Educerie
Level

Educerie · IB Diploma · Physics

Theme A Space, time and motion · A.3 Work, energy and power

Level
SL and HL. Nothing here is HL only, so every section is examinable for both.
Themes (key concepts)
energy, forces. Work is what happens when a force moves something, and it is always a transfer of energy; conservation of energy then lets you skip the forces altogether and solve a motion problem from the start and end alone.
The question this unit answers
how can the transfer of energy by forces be used to predict what a system will do, and how quickly and how usefully it does it?
Where it is examined
Paper 1A multiple choice, on work at an angle, efficiency and energy changes; Paper 1B, where energy data from a ramp, spring or motor experiment must be processed; Paper 2, where an energy method is often the fastest route through a mechanics question (3–5 marks), and efficiency, power and fuel questions run to 2–4 marks each.

What you must be able to do

You must be able toLevelWhat it looks like in the exam
State and apply the principle of conservation of energySL, HL"Outline what is meant by conservation of energy" (1–2 marks)
Explain work done by a force as a transfer of energy, and draw and read Sankey diagramsSL, HL"Sketch a Sankey diagram for the motor" (2 marks)
Calculate work done by a constant force with W = Fs cos θ, and as the area under a force–displacement graphSL, HL"Calculate the work done by the tension" (2 marks)
Use the result that work done by the resultant force equals the change in energy of the systemSL, HLBraking-distance and stopping-force calculations (3 marks)
Use kinetic, gravitational potential and elastic potential energy, and their sum, mechanical energySL, HLEₖ = ½mv² = p² ÷ 2m, ΔEₚ = mgΔh, ½k(Δx)² in a 3–4 mark calculation
Use conservation of mechanical energy when there are no resistive forces, and account for non-conservative forces when there areSL, HL"Determine the average frictional force" (3 marks)
Calculate power from P = ΔW ÷ Δt and P = FvSL, HLA lift, a car or a cyclist at constant speed (2–3 marks)
Calculate efficiency from energies or from powersSL, HL"Determine the efficiency of the motor" (2 marks)
Use the energy density of fuel sourcesSL, HL"Calculate the mass of fuel needed" (2–3 marks)

Before you start

You need A.1 (the equations of motion and projectiles) and A.2 (resultant force, friction, Hooke's law and the free-body diagram). The unit of energy is the joule (J): the energy transferred when a force of 1 N moves its point of application 1 m in the direction of the force, so 1 J = 1 N m.


1The idea in one paragraph

Energy is a quantity that is never created or destroyed; it only moves from one store to another. The way a force moves energy is by doing work: push something along and the energy you transfer is the force times the distance moved in the direction of the force. When nothing resists the motion, a moving body simply trades kinetic energy for potential energy and back, and the total, its mechanical energy, stays fixed. When friction or drag acts, the mechanical energy falls by exactly the work they do, and that energy ends up as internal energy of the surroundings. Power is how fast the transfer happens, efficiency is what fraction of it was useful, and energy density says how much a kilogram of fuel can supply.

2Conservation of energy, and work as a transfer

Principle of conservation of energy: energy cannot be created or destroyed. The total energy of an isolated system is constant; energy can only be transferred from one store to another.

A system is the set of bodies you choose to look at. An isolated system exchanges no energy with anything outside it. Every energy calculation in this subtopic is the same sum: energy in the stores at the start, plus energy put in, equals energy in the stores at the end, plus energy that left.

Energy is moved from one store to another in a few ways, and in mechanics the way is work. When a force moves a body, the force does work on it, and the work done equals the energy transferred. A crane lifting a crate does work on the crate, and the crane's electrical energy becomes the crate's gravitational potential energy. Friction slowing a sliding book does (negative) work on it, and the book's kinetic energy becomes internal energy of the book and the table, which get very slightly warmer.

A Sankey diagram draws those transfers to scale. The input is an arrow whose width stands for the energy supplied; it splits into arrows for each output, the useful one carried straight on, the wasted ones bent away, and the widths always add back up to the input. Figure 1 is an electric motor lifting a load.

Figure 1 · A Sankey diagram for an electric motor lifting a load Figure 1 · A Sankey diagram for an electric motor lifting a load electrical energy in 500 J gravitational potential energy of the load 320 J internal energy of the motor and surroundings 150 J sound 30 J Arrow widths are to scale. Useful 320 J out of 500 J in: the efficiency is 0.64, or 64%.
Figure 1 · A Sankey diagram for an electric motor lifting a load

Of the 500 J supplied, 320 J becomes gravitational potential energy of the load. The rest is dissipated: 150 J heats the motor's windings and bearings and the air around them, and 30 J leaves as sound. Energy that has spread out into the surroundings as internal energy is still there; it is just no longer useful to anyone. Check any Sankey diagram you draw by adding: 320 + 150 + 30 = 500.

3Work done by a constant force

When a constant force F moves a body through a displacement s, only the part of the force along the displacement does work. Figure 2 shows a suitcase pulled by its handle.

Figure 2 · Only the part of the force along the displacement does work Figure 2 · Only the part of the force along the displacement does work case F F cos θ F sin θ θ displacement s W = Fs cos θ. The upward part, F sin θ, is at right angles to the motion and does no work.
Figure 2 · Only the part of the force along the displacement does work

The force splits into F cos θ along the ground and F sin θ upwards. The upward part does not move the case upwards, so it does no work. Only F cos θ, acting through s, transfers energy:

W = Fs cos θ

where θ is the angle between the force and the displacement. Three cases are worth knowing by heart.

  • θ = 0°, force along the motion: cos θ = 1 and W = Fs, the largest possible.
  • θ = 90°, force at right angles to the motion: cos θ = 0 and no work is done. The normal force on a car driving along a level road does no work, and neither does the centripetal force on a body moving in a circle, which is why a body in uniform circular motion keeps the same kinetic energy.
  • θ = 180°, force against the motion: cos θ = −1 and the work done is negative. Friction and drag do negative work: they take energy out of the moving body.

Worked example 1. A traveller pulls a suitcase 40 m along a level platform with a force of 60 N along the handle, at 35° to the horizontal. How much work does she do on the case?

W = Fs cos θ = 60 × 40 × cos 35°
W = 2.0 × 103 J

Work is a scalar. It has a sign (positive when the force helps the motion, negative when it opposes it) but no direction.

Work as an area. When the force changes as the body moves, W = Fs cos θ cannot be used directly. Plot the force along the displacement against the displacement; the work done is the area under the graph, as Figure 3 shows.

Figure 3 · Work done is the area under a force–displacement graph Figure 3 · Work done is the area under a force–displacement graph (a) Constant force Force F / N Displacement s / m F W = Fs s (b) Stretching a spring Force F / N Extension x / m F = kx kx x area = ½ × x × kx = ½kx² (a) A constant force: a rectangle, Fs. (b) Stretching a spring: a triangle, ½kx², stored as elastic energy.
Figure 3 · Work done is the area under a force–displacement graph

For a constant force, panel (a), the area is a rectangle, Fs. For a spring, panel (b), the force grows in proportion to the extension, F = kx, so the area is a triangle, ½ × x × kx = ½kx². That work is stored in the stretched spring as elastic potential energy. Stretching a spring of spring constant 200 N m⁻¹ by 0.15 m stores ½ × 200 × 0.15² = 2.3 J (2.25 J).

4The three stores of mechanical energy

Mechanical energy is the sum of three stores: kinetic energy, gravitational potential energy and elastic potential energy. The data booklet gives each.

Kinetic energy is the energy of translational motion:

Ek = ½mv2 = p2 ÷ (2m)

The second form uses momentum p = mv, and it is the one to reach for when a question gives you momentum, as in explosions in A.2. Note the square: double the speed and the kinetic energy is four times as large.

Gravitational potential energy changes when a body moves up or down near the Earth's surface:

ΔEp = mgΔh

Only the change in height matters, so you can put the zero of height wherever is convenient, usually the lowest point. The equation assumes g is constant, which is true only close to the surface; far from the Earth, D.1 replaces it.

Elastic potential energy is stored in a stretched or compressed body that obeys Hooke's law:

EH = ½k(Δx)2

where Δx is the extension or compression and k the spring constant, exactly the area in Figure 3(b).

5Work done by the resultant force

The work done by the resultant force on a body equals the change in its energy. For a body on a level surface, where only its kinetic energy changes, this is the most useful tool in mechanics after F = ma, and it follows straight from it: the resultant force F gives an acceleration a = F ÷ m, and over a displacement s, v² = u² + 2as. Multiply through by ½m and

Fs = ½mv2 − ½mu2

The work done by the resultant force is the change in kinetic energy.

Worked example 2 · braking distance. A 1500 kg car travelling at 20 m s⁻¹ brakes with a constant resultant force of 6000 N. How far does it travel before stopping? How far if it was going twice as fast?

work done by the brakes = kinetic energy removed
6000 × s = ½ × 1500 × 202 = 3.0 × 105 J
s = 3.0 × 105 ÷ 6000 = 50 m
at 40 m s⁻¹: 6000 × s = ½ × 1500 × 402 = 1.2 × 106 J → s = 200 m

Twice the speed, four times the kinetic energy, four times the braking distance with the same brakes. That square is the physics behind every speed limit near a school.

6Conservation of mechanical energy, and when it fails

If no friction, drag or other resistive force acts, the total mechanical energy of a system is conserved. Energy moves between kinetic, gravitational and elastic stores, and the work done by gravity or by a spring is exactly the amount transferred from one mechanical store to another. Figure 4(a) shows a ball falling: the gravitational potential energy it loses is the kinetic energy it gains, and the sum is a flat line.

Figure 4 · Energy of a falling ball against its height Figure 4 · Energy of a falling ball against its height (a) No air resistance Energy / J Height h / m Ep Ek total mechanical energy start (b) With air resistance Energy / J Height h / m Ep Ek total without air resistance lost to the air start (a) No air resistance: Ep + Ek stays constant. (b) Air resistance: the total falls as the ball falls, by the work done against the air, which becomes internal energy of the ball and the air.
Figure 4 · Energy of a falling ball against its height

This is how energy solves kinematics problems. Take Worked example 2 of A.1 again: a ball is thrown at 12.0 m s⁻¹ from a roof 20.0 m high. Mechanical energy is conserved, so

½mv2 = ½mu2 + mgΔhenergy at the ground = energy at the roof
v2 = 12.02 + 2 × 9.8 × 20.0 = 536
v = 23.2 m s⁻¹

The mass cancels, and so does the direction of the throw. Thrown up, down or sideways at 12.0 m s⁻¹, the ball lands at 23.2 m s⁻¹, as the A.1 calculation found the long way. When a question asks only for a speed, and gives heights, an energy method is shorter and safer than the equations of motion. It will not give you a time or a direction; for those you still need A.1.

Worked example 3 · a spring launcher. A toy launcher has a spring of spring constant 400 N m⁻¹, compressed by 5.0 cm. It fires a 12 g dart horizontally. Find the launch speed, assuming all the elastic energy becomes kinetic energy.

EH = ½k(Δx)2 = ½ × 400 × 0.0502 = 0.50 J
½ × 0.012 × v2 = 0.50
v = √(2 × 0.50 ÷ 0.012) = 9.1 m s⁻¹

The same idea gave you the speed at the bottom of the vertical circle in A.2: 4.0 m s⁻¹ at the top of a circle of radius 0.80 m becomes √(4.0² + 2 × 9.8 × 1.6) = 6.9 m s⁻¹ after falling 1.6 m.

When resistive forces act. Friction and drag are non-conservative forces: the work they do does not come back when the motion reverses. The energy leaves the mechanical stores and becomes internal energy of the body and its surroundings. So the rule becomes a bookkeeping rule.

Change in total mechanical energy = work done on the system by non-conservative forces (negative for friction and drag).

Figure 4(b) is the falling ball with air resistance. The total falls as the ball falls, slowly at first, while the ball is slow and the drag is small, and faster lower down.

Worked example 4 · a skateboarder. A 55 kg skateboarder starts from rest at the top of a ramp 4.0 m high. The ramp is 12 m long. Without friction she would reach the bottom at 8.9 m s⁻¹; she actually reaches it at 7.5 m s⁻¹. Find the average resistive force on her.

Ep at the top = mgh = 55 × 9.8 × 4.0 = 2156 J
without friction: ½ × 55 × v2 = 2156 → v = 8.9 m s⁻¹
Ek at the bottom (real) = ½ × 55 × 7.52 = 1547 J
work done against resistance = 2156 − 1547 = 609 J
average resistive force = 609 ÷ 12 = 51 NW = Fs, force along the ramp

Figure 5 shows where the energy went.

Figure 5 · A skateboarder on a 4.0 m ramp: where the energy goes Figure 5 · A skateboarder on a 4.0 m ramp: where the energy goes Ep 2156 J top of ramp at rest Ek 2156 J bottom, no friction v = 8.9 m s⁻¹ Ek 1547 J internal 609 J bottom, real ramp v = 7.5 m s⁻¹ Friction and air resistance do 609 J of work: the mechanical energy falls by exactly that.
Figure 5 · A skateboarder on a 4.0 m ramp: where the energy goes

The 609 J is not destroyed. The wheels, bearings, ramp and air are warmer by that much in total, which is too spread out to notice or to use.

7Power

Power is the rate of doing work, or the rate of transferring energy:

P = ΔW ÷ Δt = Fv

It is measured in watts, W, where 1 W = 1 J s⁻¹. (Careful: W in italics is work, W upright is the watt.) The second form follows from the first: work Fs done in time t gives power Fs ÷ t, and s ÷ t is the speed. Use P = Fv when a force keeps something moving at constant velocity.

Worked example 5 · a lift. A motor raises an 800 kg lift at a steady 2.0 m s⁻¹. At constant velocity the resultant force is zero, so the cable tension equals the weight.

F = mg = 800 × 9.8 = 7840 N
P = Fv = 7840 × 2.0 = 1.6 × 104 W

The same logic applies to a car on a level road. At a steady 30 m s⁻¹, the driving force equals the total resistive force, say 600 N, so the engine delivers P = 600 × 30 = 18 kW to the wheels. The power goes into pushing air aside and flexing tyres, not into speeding the car up. The steady speed needs power even though the kinetic energy is not changing.

8Efficiency

Efficiency is the fraction of the energy supplied that ends up where you wanted it:

η = useful work out ÷ total work in = useful power out ÷ total power in

It is a ratio of like quantities, so it has no unit, and it is written as a decimal (0.64) or a percentage (64%). It is never greater than 1: a device cannot put out more energy than it takes in. An answer above 1 means the ratio is upside down.

The motor in Figure 1 has η = 320 ÷ 500 = 0.64. If the lift motor of Worked example 5 draws 25 kW of electrical power from the supply, its efficiency is 15.7 ÷ 25 = 0.63, and 9.3 kW is dissipated in the motor, the cables and the pulleys. Using powers and using energies give the same answer, because both measure the same transfer over the same time.

9Energy density of fuel sources

The energy density of a fuel is the energy it releases per kilogram, in J kg⁻¹ (in practice MJ kg⁻¹, where 1 MJ = 10⁶ J). It decides how heavy a vehicle's fuel or battery must be for a given journey, and it is one reason the world's energy supply looks the way it does. Figure 6 compares typical, rounded values.

Figure 6 · Energy density of some energy sources (rounded typical values) Figure 6 · Energy density of some energy sources (rounded typical values) lithium-ion battery 0.6 wood (dry) 16 coal 30 ethanol 30 petrol 46 natural gas (methane) 55 hydrogen 140 0 50 100 150 Energy density / MJ kg⁻¹ uranium-235, fully fissioned: about 80 000 000 MJ kg⁻¹ Energy released per kilogram. Nuclear fuel is off this scale by a factor of about a million.
Figure 6 · Energy density of some energy sources (rounded typical values)

Some tables quote energy density per cubic metre instead, in J m⁻³, which matters for a fuel tank's size rather than its mass. The unit tells you which is meant; read it before you divide.

Worked example 6 · fuel for a journey. A car needs 18 kW at the wheels for one hour. Its petrol engine is 25% efficient and petrol has an energy density of 46 MJ kg⁻¹. What mass of petrol does it burn?

useful energy = P × t = 18 × 103 × 3600 = 6.48 × 107 J
energy from the fuel = 6.48 × 107 ÷ 0.25 = 2.59 × 108 Jdivide by η: the input is bigger
mass = 2.59 × 108 ÷ 46 × 106 = 5.6 kg

The step that goes wrong is the efficiency. The fuel must supply more than the useful energy, so you divide by 0.25; multiplying would make the engine a source of energy.

Two things in Figure 6 are worth noticing. Fossil fuels pack roughly 30 to 55 MJ into each kilogram, while a lithium-ion battery holds around 0.6 MJ, less than a seventieth of petrol, which is why electric cars carry batteries weighing hundreds of kilograms, a disadvantage partly offset by the fact that electric motors are far more efficient than petrol engines. And nuclear fuel, with an energy density about a million times that of any chemical fuel, sits off the chart altogether. E.4 explains where that energy comes from.

10Where marks are lost

Forgetting cos θ, or using the wrong angle. θ is the angle between the force and the displacement, not between the force and the vertical. A force at right angles to the motion does no work at all.

Squaring the wrong thing, or not squaring. Eₖ = ½mv²: halve after squaring. ½ × 1500 × 20² is 300 000 J; squaring the whole product, or forgetting to square, gives nonsense.

Using mgΔh with a height that is not vertical. Δh is the change in vertical height. On a 12 m ramp that is 4.0 m high, the potential energy change uses 4.0 m.

Saying energy is "lost" and leaving it there. Energy is conserved. Say where it went: internal energy of the surroundings (they warm up), or sound.

Assuming mechanical energy is conserved when friction is mentioned. If the question gives a resistive force, a real speed less than the ideal one, or a rough surface, the mechanical energy falls by the work done against resistance.

Inverting efficiency. Useful over total. If your efficiency is above 1, turn it over.

Multiplying by the efficiency when finding the input. Input = useful output ÷ η.

Thinking a body at constant speed needs no power. If resistive forces act, work must be done against them all the time, P = Fv, even though the kinetic energy is not changing.

11Draw it right

Two diagrams in this subtopic are marked on their construction: the Sankey diagram and the energy graph.

  1. A Sankey diagram has one input arrow on the left, and output arrows whose widths are to scale. State the scale or label each arrow with its value.
  2. The useful output carries straight on; the wasted outputs bend away (conventionally downwards).
  3. The output widths add up to the input width. A marker checks this first.
  4. Every arrow is labelled with the form or store of energy (electrical, kinetic, gravitational potential, internal energy of the surroundings, sound), not just "waste".
  5. On an energy–height or energy–time graph, draw the total as well as the stores. With no resistance it is a horizontal line; with resistance it slopes, and the gap is the work done against resistance.
  6. On a force–displacement graph, shade the area you are using as work and label it with its value and unit.

12Try it

Marks in brackets. Take g = 9.8 m s⁻². Answers and marker's notes are at the end.

Q1. A small mass on a string moves in a horizontal circle of radius r at constant speed. The tension in the string is T. What is the work done by the tension during one complete revolution? 1 mark

A. zero · B. 2πrT · C. πrT · D. rT

Q2. An 8.0 W LED lamp emits 2.4 W as visible light. The rest is transferred as thermal energy.

(a) Calculate the efficiency of the lamp. 1 mark

(b) Sketch a labelled Sankey diagram for the lamp. 2 marks

(c) Calculate the energy dissipated as thermal energy in one hour. 1 mark

Q3. (data-based, Paper 1B style) A student releases a trolley from rest at different heights h on a ramp and measures its speed v at the bottom with a light gate.

h / m0.100.200.300.400.50
v / m s⁻¹1.331.832.282.622.94
v² / m² s⁻²1.773.355.206.868.64

(a) Show that, if no energy is dissipated, a graph of v² against h is a straight line through the origin with gradient 2g. 2 marks

(b) Using the first and last rows, determine the gradient of the graph. 2 marks

(c) Deduce the fraction of the trolley's gravitational potential energy that becomes kinetic energy, and suggest why it is less than 1. 2 marks

Q4. A cyclist and her bicycle have a total mass of 75 kg. On a level road she rides at a steady 8.0 m s⁻¹ against a total resistive force of 25 N.

(a) Calculate her power output. 2 marks

(b) She then climbs a hill at a steady 5.0 m s⁻¹. The road rises 1.0 m for every 20 m travelled along it. Assuming the resistive force is still 25 N, determine her power output. 3 marks

Q5. A 0.50 kg ball is dropped from rest from a height of 12 m and reaches the ground at 14 m s⁻¹.

(a) Determine the energy transferred to the surroundings by air resistance. 2 marks

(b) Calculate the average air resistance on the ball. 1 mark

Q6. A car journey needs 90 MJ of useful work at the wheels.

(a) A petrol car is 25% efficient. Petrol has an energy density of 46 MJ kg⁻¹. Calculate the mass of petrol used. 2 marks

(b) An electric car is 90% efficient. Take the energy density of its battery as 0.60 MJ kg⁻¹. Calculate the mass of battery needed to store the energy for the journey. 2 marks

(c) Comment on your answers. 1 mark

13In one breath

Energy is never created or destroyed, only moved between stores, and a force moves it by doing work: W = Fs cos θ, with θ between the force and the displacement, so a force at right angles to the motion, like a centripetal force, does no work. When the force varies, the work is the area under the force–displacement graph, which for a spring is ½kx². Sankey diagrams draw the transfers to scale, useful straight on, wasted bent away, outputs adding to the input. Mechanical energy is kinetic ½mv² = p² ÷ 2m, plus gravitational mgΔh near the Earth, plus elastic ½k(Δx)²; the work done by the resultant force equals the change in energy. With no resistive forces the mechanical energy is conserved, which gives speeds from heights in one line whatever the direction of motion; with friction or drag it falls by exactly the work they do, and that energy becomes internal energy of the surroundings. Power is the rate of transfer, ΔW ÷ Δt = Fv, in watts; efficiency is useful over total, for energies or powers, never above 1; and energy density, in MJ kg⁻¹, is the energy a kilogram of fuel releases, around 46 for petrol, under 1 for a battery, and millions of times more for nuclear fuel.


Answers

Q1. A. The tension is always along the radius, at 90° to the velocity, so cos θ = 0 and it does no work at any instant; the speed, and so the kinetic energy, does not change. A only. B multiplies force by distance and ignores the angle, which is exactly the trap.

Q2. (a) η = 2.4 ÷ 8.0 = 0.30 (30%). A1, no unit.

(b) One input arrow labelled "electrical 8.0 W" splitting into a straight-on arrow "light 2.4 W" and a bent-away arrow "thermal energy 5.6 W", with widths in the ratio 8.0 : 2.4 : 5.6. 1 for correct labels and values including 5.6 W, 1 for widths roughly to scale with the outputs adding to the input.

(c) 5.6 × 3600 = 2.0 × 10⁴ J (20 160 J). A1 with unit. Using 8.0 W gives the total energy supplied, not the dissipated energy, and scores 0.

Q3. (a) If no energy is dissipated, the gravitational potential energy lost equals the kinetic energy gained: mgh = ½mv², so v² = 2gh. This is y = mx with y = v² and x = h, a straight line through the origin with gradient 2g. M1 for mgh = ½mv², R1 for the comparison with y = mx. AG, so both steps must be shown.

(b) Gradient = (8.64 − 1.77) ÷ (0.50 − 0.10) = 6.87 ÷ 0.40 = 17.2 m s⁻². M1 for the gradient calculation, A1 for 17 to 17.2 with the unit m s⁻² (m² s⁻² per m).

(c) The fraction is 17.2 ÷ (2 × 9.8) = 17.2 ÷ 19.6 = 0.88. Some of the energy is transferred to internal energy by friction in the wheels and against the ramp, and by air resistance, so less than all of the potential energy becomes kinetic energy. A1 for 0.88 (0.87 to 0.88), R1 for a resistive force named. "Energy is lost" with no mechanism scores 0 for the second mark.

Q4. (a) At constant speed the driving force equals the resistive force, 25 N. P = Fv = 25 × 8.0 = 200 W. M1 for driving force = 25 N at steady speed, A1 for 200 W.

(b) The component of weight along the slope is mg × (1.0 ÷ 20) = 75 × 9.8 × 0.050 = 36.75 N. Driving force = 25 + 36.75 = 61.75 N. P = 61.75 × 5.0 = 310 W (309 W). M1 for the weight component along the slope, M1 for adding it to the resistive force, A1 for 310 W. Alternatively, 1 for the rate of gaining potential energy, mg × 0.25 m s⁻¹ = 184 W, 1 for adding 25 × 5.0 = 125 W, 1 for 309 W.

Q5. (a) Potential energy lost = mgh = 0.50 × 9.8 × 12 = 58.8 J. Kinetic energy gained = ½ × 0.50 × 14² = 49.0 J. Energy transferred by air resistance = 58.8 − 49.0 = 9.8 J. M1 for both energies, A1 for 9.8 J.

(b) Work done = force × distance, so average force = 9.8 ÷ 12 = 0.82 N. A1. Allow error carried forward from (a).

Q6. (a) Energy from fuel = 90 ÷ 0.25 = 360 MJ. Mass = 360 ÷ 46 = 7.8 kg. M1 for dividing by the efficiency, A1 for 7.8 kg. Multiplying by 0.25 gives 0.49 kg and scores 0.

(b) Energy from battery = 90 ÷ 0.90 = 100 MJ. Mass = 100 ÷ 0.60 = 170 kg (167 kg). M1, A1 as in (a).

(c) The battery is about twenty times heavier than the petrol, even though the electric car is far more efficient, because the battery's energy density is so much lower. a comparison that links the mass difference to energy density.


Educerie · written from the published IB Diploma Programme Physics guide, first assessment 2025, section A.3 Work, energy and power. Original text, examples and questions. Diagrams drawn by Educerie. Last reviewed 25 September 2026.

Mocks: in the future, hold tight!