IB DP Physics: Work, energy, power and efficiency (DP Physics)

This guide covers sub-topic A.3, Work, energy and power, from the IB Physics guide (first assessment 2025). The content is the same for SL and HL: A.3 has no extra HL-only material in the guide. It teaches each understanding from scratch, with worked calculations. For quick recall use the revision notes, then test yourself with the practice questions. Find your gaps first with the free 10-minute diagnostics.

Related pages: the course hub, the printable checklist, kinematics, forces and momentum and the IB DP Physics syllabus guide.

What A.3 covers

Part	What you must be able to do	Tier
A.3	Apply the principle of conservation of energy	SL and HL
A.3	Explain that work done by a force is a transfer of energy	SL and HL
A.3	Represent energy transfers on a Sankey diagram	SL and HL
A.3	Use W = Fs cos θ for a constant force	SL and HL
A.3	Link the work done by the resultant force to the change in the system’s energy	SL and HL
A.3	Define mechanical energy as Ek + Eg + Ee, and know when it is conserved	SL and HL
A.3	Use Ek = ½mv² = p²/2m, ΔEp = mgΔh and Ee = ½k(Δx)²	SL and HL
A.3	Use P = ΔW/Δt = Fv	SL and HL
A.3	Use efficiency η = Eoutput/Einput = Poutput/Pinput	SL and HL
A.3	Use the energy density of fuel sources	SL and HL
A.3 guidance	Interpret a change in mechanical energy as work done by non-conservative forces	SL and HL

Conservation of energy and work as a transfer

Energy is never created or destroyed. It moves from one store to another or between objects, and the total stays the same in an isolated system. Energy is measured in joules (J).

Work is the way a force moves energy. When a force acts on an object that moves, the force transfers energy to or from it. Work is therefore measured in joules, and “work done” and “energy transferred” describe the same amount. Positive work adds energy to the object. Negative work takes energy away, for example friction on a sliding block.

A Sankey diagram

A Sankey diagram shows where input energy goes. The width of each arrow is drawn in proportion to the amount of energy it carries. One arrow enters. The useful output continues straight on, and wasted energy branches away. Nothing is lost from the picture: the output arrows together are exactly as wide as the input.

Example layout, as a table of arrow widths for a device with 500 J input:

Arrow	Energy (J)	Share of input
Input (electrical)	500	100%
Useful (kinetic)	150	30%
Wasted (thermal)	310	62%
Wasted (sound)	40	8%

The useful arrow is 30% of the input width, so the efficiency is 0.30.

Work done by a constant force

Only the part of the force along the line of the displacement does work:

W = Fs cos θ

F is the force, s the displacement and θ the angle between them. At θ = 0° the force is fully along the motion and W = Fs. At θ = 90° the force does no work because cos 90° = 0. This is why a centripetal force does no work on an object moving at constant speed in a circle, and why a hand carrying a bag at constant height does no work on the bag. If θ is greater than 90°, cos θ is negative and the force removes energy.

Worked example 1. A child pulls a sledge along level ground with a rope. The rope tension is 120 N and the rope makes 35° with the ground. The sledge moves 18 m. Find the work done by the tension.

W = Fs cos θ = 120 × 18 × cos 35° = 1769 J.

So the work done is 1.77 × 10³ J to 3 significant figures. Only 120 cos 35° = 98.3 N of the tension acts along the ground.

Work and the resultant force

The net work done by all the forces on a system gives the change in the energy of that system. For a single object whose only change is in speed, this gives:

W(resultant) = ΔEk

If the resultant force is zero, the net work is zero and the kinetic energy does not change. A single force can do work while the resultant does none, which is what happens when you lift a box at constant speed: your force does positive work and gravity does equal negative work.

Worked example 2. A 3.2 kg trolley starts from rest. A constant resultant force of 6.0 N acts along the track for 4.5 m. Find its final speed.

W = Fs = 6.0 × 4.5 = 27 J, so ΔEk = 27 J.

27 = ½ × 3.2 × v², so v² = 2 × 27 / 3.2 = 16.9 and v = 4.1 m s⁻¹.

Mechanical energy and its conservation

Mechanical energy is the sum of three terms:

E(mech) = Ek + Eg + Ee

where Eg is gravitational potential energy and Ee is elastic potential energy. In the absence of frictional or other resistive forces, the total mechanical energy of the system is conserved. Energy then simply switches between the three forms, and the work done is the amount transformed between them.

The three expressions

Quantity	Expression	Notes
Kinetic energy (translational)	Ek = ½mv² = p²/2m	p = mv, so ½mv² = p²/2m
Gravitational potential energy change	ΔEp = mgΔh	valid close to the Earth’s surface, where g is constant
Elastic potential energy	Ee = ½k(Δx)²	k is the spring constant, Δx the extension or compression

The relation Ek = p²/2m is useful when momentum is given or conserved. Check it: m = 0.80 kg moving at 2.62 m s⁻¹ has p = 2.10 kg m s⁻¹, and p²/2m = 4.40/1.6 = 2.75 J, the same as ½mv².

The mgΔh expression assumes the field strength does not change. Over heights that are large compared with the Earth’s radius, g falls and the expression no longer holds.

Worked example 3. A pendulum bob of mass 0.80 kg is held 0.35 m above its lowest point and released from rest. Ignore resistive forces. Find the speed at the lowest point. Take g = 9.81 m s⁻².

Loss of gravitational potential energy = mgΔh = 0.80 × 9.81 × 0.35 = 2.75 J.

All of it becomes kinetic energy: ½ × 0.80 × v² = 2.75, so v = 2.62 m s⁻¹.

The mass cancels: v = √(2gΔh). A bob of any mass released from the same height has the same speed.

Worked example 4. A spring of spring constant 320 N m⁻¹ is compressed by 0.060 m. It launches a 0.050 kg ball. The spring has no mass and there are no resistive forces.

Ee = ½ × 320 × 0.060² = 0.576 J.

(a) Horizontal launch: ½ × 0.050 × v² = 0.576, so v = 4.8 m s⁻¹.

(b) Vertical launch: the energy becomes gravitational potential energy, so 0.050 × 9.81 × Δh = 0.576 and Δh = 1.2 m.

Non-conservative forces

Friction and air resistance are non-conservative: they transfer mechanical energy to thermal energy. The guidance for A.3 says that any change in a system’s total mechanical energy is to be explained by the work that non-conservative forces do on it:

ΔE(mech) = W(non-conservative)

For friction, this work is negative, so mechanical energy falls. The lost energy is not destroyed: it appears as thermal energy in the surfaces and surroundings.

Worked example 5. A 2.0 kg block slides 3.0 m down a rough slope that drops 1.5 m in height. It starts from rest and reaches 4.2 m s⁻¹. Find the average frictional force.

Loss of gravitational potential energy = 2.0 × 9.81 × 1.5 = 29.4 J.

Gain in kinetic energy = ½ × 2.0 × 4.2² = 17.6 J.

Mechanical energy lost = 29.43 − 17.64 = 11.8 J, which is the work done against friction.

F × 3.0 = 11.79, so F = 3.9 N.

Power

Power is the rate of doing work, or the rate at which energy is transferred:

P = ΔW/Δt

Power is measured in watts (W), where 1 W is 1 J every second. Suppose a constant force F acts in the direction of travel of an object moving at steady speed v. W = Fs and s = vΔt, so:

P = Fv

Worked example 6. A car travels at a steady 25 m s⁻¹. The total resistive force on it is 640 N. With no acceleration, the driving force must balance the resistive force, so P = Fv = 640 × 25 = 16 000 W = 16 kW.

Efficiency

Efficiency compares useful output with input:

η = E(output)/E(input) = P(output)/P(input)

Here “output” means useful output. η has no unit. It is a fraction, or a percentage if multiplied by 100. It is always less than 1 for a real device, because part of the input always ends up heating the surroundings.

Worked example 7. A motor lifts a 240 kg load through 15 m in 20 s at constant speed. Its electrical input power is 2.4 kW.

Useful output power = mgΔh/Δt = 240 × 9.81 × 15 / 20 = 1766 W.

η = 1766 / 2400 = 0.736, or 74%.

Energy density of fuels

Energy density is the energy that can be released from a fuel per unit volume of the fuel. Where a table gives energy per unit mass, that is the specific energy. Use the unit given in the question, and combine it with efficiency to find how much fuel an engine uses.

Worked example 8. A generator burns a fuel of energy density 36 MJ per litre. Its overall efficiency is 28%. It supplies 1.5 kW of electrical power for 3.0 hours. Find the volume of fuel used.

Electrical energy = 1500 × 3.0 × 3600 = 1.62 × 10⁷ J.

Fuel energy needed = 1.62 × 10⁷ / 0.28 = 5.79 × 10⁷ J.

Volume = 5.79 × 10⁷ / 3.6 × 10⁷ = 1.6 litres.

Common errors

• Using W = Fs when the force is at an angle. Find the angle between the force and the displacement, not between the force and the horizontal unless they coincide.
• Writing the speed from mgΔh with the wrong sign, so that a body speeds up when it rises.
• Using ΔEp = mgΔh for a large change in height, where g is not constant.
• Putting the extension in centimetres into ½k(Δx)². Convert to metres.
• Forgetting that friction has to be included in the energy balance when the problem says the slope is rough.
• Dividing useful output by output instead of input when finding efficiency.
• Forgetting the efficiency step in energy density problems, so the fuel volume comes out too small.

Official syllabus

Source: Physics guide, first assessment 2025, International Baccalaureate Organization (Diploma Programme), sub-topic A.3 Work, energy and power.