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A Level Maths 9709 · Paper 4 Mechanics · 4.5 · ~5 min read

Energy, work and power

Work done by a force, kinetic and gravitational potential energy, the work–energy principle with resistances and friction, conservation of energy on smooth surfaces, and power as force × velocity for vehicles on level roads and hills.

Work doneKinetic energyPotential energyWork–energy principleWork against frictionConservation of energyPower P = FvMaximum speed
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Paper tag = where it is examinedFORMULA LIST (MF19) given in the examM1A1B1 how marks are awardedhighlight = comes up again and againTest yourself at the end

In 30 seconds

  • Work done = force × distance moved in the direction of the force: Fd cos θ.
  • Work–energy principle: change in KE + change in PE = work done by driving forces − work done against resistances.
  • Power = rate of doing work = Fv, where F is the driving force. At maximum speed, the driving force equals the total resistance.
  • Energy methods avoid finding the acceleration, and work on curved paths where suvat doesn't.

By the end you can

  • Calculate work done by a constant force, including at an angle to the motion.
  • Use kinetic energy ½mv² and gravitational potential energy mgh.
  • Use the work–energy principle, including with friction and resistances.
  • Use P = Fv, including vehicles accelerating or moving at maximum speed on level ground and on slopes.

1Work and energy

Definitions

W = Fd cos θ ; KE = ½mv² ; GPE = mgh
θ is the angle between the force and the direction of motion
  • Units: joules (J). A force perpendicular to the motion (like R on a horizontal floor) does no work.
  • Only the vertical height matters for GPE: a particle moving 10 m up a slope at angle α rises 10 sin α.
Worked example

A box is dragged 5 m along the floor by a 40 N force at 30° above the horizontal. Find the work done by the force.

  1. W = 40 cos 30° × 5 M1
  2. = 173 J A1

2The work–energy principle

Energy in and energy out

½mv² − ½mu² + mg(h₂ − h₁) = Wdriving − Wresistance
on a smooth surface with no driving force: KE + PE is constant
Worked example

A particle slides from rest down a smooth curved path, falling a vertical height of 3.2 m. Find its speed at the bottom.

  1. ½mv² = mg × 3.2 M1
  2. v² = 64 ⇒ v = 8 m s⁻¹ A1
Worked example

A particle of mass 2 kg is projected up a rough slope at 8 m s⁻¹. The slope has sin α = 0.6 and μ = 0.25. Use energy to find how far up the slope it travels.

  1. R = 20 × 0.8 = 16 N, F = 0.25 × 16 = 4 N B1
  2. ½ × 2 × 64 = 2 × 10 × (0.6d) + 4d M1 A1
  3. 64 = 12d + 4d ⇒ d = 4 m A1
Common mistakeCounting friction twice, or adding the GPE gain as work against friction. Write a single equation: KE lost = PE gained + work against friction.

3Power

P = Fv

P = Fv
F is the driving force (not the resultant); P in watts, 1 kW = 1000 W
  • Maximum speed: acceleration is zero, so driving force = resistance (+ mg sin α uphill).
Worked example

A car of mass 900 kg has maximum power 18 kW and resistance 600 N. (a) Find its maximum speed on a level road. (b) Find its acceleration when travelling at 20 m s⁻¹ at full power. (c) Find its maximum speed up a hill with sin α = 1/20.

  1. (a) 18000/v = 600 ⇒ v = 30 m s⁻¹ M1 A1
  2. (b) F = 18000/20 = 900 N; 900 − 600 = 900a ⇒ a = 0.333 m s⁻² M1 A1
  3. (c) 18000/v = 600 + 900 × 10 × 1/20 = 1050 ⇒ v = 17.1 m s⁻¹ M1 A1
Common mistakeUsing P = ma v or putting the resultant into P = Fv. F in P = Fv is the driving force. The 2024 Paper 4 report flagged this.
Linking questionThe forces come from Forces and equilibrium and Newton's laws; energy in Physics is in Work, energy and power.

Test yourself — cover the answers first

Quick check

  1. Find the KE of a 4 kg mass moving at 5 m s⁻¹.
  2. How much work does the normal reaction do on a horizontal floor?
  3. A car's engine has power 30 kW at 15 m s⁻¹. Find the driving force.
  4. At maximum speed on level ground, what is the acceleration?
  5. A 2 kg mass rises 5 m. Find the PE gained.

Exam-style

  1. A cyclist and bike (80 kg) freewheel down a hill, descending 12 m vertically over 200 m, and speed up from 4 to 12 m s⁻¹. Find the average resistance.
  2. A pump lifts 50 kg of water through 6 m each second and gives it speed 4 m s⁻¹. Find the power.
  3. A lorry of mass 12 000 kg climbs a hill with sin α = 0.05 at constant speed 15 m s⁻¹ against resistance 3000 N. Find its power.
Show answers
  1. 50 J
  2. none
  3. 2000 N
  4. zero
  5. 100 J
  6. PE lost 9600; KE gained 5120; work against resistance 4480 J ⇒ R = 22.4 N
  7. (3000 + 400) J per second = 3.4 kW
  8. F = 3000 + 6000 = 9000 N; P = 135 kW
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Independent notes written by Riyaz Sir. Not produced, endorsed or approved by Cambridge University Press & Assessment. Syllabus codes 9709 and 9702 refer to the Cambridge International AS & A Level syllabuses.