3.9—Kinetic energy A.3.10—Gravitational potential energy A.3.11—Elastic potential energy A.3.12—Power A.3.13—Efficiency A.3.14—Fuel energy density

Exam points

Trace named intermediate energy transfers: moving water or wind drives a turbine/generator before electrical output; pumping restores the water store.
In a thermal station, identify steam internal/kinetic energy and turbine kinetic energy; repeating the supplied input and output earns no credit.
For falling lead spheres that warm on impact, state both changes explicitly: gravitational potential → kinetic → internal energy.
With fixed battery energy, additional mass uphill needs more work per distance, so the maximum travel distance decreases.
Explain reduced useful output with a physical transfer or loss mechanism, such as thermal transfer, friction or energy used by station systems.
Degraded energy remains conserved but is less available for useful work; saying only that it is more spread out is insufficient.
At terminal speed, kinetic energy stays constant while gravitational potential energy transfers to the surroundings and mechanical energy decreases.
For a projectile without air resistance, reduced momentum magnitude and kinetic energy at the apex coexist with conserved total mechanical energy.

[Maximum number: 2]

This question is in two parts. Part 1 is about energy resources. Part 2 is about thermal physics.
Part 1 Energy resources
Electricity can be generated using nuclear fission, by burning fossil fuels or using pump storage hydroelectric schemes.

A mass of 0.22 kg of lead spheres is placed in a well-insulated tube. The tube is turned upside down several times so that the spheres fall through an average height of 0.45 m each time the tube is turned. The temperature of the spheres is found to increase by $8^{\circ} \mathrm{C}$ .

Discuss the changes to the energy of the lead spheres.

Exam points

Define work as energy transferred by a force acting through displacement; include the component in the direction of displacement.
Read work from the area under a force–distance graph over the stated interval, including the baseline area and excluding earlier motion.
For friction bringing an object to rest, match the dissipated energy to its initial kinetic energy or friction force × stopping distance.
An object moving horizontally at constant velocity without friction needs no work to maintain its motion, even if it already has kinetic energy.

[Maximum number: 2]

This question is in two parts. Part 1 is about the motion of a ship. Part 2 is about melting ice.

Outline the meaning of work.

work done = force × distance moved;
(distance moved) in direction of force;
or
energy transferred;
from one location to another;
or
work done $=F s \cos \theta$ ;
with each symbol defined;

Exam points

Read the width of the requested Sankey branch against a known output on the same scale; distinguish separate loss stages.
Draw a recognisable branched Sankey diagram and label the input, useful output and loss; split losses into multiple arrows only consistently.
Make arrow widths proportional and conserve the input: 300 kW input with 100 kW useful output requires 200 kW loss, giving widths 3:1:2.
Use useful output and efficiency to find the total input, then subtract useful output for the loss before scaling the Sankey arrows.

[Maximum number: 3]

On the grid, sketch a labelled Sankey diagram that represents the energy transformation in this OWC.

Diagram correct shape (but scaled incorrectly) and labelled.
Waste energy $0.32\ \text{MW}$ (allow ECF from (b)(i)).
All dimensions scaled correctly (by eye).

Figure for Answer to Question 1 — IB Physics SL

Exam points

Resolve an angled force along the displacement in W = Fs cosθ, converting kN and km before calculating the work.
Explain zero work in uniform circular motion through the force being perpendicular to displacement; a nonzero centripetal force still acts.
For constant friction stopping a block, use the kinetic-energy loss divided by stopping distance for force; work on the block is negative.
When a task prescribes an average-force model, equate its work to the kinetic-energy gain before calculating contact distance.
For braking over several turns, use turns × circumference as the friction path; report the dissipated magnitude or signed work on the system.
At terminal speed, drag balances weight and distance is vt: the positive work transferred to the air is mgvt.

[Maximum number: 2]

This question is in two parts. Part 1 is about two children on a merry-go-round. Part 2 is about electric circuits.

Part 1 Two children on a merry-go-round
Aibhe and Euan are sitting on opposite sides of a merry-go-round, which is rotating at constant speed around a fixed centre. The diagram below shows the view from above.

Aibhe is moving at speed $1.0 \mathrm{~ms}^{-1}$ relative to the ground.

Euan drags his foot along the ground to act as a brake. The merry-go-round comes to a stop after 4.0 rotations. The radius of the merry-go-round is 1.5 m . The average frictional force between his foot and the ground is 45 N . Calculate the work done.

distance travelled by Euan $=4.0 \times 2 \pi \times 1.5(=37.70 \mathrm{~m})$ ;

W\left(=F_{a v} d=45 \times 37.70\right)=1700(\mathrm{~J}) ;

Exam points

Set net work equal to ½mv² − ½mu²; include initial kinetic energy and avoid adding forces already included in a stated net force.
Use the area under a net-force–distance graph for kinetic-energy change, including triangular sections and their full bases.
A decreasing positive net force still adds kinetic energy while acceleration decreases; speed depends on accumulated work, not the peak force.
For an acceleration–distance graph, multiply the total area by mass to obtain the resultant work.
If a graph shows applied force rather than net force, subtract resisting work over the full distance before calculating final kinetic energy.
At fixed stopping distance and mass, initial speed scales with the square root of friction force; halving friction gives v/√2.
For uphill stopping, include both friction or braking and the weight component; kinetic-energy loss determines the total opposing work.
On a descending rough ramp, use the vertical drop for gravitational work and the full path for friction: ΔEk = mgd sinθ − Fd.
For a drive that acts only initially, subtract dynamic friction during the drive, then use the remaining kinetic energy for the later stopping phase.
Compare initial kinetic energy with resisting work along the actual route to decide whether an object reaches a destination.
When a frictionless descent is followed by rough horizontal stopping, equate mgh to μmgs so the initial ramp height is h = μs.

[Maximum number: 4]

This question is about the motion of a bicycle.

A cyclist is moving up a slope that is at an angle of $19^{\circ}$ to the horizontal. The mass of the cyclist and the bicycle is 85 kg .

At the bottom of the slope the cyclist has a speed of $5.5 \mathrm{~m} \mathrm{~s}^{-1}$ . The cyclist stops pedalling and applies the brakes which provide an additional decelerating force of 250 N . Determine the distance taken for the cyclist to stop. Assume air resistance is negligible and that there are no other frictional forces.

total decelerating force $=271+250(=521 \mathrm{~N})$ ;
acceleration $=(-) \frac{521}{85}\left(=-6.13 \mathrm{~ms}^{-2}\right)$ ;
$s=\frac{v^{2}-u^{2}}{2 a} ;$ $2.47(\mathrm{~m}) ;\} \begin{aligned} & \text { (signs must be consistent for this mark, ie: if acceleration assumed } \\ & \text { positive, look for negative distance) }\end{aligned}$

Allow use of g=10. Answers are $527 \mathrm{~N}, 6.2 \mathrm{~m} \mathrm{~s}^{-2}, 2.44 \mathrm{~m}$ .
or
total decelerating force $=271+250(=521 \mathrm{~N})$ ;
initial kinetic energy $=\frac{1}{2} m v^{2}=1290 \mathrm{~J}$ ;
distance $=\frac{\text { energy lost }}{\text { force }}=\frac{1290}{521}$ ;
2.47 (m);

Exam points

For an ideal descent, add the lost gravitational potential energy to any initial kinetic energy before finding the final speed.
Use the actual vertical height drop from the diagram: reaching the top of a loop leaves the car two radii above ground.
For a rebound rising without losses, kinetic energy becomes gravitational potential energy; half the starting speed gives a quarter of the height.
During a local constant-field fall from rest, kinetic energy gained equals the lost fraction of the initial gravitational potential energy.
For an ideal maximum-speed estimate of a falling club head, equate mgh to ½mv²; the mass cancels.
In an elastic spring collision, subtract the carts’ total kinetic energy at the stated instant from the initial mechanical energy for spring energy.
Keep energy-conservation stages separate: a frictionless approach may conserve mechanical energy even when the following sticking collision does not.
For a show-that speed, display the energy equation and substitution or sufficient precision; quoting the rounded target alone may lose method credit.

[Maximum number: 2]

A student strikes a tennis ball that is initially at rest so that it leaves the racquet at a speed of $64 \mathrm{~m} \mathrm{~s}^{-1}$ . The ball has a mass of 0.058 kg and the contact between the ball and the racquet lasts for 25 ms .

The student strikes the tennis ball at point P . The tennis ball is initially directed at an angle of $7.00^{\circ}$ to the horizontal.

The following data are available.

Determine the speed of the tennis ball as it strikes the ground.

ALTERNATIVE 1

Initial KE+PE= final KE /
$\frac{1}{2} \times 0.058 \times 64^{2}+0.058 \times 9.81 \times 2.80=\frac{1}{2} \times 0.058 \times v^{2}$

ALTERNATIVE 2
$\begin{aligned} & v_{v}=<\sqrt{7.8^{2}+2 \times 9.81 \times 2.8}>=10.8<\mathrm{ms}^{-1} \gg & <v=\sqrt{63.5^{2}+10.8^{2}}> & v=64.4<\mathrm{ms}^{-1} \gg \end{aligned}$

Exam points

As a stretched spring draws a block towards equilibrium on a frictionless surface, spring elastic potential energy falls and block kinetic energy rises.
Before an elastic rope becomes taut, state explicitly that the falling block’s gravitational potential energy transfers to its kinetic energy.
While a falling block stretches a rope to maximum extension, both its initial kinetic energy and further gravitational energy loss become rope elastic energy.
For an ideal swinging ball, equate the centre-of-mass gravitational energy loss to translational kinetic energy and show the requested precision.

[Maximum number: 2]

An elastic climbing rope is tested by fixing one end of the rope to the top of a crane. The other end of the rope is connected to a block which is initially at position A. The block is released from rest. The mass of the rope is negligible.

The unextended length of the rope is 60.0 m . From position A to position B, the block falls freely.

(a)

For the rope and block, describe the energy changes that take place

[ 2 ]

(i)

between A and B .

[ 1 ]

(ii)

between B and C .

[ 1 ]

Exam points

Use Ek = ½mv² and take the positive speed magnitude when rearranging; distinguish energy from power expressions such as Fv.
With constant acceleration from rest, speed is linear in time, kinetic energy is quadratic in time, and kinetic energy is linear in distance.
For vertical flight without air resistance, kinetic energy follows a nonnegative quadratic time curve with zero at the highest point.
For a particle circling at n revolutions per second, first use v = 2πrn, then Ek = ½mv² = 2π²mn²r².
Use Ek = p²/(2m) for a momentum-based energy estimate; at equal momentum, kinetic energy is inversely proportional to mass.
When mass and kinetic energy both change between objects, use p = √(2mEk) and take the square root of the product of their ratios.
For the same mass before and after a bounce, retained kinetic-energy fraction is the squared ratio of speed magnitudes.
For fractional kinetic-energy loss, subtract final energy from initial energy and divide by initial energy; the mass cancels in a same-particle ratio.
At a projectile’s peak, horizontal kinetic energy remains; add the subsequent height gain to infer its earlier post-bounce kinetic energy.
Convert net-force–distance area into kinetic energy before finding speed; include rectangular and triangular sections of the graph.
Recognise kinetic energy as a scalar: it depends on squared speed, unlike the vector quantities velocity, momentum and acceleration.

[Maximum number: 2]

A toy rocket is made from a plastic bottle that contains some water.

Air is pumped into the vertical bottle until the pressure inside forces water and air out of the bottle. The bottle then travels vertically upwards.

The air-water mixture is called the propellant.
The variation with time of the vertical velocity of the bottle is shown.

The bottle reaches its highest point at time $T_{1}$ on the graph and returns to the ground at time $T_{2}$ . The bottle then bounces. The motion of the bottle after the bounce is shown as a dashed line.

The bottle bounces when it returns to the ground.

Calculate the fraction of the kinetic energy of the bottle that remains after the bounce.

Attempt to evaluate KE ratio as $\left(\frac{v_{\text {final }}}{v_{\text {initial }}}\right)^{2}$

« $\left(\frac{4.5}{10}\right)^{2}=$ 》 0.20 OR 20 % OR $\frac{1}{5}$

Marking guidance:

Accept ± 0.5 velocity values from graph

Exam points

Use the vertical height change in ΔEp = mgΔh; during constant-speed descent, the lost gravitational energy becomes thermal energy.
If weight is supplied, multiply weight by vertical displacement directly; convert joules to kJ and round to the precision of the data.
For stored water spread over a depth, use its centre-of-mass or average height drop; specific energy is gΔh in J/kg.
In a draining tidal basin, use half the level difference for the water’s mean height loss, then multiply the gravitational energy by efficiency.
When a gravitational-energy gain rate is given, first multiply by time, then use Δh = ΔEp/(mg), converting minutes and kW consistently.

[Maximum number: 3]

This question is about a tidal power station.

A tidal power station is built for a coastal town. Sea water is stored in a tidal basin behind a dam at high tide and released in a controlled manner between high tides, so that it passes through turbines to generate electricity.

The following data are available.

Calculate the electrical energy produced between successive high and low tides.

difference in height of centre of mass of water $=\frac{1.8}{2}=0.9(\mathrm{~m})$ ;
$\Delta E_{\mathrm{P}}\left(=m g \Delta h=2.8 \times 10^{8} \times 9.8 \times 0.9\right)=2.5 \times 10^{9}(\mathrm{~J}) ;$
electrical energy $\left(=0.24 \times 2.5 \times 10^{9}\right)=5.9 \times 10^{8}(\mathrm{~J})(=590 \mathrm{MJ})$ ;

Marking guidance:

Allow ECF for [2 max] if candidate omits factor of 2 in first marking points.
Accept $g=10 \mathrm{~m} \mathrm{~s}^{-2}$ giving an answer of $6.0 \times 10^{8}(\mathrm{~J})$ .

Exam points

Use elastic energy ½kx² with extension or compression measured from natural length; rearrange as k = 2E/x².
For a linear force–extension relation, stored energy is the triangular area ½Fx, not Fx.
Work released by contraction is the decrease in stored energy: if extension becomes ax, the released work is E(1 − a²).
For a mass hanging at rest, first use kx = mg to find the extension, then calculate ½kx².
For a spring launch without losses, equate ½kx² to ½mv²; at fixed k and m, launch speed is proportional to compression.
For two identical parallel springs under the same total load, each extension halves; add both springs’ energies to get half the original total.
At maximum rope stretch, balance the elastic energy with the initial gravitational energy, or with both gravitational and kinetic energy at tautness.
When spring energy lifts a block that is still moving, include both the gravitational gain and its final kinetic energy.
Explain an elastic deformation mechanism by identifying temporary stored elastic energy and its subsequent transfer to the struck ball.

[Maximum number: 2]

The unextended length of the rope is 60.0 m . From position A to position B, the block falls freely.

The length reached by the rope at C is 77.4 m . Suggest how energy considerations could be used to determine the elastic constant of the rope.

k can be determined using EPE $=\frac{1}{2} k x^{2}$

correct statement or equation showing

GPE at A=EPE at C

OR

( GPE+KE ) at B=EPE at C

Candidate must clearly indicate the energy associated with either position A or B for MP2.

Exam points

Average power is energy transferred divided by the matching time interval; use ΔEk/Δt for acceleration and convert milliseconds to seconds.
For uniform acceleration from rest, mean driving power uses work/time or the mean speed, not the final-speed value of Fv.
When lifting while still gaining speed, include both gravitational and kinetic energy gains in the average power.
Use P = Fv with the driving or resisting force at steady speed; zero resultant force does not mean zero useful power.
At fixed uphill speed, supplied power must overcome both the ordinary resistance and the component of weight down the slope.
For angled traction, use the force component along velocity; a source’s power fraction uses the total power from all contributing sources.
When a quadratic resistance law F = kv² is supplied, steady driving power is kv³; compare steady speeds with a cube-root power ratio.
Keep the requested interval when calculating average dissipation; subtract final kinetic energy and divide by the later travel time.
Use weight × vertical height/time for power against gravity, and distinguish energy-release time from the subsequent flight time.
For water flow, convert volume per time to mass per second and multiply by g times the mean vertical drop.
A finite energy store sustains output for t = E/P; sum simultaneous generator ratings and convert specific energy to J/kg.
Multiply fuel energy per mass by mass consumed per second for input power, then apply the efficiency once for useful output.
For repeated events, convert each event’s energy to joules, multiply by events per second, and apply the stated output efficiency.
Use E = Pt for a thermal transfer; where some input warms the system, subtract that gain before finding the average loss power.
Battery energy qV divided by discharge time gives mean power; convert kC and km consistently before relating that power to resistance.
For incoming material, distinguish its kinetic-energy gain rate from the driving power Fv; pickup can also dissipate energy.
For an ideal uniform incoming air stream, derive incident power from kinetic energy per time with mass flow ρAv, giving ½ρAv³.
Keep swept blade area separate from land area per turbine; divide required output by useful power per turbine before sizing the farm.
Under the same ideal input model, compare wind powers with the cube of the speed ratio and multiply power by time for an energy estimate.
Power returned by a relaxing Hooke’s-law spring averages its released elastic energy divided by return time: ½kx²/Δt.
In vacuum fall from rest, the rate of kinetic-energy gain grows linearly with time because P = mgv and v = gt.
For constant acceleration with increasing resistance, engine force must increase as speed rises, so power does not remain constant or merely follow v.

[Maximum number: 2]

A 300 W immersion heater is placed in a beaker containing 0.25 kg of water at a temperature of $18^{\circ} \mathrm{C}$ . The heater is switched on for 120 s , after which time the temperature of the water is $45^{\circ} \mathrm{C}$ . The thermal capacity of the beaker is negligible and the specific heat capacity of water is $4.2 \times 10^{3} \mathrm{Jkg}^{-1} \mathrm{~K}^{-1}$ .

Determine the rate at which thermal energy is transferred from the water to the surroundings during the time that the heater is switched on.

energy transfer $=[300 \times 120]-\left[2.835 \times 10^{4}\right]=7.65 \times 10^{3} \mathrm{~J}$ ;
rate of transfer $=\frac{7.650 \times 10^{3}}{120}=64 \mathrm{~W}$ ;

Marking guidance:

Allow ECF from (b)(i).

Exam points

Calculate efficiency as useful output divided by total input, using energies or powers for the same system and time interval.
If wasted energy is given, it is the fraction 1 − η of input; find the total input before calculating the useful work.
Read useful and input widths on a Sankey diagram, including all relevant loss branches and avoiding useful-output divided by loss.
Match the efficiency boundary to the named device: motor output comes before downstream pulley losses, while system output reaches the load.
For steady lifting, useful power is mgv or weight × height/time; compare it with electrical input VI or the stated input power.
For a pump with negligible kinetic-energy changes, useful power is mass flow × g × depth, so required input is Rgd/η.
For solar panels, input power is irradiance × area and useful output is ηIA; convert kW, kJ/s and panel dimensions before substitution.
For photovoltaic output to an external circuit, use terminal voltage × current, then divide by incident radiation power.
For a required solar output, rearrange area = Poutput/(ηI), using the irradiance appropriate to the stated location and averaging interval.
Compare alternative systems with useful output and a quantitative area or conversion argument, following the question’s stated exclusions.
Apply efficiency and the number of generators separately before sizing each unit or counting the whole units needed to meet demand.
For a declared uniform incoming-wind model, useful output scales as efficiency × swept area × v³; apply every changed factor.
Explain a loss with its physical location or destination, such as pipe friction, bearing friction, wire heating or residual outgoing fluid energy.

[Maximum number: 3]

This question is in two parts. Part 1 is about renewable energy. Part 2 is about nuclear energy and radioactivity.

A small coastal community decides to use a wind farm consisting of five identical wind turbines to generate part of its energy. At the proposed site, the average wind speed is $8.5 \mathrm{~m} \mathrm{~s}^{-1}$ and the density of air is $1.3 \mathrm{~kg} \mathrm{~m}^{-3}$ . The maximum power required from the wind farm is 0.75 MW . Each turbine has an efficiency of 30 %.

Determine the diameter that will be required for the turbine blades to achieve the maximum power of 0.75 MW .

total wind power required $=\frac{750000}{0.3}$ ;

Marking guidance:

maximum wind power required per turbine, $P=\frac{750000}{5 \times 0.3}(=500 \mathrm{~kW})$ ;
$d=\left(\frac{8 P}{\rho \pi v^{3}}=\right)^{\frac{1}{2}} 40(\mathrm{~m}) ;$
[3]
Award [1 max] for an answer of 48.9(m) as it indicates 5 and 0.3 ignored.
Award [2 max] for 22(m) as it indicates 0.3 ignored.
Award [2 max] for 89(m) as it indicates 5 ignored.

Exam points

Read the stated units: specific energy is energy per mass in J/kg, while volumetric energy density is in J/m³; older papers may use looser names.
Convert mass-specific energy to volumetric energy density by multiplying by mass density: u = ρε; the ratio ε/u is 1/ρ.
With volumetric energy density u, fuel volume is E/u and mass is ρE/u; check that the final units match the requested quantity.
Fuel input power is specific energy × mass flow, or volumetric energy density × volume flow; useful output is the input multiplied by efficiency.
For steady vehicle motion, fuel volume consumed per second is useful thrust power Fv divided by efficiency and volumetric fuel energy density.
For fuel consumption over a time interval, first calculate output energy Pt, divide by efficiency for input energy, then divide by energy per kg.
A minimum fuel-saving estimate may assume ideal conversion; state that boundary rather than inventing an unstated plant efficiency.
Distinguish high fuel energy density from renewability: the conventional nuclear-fuel example has high energy density but a non-renewable supply.

[Maximum number: 2]

This question is in two parts. Part 1 is about renewable energy. Part 2 is about nuclear energy and radioactivity.

Currently, a nearby coal-fired power station generates energy for the community. Less coal will be burnt at the power station if the wind farm is constructed.

The energy density of coal is $35 \mathrm{MJ} \mathrm{kg}^{-1}$ . Estimate the minimum mass of coal that can be saved every hour when the wind farm is producing its full output.

mass of coal per second $(=0.0214 \mathrm{~kg})$ ;
$77.1(\mathrm{~kg})$ ;
or
energy saved per hour $=0.75 \times 3600\left(=2700 \mathrm{MJ} \mathrm{h}^{-1}\right)$ ;
mass of coal saved $=\left(\frac{2700}{35}=\right) 77.1(\mathrm{~kg})$ ;

Marking guidance:

Award [2] for a bald correct answer.

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