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

These are condensed revision notes for sub-topic A.3, Work, energy and power, in the IB Physics guide (first assessment 2025). SL and HL study the same content here, because A.3 has no extra HL-only material in the guide. For full explanations and longer worked examples, read the study guide. When you are ready, try the practice questions, or take a free 10-minute diagnostic to find weak spots.

Also useful: the course hub, the printable checklist, forces and momentum and kinematics.

Core ideas

• Energy is conserved. It is transferred between stores and objects, and the total in an isolated system stays constant.
• Work done by a force is a transfer of energy. Positive work adds energy to the object, negative work removes it.
• A Sankey diagram draws energy flow as arrows whose widths are proportional to the energy. Outputs add up to the input.
• The net work of all forces acting on a system gives the change in that system’s energy.
• Mechanical energy is the sum of kinetic, gravitational potential and elastic potential energy.
• With no frictional or other resistive forces, mechanical energy is conserved.
• With non-conservative forces acting, the change in mechanical energy equals the work done by those forces.

Formulas

Quantity	Formula	Unit	Conditions
Work	W = Fs cos θ	J	constant force; θ between force and displacement
Work by resultant force	W = ΔEk	J	for a body changing only its speed
Kinetic energy	Ek = ½mv² = p²/2m	J	translational motion; p = mv
Gravitational potential energy	ΔEp = mgΔh	J	close to the Earth’s surface
Elastic potential energy	Ee = ½k(Δx)²	J	k = spring constant, Δx = extension or compression
Mechanical energy	Ek + Eg + Ee	J	conserved if no resistive forces
Power	P = ΔW/Δt = Fv	W	Fv for constant force along constant velocity
Efficiency	η = Eoutput/Einput = Poutput/Pinput	none	useful output over total input
Energy density	energy released per unit volume of fuel	e.g. MJ per litre	use the units given

Methods in steps

Work by an angled force

1. Mark the angle between the force and the displacement.
2. Calculate Fs cos θ.
3. Give the sign: negative if θ is greater than 90°.

Conservation of mechanical energy

1. Choose a start point and an end point.
2. Write Ek + Eg + Ee at each point.
3. Set them equal if there are no resistive forces.
4. Solve. Check that the mass cancels if only Ek and Eg are involved.

Energy with friction or air resistance

1. Find total mechanical energy at the start and the end.
2. The difference is the work done against the resistive force.
3. Divide by the distance travelled to get an average force.

Fuel problems

1. Work out the useful output energy: power × time, or force × distance.
2. Divide by efficiency to get the input energy.
3. Divide by the energy density to get the fuel volume.

Small worked reminders

Angled force. A 20 N force acts at 30° to a 4.0 m displacement. W = 20 × 4.0 × cos 30° = 69.3 J.

Spring. k = 25 N m⁻¹ compressed 0.040 m. Ee = ½ × 25 × 0.040² = 0.020 J.

Lifting. A 12 kg object is raised 2.5 m. Work done = mgΔh = 12 × 9.81 × 2.5 = 294 J. If the lift takes 5.0 s, P = 294 / 5.0 = 59 W.

Friction on a slope. A 0.60 kg block slides down a rough slope that drops 0.90 m and reaches 4.0 m s⁻¹ from rest. Loss of Eg = 0.60 × 9.81 × 0.90 = 5.30 J. Gain of Ek = ½ × 0.60 × 4.0² = 4.8 J. The difference, 0.50 J, is the work done against friction, which becomes thermal energy.

Reading a Sankey diagram. The input arrow is 80 J. The useful arrow is 52 J, one waste arrow is 20 J and the other is 8 J, so 52 + 20 + 8 = 80 J. The efficiency is 52 / 80 = 0.65, or 65%.

Fuel volume. A 1.0 kW electrical load runs for 1.0 hour from a generator that is 25% efficient. The fuel has an energy density of 32 MJ per litre. Output energy = 1000 × 3600 = 3.6 × 10⁶ J. Input energy = 3.6 × 10⁶ / 0.25 = 1.44 × 10⁷ J. Volume = 1.44 × 10⁷ / 3.2 × 10⁷ = 0.45 litres.

Units to watch

Given	Convert to	How
cm or mm	m	divide by 100 or 1000
g	kg	divide by 1000
kJ, MJ	J	multiply by 10³ or 10⁶
kW	W	multiply by 1000
hours	s	multiply by 3600
MJ per litre	J per litre	multiply by 10⁶

Always convert before substituting. A power in kW combined with a time in hours does not give energy in joules.

Must-know distinctions

• Work and power. Work is energy transferred, in joules. Power is the rate, in watts.
• Conservative and non-conservative forces. Gravity and the spring force can be stored as potential energy. Friction and air resistance transfer energy to thermal energy, so mechanical energy is not conserved.
• Energy conserved and mechanical energy conserved. Total energy is always conserved. Mechanical energy is conserved only without resistive forces.
• Output and input. Efficiency uses useful output over total input, never the reverse.
• Energy density and efficiency. Energy density says how much energy the fuel can release per unit volume. Efficiency says how much of that becomes useful output.
• Sign of work. A force along the motion does positive work. A force against it does negative work. A force at 90° does none.

Quick self-test

1. A 30 N force acts at 60° to a 3.0 m displacement. How much work does it do?
2. A force acts at exactly 90° to the displacement of an object. How much work does it do?
3. Find the kinetic energy of a 0.40 kg object moving at 6.0 m s⁻¹.
4. A spring (k = 400 N m⁻¹) is compressed by 0.030 m and then by 0.060 m. Find Ee both times. What is the ratio?
5. A 3.0 kg object has momentum 6.0 kg m s⁻¹. Find its kinetic energy using p²/2m.
6. A 300 N force drives an object at a steady 12 m s⁻¹ in its own direction. Find the power.
7. A device takes in 1200 J and gives 420 J of useful output. Find its efficiency.
8. An engine must deliver 2.0 × 10⁸ J of useful work at 30% efficiency. The fuel has an energy density of 40 MJ per litre. What volume of fuel is needed?
9. A 1.5 kg block slows from 5.0 m s⁻¹ to 3.0 m s⁻¹ on a rough horizontal surface. What is the work done by friction?
10. A 2.0 kg object is 1.5 m above the ground and moving at 3.0 m s⁻¹. Find its mechanical energy relative to the ground, taking g = 9.81 m s⁻².
11. Why can ΔEp = mgΔh not be used for a large change of height?

Answers

1. W = 30 × 3.0 × cos 60° = 45 J.
2. Zero, because cos 90° = 0.
3. Ek = ½ × 0.40 × 6.0² = 7.2 J.
4. Ee = 0.18 J and 0.72 J. The ratio is 4, because Ee is proportional to (Δx)².
5. Ek = 6.0² / (2 × 3.0) = 6.0 J.
6. P = Fv = 300 × 12 = 3600 W, or 3.6 kW.
7. η = 420 / 1200 = 0.35, or 35%.
8. Input energy = 2.0 × 10⁸ / 0.30 = 6.7 × 10⁸ J. Volume = 6.67 × 10⁸ / 4.0 × 10⁷ = 17 litres.
9. Work by friction = ΔEk = ½ × 1.5 × 3.0² − ½ × 1.5 × 5.0² = −12 J. Friction does 12 J of negative work.
10. Ek = ½ × 2.0 × 3.0² = 9.0 J. Eg = 2.0 × 9.81 × 1.5 = 29.4 J. Mechanical energy = 38 J.
11. The expression assumes a constant gravitational field strength. Over a large height change g decreases, so the expression overestimates the change.

Where marks are usually lost

• Using the angle to the horizontal in W = Fs cos θ when the displacement is not horizontal.
• Leaving the answer positive when the force opposes the motion and the work is negative.
• Not squaring the speed in Ek, or the extension in Ee.
• Leaving the compression in centimetres or the mass in grams.
• Equating mechanical energy at two points when the question mentions friction or air resistance.
• Using total input as the output when calculating efficiency from a Sankey diagram.
• Dropping the efficiency step in a fuel-volume problem.
• Quoting Fv for power when the force is not along the velocity or the velocity is changing.
• Not giving a unit, or giving J where W is needed.

Official syllabus

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