OCR A Level Physics A: Work, energy, power and efficiency (H556) – Revision Notes

These notes condense section 3.3 (Work, energy and power) of the OCR Advanced GCE in Physics A specification (H556), Version 3.0 (March 2026), for first assessment in 2017. The study guide has the full explanations, and these notes are for recall. Afterwards use the practice questions.

Links: OCR A Level Physics hub, printable checklist and free diagnostics. Use g = 9.81 N kg⁻¹ unless told otherwise.

Formulas and definitions

Statement	Result	Unit
3.3.1(a)	Work done = energy transferred by a force	joule, J = N m
3.3.1(b)	W = Fx cos θ, with θ between F and the displacement	J
3.3.2(a)	Ek = ½mv²	J
3.3.2(b)	Ep = mgh (ΔEp = mgΔh) in a uniform field	J
3.3.3(a)	P = W/t	watt, W = J s⁻¹
3.3.3(b)	P = Fv	W
3.3.3(c)	efficiency = useful output energy / total input energy × 100%	none

Definition to learn: the joule is the work done by a force of one newton that moves its point of application one metre along the line of the force. One watt is one joule transferred per second.

3.3.1 Work and conservation of energy

• Work done is a scalar. Only the component of the force along the displacement does work.
• θ = 0°: work is Fx. θ = 90°: work is zero. θ above 90°: work is negative, so the force removes energy from the object.
• Principle of conservation of energy: the total energy of an isolated system is constant, because energy is only ever moved between forms.
• Energy transferred = work done by the force.
• Resistive forces transfer energy to thermal energy of the surroundings. The energy is dissipated, not destroyed.

Method for an energy-balance problem:

1. List every force that does work, with its sign.
2. Write: work done by driving forces = gain in Ek + gain in Ep + energy dissipated against resistive forces.
3. Work out the vertical height gained (not the distance along a slope).
4. Solve for the unknown and check the sign and units.

3.3.2 Kinetic and potential energy

Derive Ek = ½mv²

1. F = ma.
2. From rest: v² = 2ax, so x = v²/2a.
3. W = Fx = ma × v²/2a = ½mv².
4. Work done on the object = energy transferred, so Ek = ½mv².

Derive Ep = mgh

1. Lift at constant speed, so the lifting force F equals the weight mg.
2. Distance moved in the direction of the force = h.
3. W = Fh = mgh = energy stored, so Ep = mgh.

Exchange of Ep and Ek

• With no resistive forces: loss of Ep = gain of Ek, so mgΔh = ½mv² (when starting at rest), giving v = √(2gΔh).
• The mass cancels. The speed gained by a falling object (ignoring resistance) does not depend on its mass.
• With resistive forces: loss of Ep = gain of Ek + energy dissipated.
• Doubling the speed multiplies Ek by four. Doubling the mass doubles Ek.
• The equation Ep = mgh applies only where g is constant (a uniform field).

3.3.3 Power and efficiency

Derive P = Fv

1. W = Fx, so P = W/t = Fx/t.
2. At constant speed, x/t = v.
3. P = Fv.

• A vehicle at constant speed: driving force = resistive force, so power = resistive force × speed.
• Useful power of a lifting machine at constant speed = mgv.
• Efficiency can use energy or power, because the time cancels. It is never more than 100%.
• Efficiencies of machines in series multiply (as fractions): 0.80 × 0.90 = 0.72.
• 1 kW h = 3.6 × 10⁶ J. It is a unit of energy.

Must-know distinctions

Pair	Difference
Work and energy	Work is the process of transferring energy by a force. Energy is the stored quantity. Both are in joules.
Energy and power	Power is the rate of energy transfer: joules per second.
kW and kW h	kW is power. kW h is energy, 3.6 MJ.
Ep and weight	Weight is a force in newtons. Ep is an energy in joules.
Dissipated and destroyed	Dissipated energy goes to a less useful form. Total energy is conserved.
Efficiency and power	Efficiency is a ratio with no unit.

Worked reminders

Work at an angle. A lawnmower is pushed 9.0 m along level grass by a force of 120 N directed 25° below the horizontal. The work done by the push is W = Fx cos θ = 120 × 9.0 × cos 25° = 979 J. Only the horizontal component moves the mower along the ground.

Energy dissipated. A ball of mass 0.20 kg is dropped from 3.0 m and rebounds to 2.1 m. Ep at release = 0.20 × 9.81 × 3.0 = 5.89 J. Ep at the top of the rebound = 0.20 × 9.81 × 2.1 = 4.12 J. The energy dissipated in the bounce and through the air is 5.886 − 4.120 = 1.77 J, which is 30% of the original Ep. It has been transferred to thermal energy and sound, not destroyed.

Efficiency. A machine takes in 800 J and delivers 560 J of useful work. Efficiency = 560 ÷ 800 × 100% = 70%. The wasted energy is 800 − 560 = 240 J. If this machine drives a second machine that is 50% efficient, the overall efficiency is 0.70 × 0.50 = 0.35, or 35%.

Choosing the equation. If a question gives a force and a distance, use W = Fx cos θ. If it gives a speed and a mass, think Ek. If it gives a change in height, think Ep. If it gives a time or a rate, think P = W/t, and if a steady force and a steady speed are given, P = Fv.

Quick self-test

1. A 30 N force is directed at 60° to an object’s path, and the object moves 4.0 m. Calculate the work done.
2. A 4.0 kg cart moves at 2.5 m s⁻¹. What is its kinetic energy?
3. A ball of mass 0.060 kg moves at 30 m s⁻¹. Calculate its kinetic energy.
4. A 0.50 kg object has 100 J of kinetic energy. Calculate its speed.
5. Calculate the gain in gravitational potential energy when a 5.0 kg bag is lifted through 0.50 m.
6. A pendulum bob is released from rest at a height of 0.45 m above its lowest point. Ignoring resistive forces, calculate its speed there.
7. A 40 kg load is raised 2.0 m at constant speed in 5.0 s. Calculate the useful power.
8. A lorry travels at a steady 25 m s⁻¹ with a driving force of 3000 N. Calculate the power delivered to the wheels.
9. A machine gives 45 W of useful output from an input of 150 W. Calculate its efficiency.
10. A 2.5 kW heater runs for 4.0 hours. State the energy transferred in kW h and in MJ.
11. Explain why the tension in a pendulum string does no work.

Answers

1. W = 30 × 4.0 × cos 60° = 60 J.
2. ½ × 4.0 × 2.5² = 12.5 J.
3. ½ × 0.060 × 30² = 27 J.
4. v = √(2 × 100 ÷ 0.50) = 20 m s⁻¹.
5. 5.0 × 9.81 × 0.50 = 24.5 J (24.525 J).
6. v = √(2 × 9.81 × 0.45) = 2.97 m s⁻¹.
7. Work = 40 × 9.81 × 2.0 = 785 J. Power = 785 ÷ 5.0 = 157 W.
8. P = Fv = 3000 × 25 = 7.5 × 10⁴ W (75 kW).
9. 45 ÷ 150 × 100% = 30%.
10. 2.5 × 4.0 = 10 kW h. 10 × 3.6 = 36 MJ.
11. The tension acts along the string, at 90° to the bob’s displacement at every instant, and cos 90° = 0, so no work is done.

Where marks are usually lost

• Writing W = Fx without cos θ when the force is at an angle, or using the wrong angle (it must be the angle to the displacement, not to the vertical or the surface).
• Using the length of the slope instead of the vertical height in mgh.
• Forgetting that the mass cancels in the speed from mgh = ½mv², then using a wrong mass.
• Omitting the energy dissipated by friction in an energy-balance question.
• Squaring only v (or the product mv) in ½mv².
• In a derivation, not stating the starting condition (from rest, or constant speed) or the line that equates work to energy transferred.
• Mixing up kW and kW h, or not converting to joules and watts before substituting.
• Quoting efficiency above 100%, or giving no unit or percentage sign.
• Saying energy is “used up” or “lost” rather than transferred to a less useful form.
• Rounding too early: carry full values and round the final answer to a suitable number of significant figures.

Official syllabus

OCR Advanced GCE in Physics A (H556) specification, Version 3.0 (March 2026), for first assessment in 2017, published by OCR. Section 3.3 Work, energy and power.