Work and energy give physics its 'accounting system' — a way of tracking how much a force can achieve, and how that capability moves between kinetic, gravitational and elastic stores as a system changes. In this workbook you will calculate work done by forces at an angle, connect work to energy transfer, use Sankey diagrams to visualise where energy goes, apply the conservation of mechanical energy to falling and sliding objects, meet Hooke's law and elastic potential energy, and use power and efficiency to compare how quickly — and how well — that energy is put to use.
By the end of this workbook you should be able to:
calculate the work done by a constant force, including when it acts at an angle to the displacement, W = Fs cosθ
explain work done as a transfer of energy, and represent energy transfers on a Sankey diagram
calculate kinetic energy, gravitational potential energy and elastic potential energy, and identify mechanical energy as their sum
state and apply Hooke's law and derive elastic potential energy from a force–extension graph
apply the conservation of mechanical energy to systems with negligible friction, and explain what happens when friction is present
calculate power as the rate of energy transfer, P = E/t = Fv, and efficiency as a ratio of useful to total energy (or power)
use the energy density of fuels to compare how much energy a given volume of fuel can release
1. Work done by a force
In physics, work has a much more precise meaning than in everyday language. A force does work on an object whenever it moves the object in the direction the force acts. A weightlifter holding a barbell motionless above their head is applying a large force, but because nothing is moving, no mechanical work is being done on the barbell at that instant.
work done, W = Fs (when F acts in the direction of motion)
Joule, J: the SI unit of work and energy. One joule is the work done when a force of 1 N moves its point of application by a distance of 1 m. Work done is a scalar quantity — it has no direction.
A force often acts at an angle to the direction an object moves. In that case, only the component of the force along the direction of displacement contributes to the work done — the component perpendicular to the motion does no work at all.
Fig. 1.1 A force F acting at angle θ to the displacement, s. Only the component F cosθ, along the direction of motion, does work; the perpendicular component F sinθ does none.
work done, W = Fs cosθ
Worked example 1.1
A dog pulls a load with a force of 250 N at an angle of 10° to the horizontal. Calculate the work done, in kJ, by the dog in moving the load horizontally by 15 m.
Answer:
W = Fs cosθ = 250 × 15 × cos10°
Work done by a varying force. The equation W = Fs cosθ assumes a constant force. When a force varies — for example while stretching a rubber band — the work done is equal to the area under a force–distance graph, found by counting squares or, for simple shapes, using the area of a triangle or trapezium.
Worked example 1.2
A force increases steadily from 0 N to 60 N as it stretches a rubber band by 0.40 m. Estimate the work done on the band.
Answer:
The force–extension graph is a straight line through the origin, so the area under it is a triangle.
W = ½ × base × height = ½ × 0.40 × 60
Check your understanding
1A weightlifter holds a barbell stationary above her head for several seconds. Explain why no mechanical work is being done on the barbell during this time, even though a large force is being applied.
Work is only done on an object while it is being displaced in the direction of the applied force. Because the barbell is stationary (its displacement is zero while held still), no work is being done on it at that moment, regardless of how large the supporting force is. (Work is, however, still being done inside the weightlifter's muscles, as chemical energy is used to maintain tension.)
2A man pulls a crate with a horizontal force of 150 N and moves it along a horizontal floor by 5.0 m. Calculate the work he has done.
W = Fs = 150 × 5.0
3A gardener pushes a lawnmower at a constant speed with a force of 70 N at an angle of 40° to the ground. Calculate the work done in moving the lawnmower a distance of 12 m.
W = Fs cosθ = 70 × 12 × cos40°
4A satellite moves in a circular orbit around the Earth at constant speed. Explain why the Earth's gravitational force on the satellite does no work on it, even though the force is not zero.
In a circular orbit, the gravitational force on the satellite always points towards the centre of the orbit, which is perpendicular to the satellite's velocity (and therefore its displacement) at every instant. Since θ = 90° and cos90° = 0, the component of the force along the direction of motion is zero, so W = Fs cosθ = 0 however large F or s is.
2. Energy and its conservation
Why can't someone push a loaded trolley along a flat road forever? Eventually they run out of energy — which suggests a useful definition: energy is the capacity, or ability, to do work. Whenever work is done on an object, energy is transferred to it from whatever supplied the force.
Key idea. When work is done, energy is transferred from a source to the object. Work done by a force is equivalent to a transfer of energy — the two describe the same physical process.
Energy exists in many forms. Some of the most important for mechanics are kinetic energy (of moving objects), gravitational potential energy (stored because of position in a gravitational field) and elastic potential energy (stored in a stretched or compressed material). Together, these three make up an object's mechanical energy.
Mechanical energy: the sum of an object's kinetic energy, gravitational potential energy and elastic potential energy — the energy a macroscopic object or system has that is able to do useful work.
Other, non-mechanical forms include chemical energy (stored in bonds), internal energy (the random kinetic and potential energy of particles within a substance), electrical energy, radiation energy, and nuclear energy. Whatever the form, one principle always holds:
Principle of conservation of energy. The total energy of an isolated system is constant. Energy cannot be created or destroyed — only transferred from one form to another, or from one object to another.
Worth remembering"Lost" energy isn't destroyed. A bouncing ball rises a little less with every bounce, and it's tempting to say it has "lost" energy. In fact, the total energy is unchanged — some of the ball's mechanical energy has been transferred to internal energy in the ball and the ground (making them very slightly warmer) and to sound. Energy transferred to the surroundings in this uncontrolled way is called dissipated (or degraded) energy: it still exists, but it can no longer be recovered to do useful work.
Check your understanding
5State the principle of conservation of energy, and explain the difference between saying energy has been "destroyed" and saying it has been "dissipated".
The principle of conservation of energy states that the total energy of an isolated system is constant — energy cannot be created or destroyed, only transferred between forms. "Destroyed" would mean the energy no longer exists anywhere, which never happens. "Dissipated" means the energy has been transferred to a form (usually internal energy in the surroundings, or sound) that has spread out and can no longer be usefully recovered — the energy still exists, but not in a useful, concentrated form any more.
6State the main energy transfer that takes place when: (a) a battery-powered toy car moves across the floor, (b) an electric kettle heats water, (c) a person climbs a flight of stairs.
(a) chemical energy (in the battery) to kinetic energy (of the car), with some to internal energy and sound. (b) electrical energy to internal (thermal) energy of the water. (c) chemical energy (from food, in the person's body) to gravitational potential energy, with some to internal energy.
7A ball is dropped and allowed to bounce several times on a hard floor. Explain why the height reached after each bounce is lower than the height it fell from.
At each bounce, some of the ball's mechanical energy (kinetic energy just before impact) is dissipated — transferred to internal energy in the ball and floor, and to sound — rather than being fully returned as kinetic energy on the way back up. With less mechanical energy available after each bounce, the ball can only reach a smaller maximum height before falling again.
3. Kinetic and gravitational potential energy
Kinetic energy is the energy a mass has because it is moving. Using the equations of motion for a mass accelerated from rest by a constant force, it can be shown that:
kinetic energy, Ek = ½mv²
Key idea. Kinetic energy is a scalar — even though it is calculated from velocity, the v² term means direction drops out. Ek can also be written as p²/2m, in terms of momentum p, which is useful when comparing kinetic energy and momentum together.
Gravitational potential energy is the energy an object has because of its position in a gravitational field. Close to the Earth's surface, where the gravitational field strength g can be treated as constant, raising an object a height Δh requires a force equal to its weight, mg, so the object gains:
change in GPE, ΔEp = mgΔh
Key idea. GPE is always calculated as a change, ΔEp, relative to some reference height that you choose (often the ground, or a table top). For this reason it is really a work done equation (force × distance). The equation ΔEp = mgΔh also only applies when g can be treated as constant, which fails for very large changes in height (such as a satellite launch).
Worked example 3.1
Calculate the kinetic energy, in kJ, of a 60 kg sprinter running at 10 m s−1.
Answer:
Ek = ½mv² = ½ × 60 × 10²
Worked example 3.2
A 1.2 kg box is raised from the floor to a table top 0.85 m higher. Calculate the gravitational potential energy transferred.
Answer:
ΔEp = mgΔh = 1.2 × 9.8 × 0.85
Check your understanding
8Calculate the kinetic energy, in MJ, of an articulated lorry of mass 40 000 kg travelling at 30 m s−1.
Ek = ½mv² = ½ × 40 000 × 30²
9A cable car rises a vertical height of 700 m. Calculate the gravitational potential energy gained by a cabin of total mass 1800 kg during the journey.
ΔEp = mgΔh = 1800 × 9.8 × 700
10A rocket launches a satellite to a height of 400 km above the Earth's surface. Explain why the equation ΔEp = mgΔh, using g = 9.8 N kg−1, cannot be used to accurately calculate the gravitational potential energy gained by the satellite.
The equation ΔEp = mgΔh assumes that g stays constant over the height change involved. This is a reasonable approximation for everyday heights, but 400 km is a large fraction of the Earth's radius, and the gravitational field strength decreases noticeably over that distance (g is smaller at 400 km altitude than at the surface). Using a single, surface value of g throughout would overestimate the true energy gained.
4. Hooke's law and elastic potential energy
A spring, or any elastic material, is a store of elastic potential energy when it is stretched or compressed away from its natural shape. Whenever a material is deformed like this, it exerts an elastic restoring force — a contact force that pulls or pushes back towards the material’s original shape. In the 17th century, the scientist Robert Hooke investigated how the size of this restoring force depends on the size of the deformation, and found a simple relationship that now carries his name.
Hooke's law. Provided a spring or elastic material is not stretched beyond its elastic limit, the restoring force is directly proportional to the extension (or compression): FH = −kΔx, where k is the spring constant (unit: N m−1) and Δx is the extension. A larger k means a stiffer spring.
Elastic and plastic behaviour: a deformation is elastic if the material returns to its original shape once the force is removed — this is the regime Hooke's law describes. Beyond a material's elastic limit, the deformation becomes plastic: some of it is permanent, FH = kΔx no longer holds, and a force–extension graph is no longer a straight line through the origin.
Worked example 4.1
A spring is 0.38 m long. When it is pulled by a force of 2.0 N, it stretches to 0.42 m. Assuming the spring behaves elastically, calculate its spring constant.
Answer:
extension, Δx = 0.42 − 0.38 = 0.04 m
FH = kΔx, so k = FH ÷ Δx = 2.0 ÷ 0.04
Live simulation: springs and an elastic band
Use the apparatus below to load a steel spring (singly, or in series/parallel) or an elastic band with masses, and watch the force–extension graph build up. Springs obey Hooke's law up to a limit, beyond which they deform permanently (plastically); an elastic band shows a curved, non-Hookean force–extension relationship, and will eventually snap. Try loading and then unloading to see the difference between elastic and plastic behaviour.
Add or remove 100 g masses, switch between a steel spring (single, series or parallel) and an elastic band, and press "New Run" to start a fresh loading line while keeping previous runs faintly visible for comparison.
Interactive simulation — open the online version of this workbook to launch it.
Simulation by Dr Dan Jones, Hookean Physics (graph redrawn locally so the simulation runs without an internet connection).
When a spring is stretched from zero extension, the force needed grows steadily from 0 up to its final value FH, so the average force used is ½FH. Using work done = average force × distance:
Fig. 4.1 A force–extension graph for a spring obeying Hooke's law. The area of the shaded triangle — ½ × F₀ × Δx — equals the work done stretching the spring, which is stored as elastic potential energy.
elastic potential energy, EH = ½FHΔx
elastic potential energy, EH = ½kΔx²
Key idea. Elastic potential energy EH is a work-done relationship (force × distance is the area under the line). There is no need for a minus sign in EH = ½kΔx² — unlike force, work done (and therefore stored energy) is not a vector, so it is always positive for a deformation in either direction.
Worked example 4.2
A spring has a spring constant of 384 N m−1 and is stretched by 2.0 cm. Calculate the elastic potential energy stored in the spring, assuming it obeys Hooke's law.
Answer:
EH = ½kΔx² = ½ × 384 × 0.020²
Check your understanding
11A spring is 15 cm long and has a spring constant of 50 N m−1. When a force is hung from the end, the spring stretches to 20 cm long. Determine the force that was added.
12Calculate the elastic potential energy stored in an elastic cord with a spring constant of 525 N m−1 that has been stretched by 15 cm.
EH = ½kΔx² = ½ × 525 × 0.15²
13Using the simulator, load the steel spring and the elastic band in turn, up to a similar extension. Describe one difference between the shapes of their force–extension graphs, and state what this tells you about whether each obeys Hooke's law.
The steel spring's force–extension line is straight, showing that force is directly proportional to extension — it obeys Hooke's law (until it is overloaded and starts to deform permanently, seen as a change in gradient and a gap opening up between the loading and unloading lines). The elastic band's line is curved rather than straight, showing that its force is not directly proportional to its extension — an elastic band does not obey Hooke's law.
14A catapult has a spring constant of 100 N m−1 and has been stretched to hold an elastic potential energy store of 12.5 J. Calculate how far it has been stretched.
Applying the principle of conservation of energy specifically to mechanical energy gives a very useful result:
Conservation of mechanical energy. In the absence of frictional or other resistive forces, the total mechanical energy of a system is conserved: kinetic energy + gravitational potential energy + elastic potential energy = constant.
If mechanical energy is conserved, then work is simply the amount of energy transformed between the different mechanical forms — kinetic, gravitational potential and elastic potential — as the system changes. Nothing is gained or lost overall; energy just moves between these three stores.
Worked example 5.1
A ball is thrown upwards with a speed of 18 m s−1. Assuming air resistance is negligible, calculate the maximum height it reaches.
Answer:
At maximum height, all of the ball's initial kinetic energy has been transformed to gravitational potential energy:
½mu² = mgΔh, so Δh = u² ÷ (2g) = 18² ÷ (2 × 9.8)
Worked example 5.2
A 4.7 kg box slides down a frictionless slope of vertical height 0.80 m, starting from rest. Calculate the speed of the box at the bottom of the slope.
Answer:
Loss in GPE = gain in KE:
mgΔh = ½mv²
v = √(2gΔh) = √(2 × 9.8 × 0.80)
Live simulation: energy conservation in a valley
Set a release height and a mass, then press play to watch a block slide down into a smooth valley and back up the other side. The bar chart shows kinetic energy, gravitational potential energy and total mechanical energy as the block moves. Switch on friction to see what happens to the total when energy is dissipated.
Fig. 5.1 A block released from rest slides through a frictionless (or, with friction switched on, a lossy) valley. With no friction, total mechanical energy stays constant as it is exchanged between kinetic and gravitational potential energy.
Worth rememberingReal slopes have friction. A real block sliding down a real slope will arrive slower than the frictionless prediction, because some mechanical energy is dissipated as internal energy (and a little sound) at the sliding surfaces. The conservation of mechanical energy equation still gives a very useful ideal prediction, and the gap between the ideal and measured speeds tells you how much energy was dissipated.
Check your understanding
15A ball is thrown upwards with a speed of 23 m s−1. Calculate the maximum height it can reach, assuming air resistance is negligible.
u² = 2gΔh
Δh = u² ÷ (2g) = 23² ÷ (2 × 9.8)
16A skateboarder of mass 55 kg, starting from rest, rolls down a smooth ramp of vertical height 1.6 m. Assuming resistive forces are negligible, calculate her speed at the bottom.
mgΔh = ½mv²
v = √(2gΔh) = √(2 × 9.8 × 1.6)
17Using the simulator with friction switched off, release the block and describe what happens to the kinetic energy and gravitational potential energy readouts as it swings from side to side. Use the readout to confirm that the total mechanical energy is conserved.
As the block descends, its gravitational potential energy decreases while its kinetic energy increases by the same amount; at the bottom of the valley, GPE is at a minimum and KE is at a maximum. As the block rises up the far side, kinetic energy decreases back to zero while gravitational potential energy increases again, reaching (almost) the same height as the release point. Throughout the motion, the total mechanical energy readout (KE + GPE) stays constant, confirming that mechanical energy is conserved when there is no friction.
18Switch friction on in the simulator and release the block again. Explain what you observe happening to the maximum height reached on each successive swing, and where the missing mechanical energy has gone.
With friction switched on, the block reaches a lower height on each successive swing — the total mechanical energy readout decreases over time rather than staying constant. This is because friction is a resistive (non-conservative) force: it dissipates some mechanical energy into internal energy (and a small amount of sound) at the surface with every pass through the valley, so it is not available to be converted back into kinetic or gravitational potential energy.
6. Sankey diagrams
A Sankey diagram is a flow diagram used to visualise energy transfers: the width of each arrow is proportional to the amount of energy (or power) it represents. They are named after Captain Matthew Henry Sankey, who in 1898 first published a diagram of this kind, comparing the energy flow in a real steam engine with an ideal one.
Fig. 6.1 A simple Sankey diagram for a lamp: of 100 J of electrical energy supplied, only 5 J leaves as useful light energy — the remaining 95 J is dissipated as heat. The height of the input bar and the thickness of each horizontal arrow are both proportional to the amount of energy they carry, so the two arrows' thicknesses add up to the height of the input bar.
Worked example 6.1
An electric kettle has a power rating of 2000 W and is switched on for 90 s. During this time, 20 kJ of energy is dissipated to the surroundings by the kettle. Determine the total energy supplied, and the useful energy transferred to the water, ready to draw a Sankey diagram for this transfer.
Answer:
total energy supplied = power × time = 2000 × 90 = 180 000 J = 180 kJ
useful energy to heat the water = total − dissipated = 180 − 20
Live simulation: build a Sankey diagram
Set a total input energy and a useful-output percentage, and watch the Sankey diagram redraw to scale. This is the same idea used in Section 8 to picture efficiency.
Sankey diagram builder
useful energy: 120 J | wasted energy: 80 J
Fig. 6.2 An interactive Sankey diagram. The height of the input bar scales with total input energy; the useful and wasted arrows' thicknesses scale with their share of that energy, and always add up to the bar's height.
Check your understanding
19In a petrol-powered car, 35% of the chemical energy in the fuel is transferred to kinetic energy of the car. Heating the exhaust gases accounts for 15% of the energy from the fuel, and the remainder is dissipated in the engine, gearbox and wheels. For 1000 J of chemical energy supplied by the fuel, calculate the energy transferred to (a) kinetic energy of the car, (b) heating the exhaust gases, (c) the engine, gearbox and wheels.
(a) 35% of 1000 J = 0.35 × 1000
(b) 15% of 1000 J = 0.15 × 1000
(c) remainder = 1000 − (answer a) − (answer b), i.e. (100 − 35 − 15)% of 1000 J = 0.50 × 1000
20Using the simulator, set the total input energy to 300 J and the useful output to 25%. State the useful energy and wasted energy shown by the readout, and describe how the shape of the diagram has changed compared with a high useful-output percentage.
useful energy = 25% of 300 J = 75 J; wasted energy = 300 − 75 = 225 J. With a low useful-output percentage, the arrow continuing to the right (useful output) is narrow, while the arrow branching downward (wasted energy) is wide — the opposite of what is seen at a high useful-output percentage, where the rightward arrow is wide and the downward branch is narrow.
21Explain what determines the width of each arrow in a Sankey diagram, and why the combined width of all the arrows leaving a Sankey diagram must always equal the width of the arrow entering it.
The width of each arrow is proportional to the amount of energy (or power) that it represents. The combined width leaving must equal the width entering because of the principle of conservation of energy — none of the input energy can disappear, so the useful and wasted (dissipated) outputs must together account for all of it.
7. Power
Power is the rate at which work is done, or equivalently, the rate at which energy is transferred. If the same amount of useful work is done by two machines (or two people), the one that does it in less time is more powerful.
power, P = E ÷ t = W ÷ t (SI unit: watt, W; 1 W = 1 J s−1)
Combining P = W ÷ t with W = Fs gives a second useful form. Since s ÷ t is speed, v:
power needed to maintain a constant velocity, P = Fv
Key idea. P = Fv applies when a resultant forward force F is needed to maintain a constant speed v against a resistive force of the same magnitude — for example, a car's engine working against air resistance and friction.
Worked example 7.1
Calculate the average power output of a 65 kg climber moving up a height of 40 m in 3.0 minutes.
A car maintains a constant speed of 25 m s−1 against a constant resistive force of 2300 N. Calculate the power needed to maintain this speed.
Answer:
P = Fv = 2300 × 25
Check your understanding
22Explain why two people who lift identical boxes to the same height, but in different times, have done the same amount of work but not the same amount of power.
Work done depends only on the force and the distance moved (W = Fs), which are the same for both people since the boxes and heights are identical — so both people do the same amount of work. Power is the rate at which that work is done, P = W ÷ t. Since the two people take different times, the one who lifts the box in less time has a greater power output, even though the work done is the same.
23A 1600 kg car accelerates from rest to 25 m s−1 in 12.0 s. Calculate the average power needed to produce this change in kinetic energy.
P = ΔW ÷ Δt = kinetic energy gained ÷ time taken = (½ × 1600 × 25²) ÷ 12.0
24A small boat is powered by an outboard motor with a maximum output power of 40 kW. At its greatest speed of 14 m s−1, determine the magnitude of the forward force provided by the motor.
P = Fv
40 000 = F × 14
F = 40 000 ÷ 14
8. Efficiency
No real energy transfer is completely useful — some energy is always dissipated. A process that produces a greater useful output for a given input is described as more efficient.
efficiency, η = useful energy output ÷ total energy input = Eoutput ÷ Einput
Key idea. Dividing both energies by the same time interval gives an equivalent power form: η = Poutput ÷ Pinput. Because it is a ratio of two energies (or powers) with the same unit, efficiency has no unit, and is often expressed as a percentage. Because of the conservation of energy, efficiency can never reach or exceed 1 (100%).
Worked example 8.1
Electrical power is supplied to a small motor at a rate of 0.72 W. In 2.46 s, the motor raises a mass through 1.12 m, transferring 0.27 J of gravitational potential energy to it. Determine the efficiency of the motor.
Answer:
useful energy output = 0.27 J
total energy input = P × t = 0.72 × 2.46
η = Eoutput ÷ Einput = 0.27 ÷ (0.72 × 2.46)
Worth rememberingEfficiency isn't fixed. The efficiency of a machine or engine usually changes with how it is being used — a car, for example, typically has its highest efficiency at a particular cruising speed (often around 100 km h−1), and becomes less efficient if driven faster or slower.
Check your understanding
25A natural gas power station has a power output of 540 MW and an efficiency of 48%. Calculate the total input power required.
26Use the principle of conservation of energy to explain why no machine can ever be 100% efficient.
Conservation of energy means the total energy output of a machine (useful output plus dissipated energy) must exactly equal the total energy input — energy cannot appear from nowhere or vanish. In every real macroscopic process, some energy is inevitably dissipated (for example through friction, air resistance, or heating), so the useful output is always less than the total input, making η = Eoutput ÷ Einput always less than 1 (100%).
27Using your answer to Question 19 (energy transfers in a petrol car), calculate the efficiency of the car as a percentage.
The useful output of a car is the kinetic energy it gains, which was 35% of the input energy.
η = Eoutput ÷ Einput = 350 ÷ 1000 = 0.35 = 35%
9. Energy density
One advantage of liquid fossil fuels is how much energy can be released from a small volume of them. The energy density of a fuel is the amount of energy that can be transferred from each cubic metre of it.
Energy density is the amount of energy that can be transferred from each unit volume of a fuel. SI unit: J m−3 (commonly quoted in MJ m−3).
Fuel
Energy density / MJ m−3
Ethanol
24 000
Petrol (gasoline)
36 000
Diesel
38 000
Worked example 9.1
Estimate the energy released by completely burning 0.50 m³ of diesel.
Answer:
energy = energy density × volume = 38 000 × 0.50 (MJ)
Check your understanding
28Energy density is the amount of energy produced when 1 m³ of a liquid fuel is completely burned. The table above gives energy density values for ethanol, diesel and petrol. (a) Using the table, state which of the three fuels releases the most energy per m³ burned. (b) A car's fuel tank has a fixed volume. Using your answer to (a), explain which of the three fuels would let the car travel furthest on one full tank, assuming the engine's efficiency is the same whichever fuel is used. (c) Each of these three fuels also has a different mass density (in kg m−3). Explain whether you need to know these mass density values to answer part (b), and why the energy density given in the table is already the quantity that matters for a fixed-volume tank. (d) A friend argues that, because diesel has the highest energy density, a diesel engine must be more efficient than a petrol engine. Use the definition of efficiency from Section 8 to explain why this reasoning is incorrect.
(a) Diesel releases the most energy per m³ (38 000 MJ m−3), followed by petrol (36 000 MJ m−3), then ethanol (24 000 MJ m−3).
(b) Since the tank volume is the same for all three fuels, and diesel releases the most chemical energy per m³, a full tank of diesel would supply the most total chemical energy. With the same engine efficiency, more useful (kinetic) energy would be available, so the car could travel furthest on diesel.
(c) No extra information about mass density is needed. Energy density (MJ per m³) is already defined per unit volume, so it automatically takes into account how much mass of fuel fits into that volume (a denser fuel packing more mass, and therefore more chemical energy, into the same space) — comparing energy density values directly answers the question for a tank of fixed volume, without needing to consider mass density separately.
(d) Energy density describes how much chemical energy is stored in the fuel itself, before any of it is used. Efficiency, η = Eoutput ÷ Einput, describes how much of whatever chemical energy is supplied to an engine is actually converted to useful kinetic energy, rather than being dissipated. A fuel can store more energy per m³ without the engine that burns it being any better at converting that energy usefully — the two quantities describe different stages of the energy transfer, so a higher energy density does not imply a higher efficiency.
Glossary
Work, W
The transfer of energy that occurs when an object is moved by a force; W = Fs cosθ, measured in joules.
Joule, J
The SI unit of work and energy; 1 J is the work done when a 1 N force moves its point of application by 1 m.
Energy
The capacity to do work.
Principle of conservation of energy
The total energy of an isolated system is constant; energy cannot be created or destroyed, only transferred between forms.
Mechanical energy
The sum of an object's kinetic energy, gravitational potential energy and elastic potential energy.
Kinetic energy, Ek
The energy a mass has because of its motion, Ek = ½mv².
Gravitational potential energy, Ep
The energy an object has because of its position in a gravitational field; close to the Earth's surface, ΔEp = mgΔh.
Hooke's law
The restoring force of a spring or elastic material is directly proportional to its extension (or compression), FH = −kΔx, provided the elastic limit is not exceeded.
Elastic potential energy, EH
The energy stored in a deformed elastic material or spring, EH = ½kΔx².
Dissipated (degraded) energy
Energy that has spread out into the surroundings, usually as internal energy or sound, and can no longer be recovered to do useful work.
Sankey diagram
A flow diagram representing energy transfers, in which the width of each arrow is proportional to the amount of energy (or power) it represents.
Power, P
The rate of doing work, or the rate of energy transfer, P = E ÷ t = Fv, measured in watts.
Efficiency, η
The ratio of useful energy (or power) output to total energy (or power) input; always less than 1 (100%).
Energy density
The amount of energy that can be transferred from each unit volume of a fuel, measured in J m−3.