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Edexcel A-level Further Maths revision notes

Work, energy and power

Section 2
Year 1
Year 1: this is the AS subject content the exam board publishes, which is what most schools teach in Year 12.
1 specification point

Notes and three levels of exam-style practice for each registered specification point in this section.

Checked against Edexcel 9FM0 section 2

Checked against Edexcel 9FM0 section 2. Review basis: the qualification registry sourced from the Pearson Edexcel Level 3 Advanced GCE in Further Mathematics (9FM0) specification; registry verification recorded 17 July 2026.

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In the exam: Formulae booklet provided · calculator allowed in every paper

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Year 1
Year 1: this is the AS subject content the exam board publishes, which is what most schools teach in Year 12.

Kinetic and potential energy, work and power. The work-energy principle. The principle of conservation of mechanical energy.

Notes
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Explanation

  • Kinetic energy is 12mv2\tfrac12mv^2, while gravitational potential energy changes by mgΔhmg\Delta h. Work is force times displacement for a constant parallel force and ∫F dx\int F\,dx when the force varies with position.
  • The work-energy principle states that work done by the resultant force equals the change in kinetic energy.
  • Mechanical energy is conserved only when no non-conservative force does work; driving forces and resistance must otherwise appear through their work.
  • Power is P=dW/dtP=dW/dt and equals FvFv only for the component of force parallel to velocity.
  • Examiners expect a consistent energy balance, correct signs for work and resolved forces on inclined planes.
A body moving up an incline with driving force, resistance and weight shown.
Worked example

A 900 kg900\,\text{kg} car travels up a slope of angle 4∘4^\circ at constant speed 18 m s−118\,\text{m s}^{-1}. Resistance is 350 N350\,\text{N}. Taking g=9.8 m s−2g=9.8\,\text{m s}^{-2}, find the engine power.

  1. 1.Constant speed gives zero resultant force along the slope.
  2. 2.The driving force is F=900(9.8)sin⁡4∘+350=965.2 NF=900(9.8)\sin4^\circ+350=965.2\,\text{N} to the stated accuracy.
  3. 3.P=Fv=965.2(18)=1.737×104 WP=Fv=965.2(18)=1.737\times10^4\,\text{W}.

Answer: The engine power is 17.4 kW17.4\,\text{kW} to 33 significant figures.

Common mistakes

  • Don't use mgcos⁡αmg\cos\alpha instead of mgsin⁡αmg\sin\alpha for the component of weight down the slope.
  • Don't conserve mechanical energy despite a stated resistance force.
  • Don't use the full force magnitude in P=FvP=Fv when force and velocity are not parallel.

Exam tip

State whether the equation is work-energy, conservation of energy or power before substituting values.

Tier 1 · Easy

ORIGINAL

1.

A 1200 kg1200\,\text{kg} car increases its speed from 10 m s−110\,\text{m s}^{-1} to 14 m s−114\,\text{m s}^{-1}. Calculate the net work done on the car.

(2)

(Total for Question 1 is 2 marks)

Tier 2 · Standard

ORIGINAL

1.

A 1200 kg1200\,\text{kg} car climbs a road inclined at 5∘5^\circ to the horizontal at a constant speed of 20 m s−120\,\text{m s}^{-1}. The resistance is 400 N400\,\text{N}. Taking g=9.8 m s−2g=9.8\,\text{m s}^{-2}, find the engine power.

(4)

(Total for Question 1 is 4 marks)

Tier 3 · Hard

ORIGINAL

1.

A 2 kg2\,\text{kg} particle moves along a horizontal line. A constant driving force of 20 N20\,\text{N} acts forwards while the resistance after displacement x mx\,\text{m} is (2+0.5x) N(2+0.5x)\,\text{N}. Its speed at x=0x=0 is 3 m s−13\,\text{m s}^{-1}. Find its speed when x=8x=8.

(5)

(Total for Question 1 is 5 marks)

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