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Energy, work and power

A-Level Maths 9709 · Topic 4.5 revision notes

Everything from the 3 lessons of this topic on a few pages: the key idea of each lesson, the diagram to remember, the points to learn and the traps examiners set. Read it, cover it, test yourself.

  1. Work done by a force
  2. Kinetic and potential energy, the work-energy principle
  3. Power and driving force

1Work done by a force

Syllabus 4.5

Work done by a constant force \(= Fd\cos\theta\), where \(d\) is the distance moved and \(\theta\) the angle between the force and the direction of motion. Work is measured in joules (\(1\ \text{J} = 1\ \text{N m}\)). Work against gravity \(= mgh\); work against friction \(= Fd\) with \(F = \mu R\).

25° 60 N 60 cos 25° = 54.4 N displacement 8 m W = 60 × 8 × cos 25° = 435 J
Fig. 1: a 60 N pull at 25° to the horizontal moves a crate 8 m. Only the 54.4 N component along the motion does work (angle drawn to scale)

Worked example: A rope at an angle (Fig. 1)

A crate is pulled 8 m along a horizontal floor by a rope with tension 60 N inclined at \(25^\circ\) above the horizontal. Find the work done by the tension.
  1. Component of the tension along the motion \(= 60\cos 25^\circ = 54.38\) N.
  2. \(W = 54.38 \times 8\).
Answer: 435 J (3 s.f.)

Resolve along the motion, times distance.

  • Work \(= Fd\cos\theta\), \(\theta\) between force and motion.
  • Perpendicular forces (normal reaction on a level floor) do no work.
  • Gravity: \(mg \times\) vertical height.
  • Friction: find \(R\) first, then \(\mu R \times d\).
  • Constant speed: work done by the driving force \(=\) total work done against the resistances.

Watch out!

Do not use the full force when it is at an angle: \(60 \times 8 = 480\) J is the work only if the rope is horizontal.

Work done by the weight on a slope uses the VERTICAL height, \(mg \times d\sin\alpha\), not \(mg \times d\). The distance along the slope only multiplies the component \(mg\sin\alpha\).

With the rope at an angle, the normal reaction is NOT \(mg\). Using \(\mu mg\) for friction in the first example gives 800 J instead of 650 J.

Revisit it: Work done by a force lesson · worked practice

2Kinetic and potential energy, the work-energy principle

Syllabus 4.5

\(\text{KE} = \tfrac{1}{2}mv^2\) and gravitational \(\text{PE} = mgh\). If only gravity does work, \(\text{KE} + \text{PE}\) is constant. Otherwise use the work-energy principle: initial energy \(+\) work done by driving forces \(=\) final energy \(+\) work done against resistances.

4 m at rest: PE = mg(4), KE = 0 KE = ½mv², v = √80 = 8.94 m s⁻¹ smooth slide
Fig. 1: a child slides from rest down a smooth curved slide that drops 4 m. Only gravity does work, so mg(4) = ½mv² whatever the shape of the slide

Worked example: A smooth curved slide (Fig. 1)

A child of mass 30 kg starts from rest at the top of a smooth slide and descends a vertical height of 4 m. Find the speed at the bottom.
  1. PE lost \(= 30 \times 10 \times 4 = 1200\) J.
  2. KE gained: \(\tfrac{1}{2}(30)v^2 = 1200\), so \(v^2 = 80\).
  3. \(v = 8.94\ \text{m s}^{-1}\).
Answer: \(8.94\ \text{m s}^{-1}\). The mass cancels: \(v = \sqrt{2gh}\).

Start + work in = end + work out.

  • KE \(= \tfrac{1}{2}mv^2\) at the start and the end.
  • PE change \(= mg \times\) vertical height change (not distance along a slope).
  • Work in: driving forces. Work out: friction and resistance (\(\mu R \times d\)).
  • Equation: initial KE \(+\) PE \(+\) work in \(=\) final KE \(+\) PE \(+\) work out.
  • Curved paths: energy is the only method you need.

Watch out!

Adding speeds instead of energies. The ball thrown at 12 and gaining 10 from the fall does NOT land at 22: energies add, so \(v^2 = 144 + 100\).

\(\tfrac{1}{2}m(25 - 15)^2\) is NOT the change in KE. Square each speed first: \(\tfrac{1}{2}m(25^2 - 15^2)\).

Watch the sign of the KE change. Here the cyclist SLOWS down, so the lost KE (1560 J) helps: \(W = 18\,000 - 1560\), not \(18\,000 + 1560\).

Revisit it: Kinetic and potential energy, the work-energy principle lesson · worked practice

3Power and driving force

Syllabus 4.5

Power \(= \dfrac{\text{work done}}{\text{time taken}}\), in watts (\(1\ \text{W} = 1\ \text{J s}^{-1}\)). For a force in the direction of motion, \(P = Fv\), so the driving force is \(F = \dfrac{P}{v}\). At maximum speed the acceleration is zero and the driving force equals the total resistance.

6000 2000 1200 15 25 40 force (N) v (m s⁻¹) F = 30 000 / v 800 N top speed 25 resistance 1200 N
Fig. 1: driving force F = P/v for a 30 kW engine (to scale) against a constant 1200 N resistance. At 15 m s⁻¹ there is an 800 N resultant; at 25 m s⁻¹ the forces balance and the car can go no faster

Worked example: Average power

A motor lifts a load of mass 50 kg through a vertical height of 12 m at constant speed in 8 s. Find the power of the motor.
  1. Work done against gravity \(= 50 \times 10 \times 12 = 6000\) J.
  2. Power \(= \dfrac{6000}{8}\).
Answer: 750 W

F = P/v, then F = ma.

  • Units: power in watts (kW \(\times 1000\)).
  • Driving force: \(F = \dfrac{P}{v}\) at the speed in question.
  • Hill: \(mg\sin\alpha\) opposes motion uphill and helps downhill.
  • Newton's second law: driving force \(-\) resistances \(= ma\).
  • Top speed: set \(a = 0\).

Watch out!

Power must be in WATTS before you use \(F = P/v\). An engine of 30 kW gives \(F = 30\,000/v\), not \(30/v\).

\(P = Fv\) uses the DRIVING force. Writing \(P = mav\) (the resultant) at 15 m/s gives \(a = 1.67\) instead of 0.667.

Downhill the weight component is on the SAME side as the driving force. Adding it to the resistance gives a top speed of 21.4 m/s for the lorry instead of 50 m/s.

Revisit it: Power and driving force lesson · worked practice
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