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

This guide covers 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: 3.3.1 work and conservation of energy, 3.3.2 kinetic and potential energies, and 3.3.3 power. H556 is the full A level.

See the OCR A Level Physics hub and the printable checklist. Pair this page with the revision notes and the practice questions, and test yourself with the free diagnostics. The equations of motion used in the derivations are in Kinematics, equations of motion and projectiles, and resolving a force into components is in Foundations of Physics.

Where no value of g is given, use 9.81 N kg⁻¹.

What you must be able to do

Statement	Content
3.3.1(a)-(b)	Work done by a force; the joule; W = Fx cos θ
3.3.1(c)-(e)	Conservation of energy; forms of energy and their transfer; energy transferred equals work done
3.3.2(a)-(b)	Ek = ½mv² and Ep = mgh: recall them and derive them from first principles
3.3.2(c)	Exchange between gravitational potential energy and kinetic energy
3.3.3(a)-(b)	Power and the watt; P = W/t; P = Fv, derived from first principles
3.3.3(c)	Efficiency = useful output energy / total input energy × 100%

3.3.1(a)-(b) Work done by a force

Work is done whenever a force moves something along a line it has a component on. The work done is the energy that force transfers, and it is a scalar. One joule (J) is the work done when a force of one newton moves its point of application one metre in the direction of the force, so 1 J = 1 N m.

When the force F points partly across the displacement x, count only the part of the force that lies along the line of motion:

W = Fx cos θ

Here θ is the angle between the force and the direction of the displacement.

Angle θ	cos θ	Work
0°	1	Fx, the largest value
0° to 90°	between 0 and 1	Fx cos θ
90°	0	zero
more than 90°	negative	negative: the force removes energy from the object

A force at 90° to the motion does no work. This is why the weight of a trolley rolling along a level track, or the pull of a string on a swinging pendulum bob, transfers no energy. Friction acts against the motion, so the work it does is negative: it takes energy from the moving object.

Worked example. A suitcase is pulled 18 m along a level floor by a strap with tension 64 N at 35° above the horizontal. Find the work done by the strap.

W = Fx cos θ = 64 × 18 × cos 35° = 944 J (3 s.f.).

Only the horizontal component, 64 cos 35° = 52.4 N, moves the case along the floor. The vertical component does no work because there is no displacement in that direction.

3.3.1(c)-(e) Conservation of energy

The principle of conservation of energy states that the total energy of an isolated system stays constant. Energy is never made or lost, only moved between forms.

Forms you must be able to use: kinetic, gravitational potential, elastic potential, thermal (internal), electrical, chemical, light and sound. “Lost” energy is not destroyed. Friction and air resistance transfer it to thermal energy of the surroundings, where it is no longer useful.

The link to work is the key idea of this section:

energy transferred = work done

When a force does work on an object, it transfers energy to it (or from it, if the work is negative). The energy transferred equals the work done by the force.

Worked example. A trolley of mass 3.2 kg starts at rest at the bottom of a ramp 2.5 m long, inclined at 15° to the horizontal. A steady pull of 14 N parallel to the ramp moves it to the top. Friction is a constant 5.0 N. Find its speed at the top.

Work done by the pull = 14 × 2.5 = 35.0 J. Work done against friction = 5.0 × 2.5 = 12.5 J, which becomes thermal energy. Gain in gravitational potential energy = mgh = 3.2 × 9.81 × (2.5 sin 15°) = 20.3 J. Gain in kinetic energy = 35.0 − 12.5 − 20.3 = 2.19 J. ½ × 3.2 × v² = 2.19, so v = √(2 × 2.19 ÷ 3.2) = 1.17 m s⁻¹.

The height gained is 2.5 sin 15° = 0.647 m. Using the length of the ramp instead of the vertical height in mgh is a common slip.

3.3.2(a) Kinetic energy and its derivation

A moving object carries kinetic energy:

Ek = ½mv²

You must recall this and derive it. Take an object of mass m, initially at rest, accelerated uniformly by a constant resultant force F through a distance x until its speed is v.

1. Newton’s second law: F = ma.
2. Equation of motion with u = 0: v² = 2ax, so x = v²/2a.
3. Work done by the resultant force: W = Fx = ma × v²/2a = ½mv².
4. This work is transferred to the object as kinetic energy, so Ek = ½mv².

Because Ek depends on v², doubling the speed multiplies the kinetic energy by four. A lorry of mass 7500 kg at 18 m s⁻¹ has Ek = ½ × 7500 × 18² = 1.2 × 10⁶ J. At 36 m s⁻¹ it has 4.9 × 10⁶ J, four times as much. Kinetic energy is a scalar and never negative.

3.3.2(b) Gravitational potential energy and its derivation

In a uniform gravitational field, the gravitational potential energy gained when an object of mass m is raised through a height h is:

Ep = mgh (more generally, ΔEp = mgΔh)

To derive it, raise the object at constant speed so that the resultant force is zero.

1. The lifting force equals the weight: F = mg.
2. The displacement in the direction of the force is h.
3. Work done = Fh = mgh. This energy is stored as gravitational potential energy, so Ep = mgh.

The equation applies for heights small compared with the radius of the Earth, where g is effectively constant. Only changes in height matter, so you choose where h = 0 for convenience.

3.3.2(c) Exchange between Ep and Ek

When only gravity does work (no resistive forces), kinetic and potential energy are exchanged and their sum stays constant. Falling: Ep decreases and Ek increases by the same amount.

Worked example (no resistive forces). A pendulum bob of mass 0.35 kg is released from rest at a height of 0.18 m above its lowest point. Find its speed at the lowest point.

Loss of Ep = gain of Ek: mgh = ½mv². The mass cancels: v = √(2gh) = √(2 × 9.81 × 0.18) = 1.88 m s⁻¹.

The tension in the string is always at 90° to the motion, so it does no work, and the mass does not affect the speed.

Worked example (with resistive forces). A car of mass 450 kg on a track is released from rest at a point 22 m above the ground. At a point 6.0 m above the ground its speed is measured as 15.0 m s⁻¹. How much energy has been dissipated?

Loss of Ep = 450 × 9.81 × (22 − 6.0) = 70 600 J. Gain of Ek = ½ × 450 × 15.0² = 50 600 J. Energy dissipated = 70 632 − 50 625 = 2.0 × 10⁴ J (2 s.f.), mainly as thermal energy.

3.3.3(a) Power

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

P = W/t

The unit is the watt (W). One watt is a transfer of energy of one joule per second, so 1 W = 1 J s⁻¹. You will also meet the kilowatt (kW) and megawatt (MW).

The kilowatt-hour (kW h) is a unit of energy, not power: 1 kW h = 3.6 × 10⁶ J, the energy transferred by 1000 W running for 3600 s.

3.3.3(b) P = Fv and its derivation

When a constant force F moves an object at constant velocity v in the direction of the force, the power delivered is:

P = Fv

Derivation:

1. Work done in time t: W = Fx.
2. Power: P = W/t = Fx/t.
3. The object moves at constant speed, so x/t = v.
4. Therefore P = Fv.

For a vehicle at constant speed, the driving force equals the resistive force, so the power needed to keep going is resistive force × speed.

Worked example. A car travels at a steady 28 m s⁻¹ on a level road against total resistive forces of 650 N. Find the power its engine delivers to the wheels.

At constant speed, driving force = resistive force = 650 N. P = Fv = 650 × 28 = 1.82 × 10⁴ W (18.2 kW).

The same equation shows why a car’s top speed is limited: resistive forces rise with speed, so the power needed rises faster still.

3.3.3(c) Efficiency

Efficiency of a mechanical system compares useful output with total input:

efficiency = (useful output energy / total input energy) × 100%

Power can be used in the same way, because both quantities are divided by the same time: efficiency = (useful output power / total input power) × 100%. Efficiency has no unit. It is never more than 100%, because that would create energy. If you quote it as a fraction, it is never more than 1.

Worked example. An electric hoist draws 2.4 kW and lifts a load of mass 150 kg at a steady 0.90 m s⁻¹. Find its efficiency.

Useful output power = Fv = mgv = 150 × 9.81 × 0.90 = 1324 W. Efficiency = 1324 ÷ 2400 × 100% = 55% (2 s.f.).

The other 45% of the input appears as thermal energy and sound in the motor and the pulleys.

When machines are linked in series, multiply the efficiencies as fractions. A motor of 80% efficiency driving a gearbox of 90% efficiency gives 0.80 × 0.90 = 0.72, an overall 72%.

When you evaluate a design, link efficiency to its effect: a more efficient machine wastes less energy for the same useful output, which lowers running cost and the thermal energy released to the surroundings.

Common errors

• Putting the slope length into mgh where the vertical height belongs.
• Using W = Fx when the force is at an angle. Take the component along the displacement.
• Forgetting that the energy dissipated by friction is the force multiplied by the distance moved against it.
• Writing Ek = ½mv instead of ½mv², or squaring only part of the expression.
• Stating that energy is “used up”. Say it is transferred to a less useful form.
• Confusing kW (power) with kW h (energy).
• Giving efficiency above 100%, or leaving it as a bare number without saying whether it is a percentage or a fraction.
• Skipping the line that equates the work done to the energy gained in a derivation. State it.

Next steps

Compress this guide into the revision notes, and then work through the practice questions. Tick off 3.3 on the checklist.

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: 3.3.1, 3.3.2 and 3.3.3.