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A Level Physics 9702 · AS Level · Topic 5 · ~7 min read

Work, energy and power

Work done by a force, conservation of energy, efficiency, power and deriving P = Fv, and deriving the formulas for gravitational potential energy and kinetic energy, with motion against resistive forces.

Work doneConservation of energyEfficiencyPowerP = FvΔEp = mgΔhEk = ½mv²Energy with friction
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Paper tag = where it is examinedDATA & FORMULAE given in the examC1A1B1 how marks are awardedhighlight = comes up again and againTest yourself at the end

In 30 seconds

  • Work done = force × displacement in the direction of the force. If the force is at angle θ to the motion, W = Fs cos θ.
  • Energy is conserved: it is transferred between stores, never created or destroyed. Work done is energy transferred.
  • Power is work done per unit time, P = W/t. For a force moving an object at speed v, P = Fv.
  • ΔEp = mgΔh (uniform field) and Ek = ½mv². You must be able to derive both.
  • Efficiency = useful energy output ÷ total energy input (or the same with power).

By the end you can

  • Define work and calculate work done, including forces at an angle.
  • Apply conservation of energy, including energy lost to resistive forces.
  • Define power and efficiency, derive P = Fv and use it.
  • Derive ΔEp = mgΔh and Ek = ½mv² and use them.

1Work and energy

Definitions

Work done — the product of the force and the displacement in the direction of the force.

Principle of conservation of energy — energy cannot be created or destroyed; it can only be transferred from one form to another. The total energy of a closed system is constant.

W = Fs LEARN IT
s = displacement in the direction of F; for a force at θ to the motion, W = Fs cos θ
  • Unit: joule, J = N m = kg m² s⁻². 1 J is the work done when a force of 1 N moves its point of application 1 m in the direction of the force.
  • No work is done by a force perpendicular to the motion (e.g. the normal contact force on a horizontal surface, or the centripetal force).
Worked example

A sledge is pulled 25 m across horizontal snow by a rope at 30° to the horizontal, with tension 40 N. Find the work done by the tension.

  1. W = Fs cos θ = 40 × 25 × cos 30° C1
  2. = 870 J A1

Deriving the energy formulas

  • Gravitational PE: lifting mass m at constant speed through Δh needs a force equal to the weight mg. Work done = force × distance = mg × Δh, so ΔEp = mgΔh (in a uniform field, near Earth's surface).
  • Kinetic energy: a resultant force F accelerates mass m from rest through distance s. From v² = u² + 2as with u = 0, s = v²/(2a). Work done = Fs = ma × v²/(2a) = ½mv².
ΔEp = mgΔh ; Ek = ½mv² LEARN IT
both must be derived and recalled
Worked example

A ball is dropped from 5.0 m. Ignoring air resistance, find its speed just before it hits the ground.

  1. mgh = ½mv² ⇒ v = √(2gh) C1
  2. v = √(2 × 9.81 × 5.0) = 9.9 m s⁻¹ A1 (independent of mass)
Worked example

A 60 kg skier starts from rest and descends a slope 120 m long, dropping 25 m vertically. Her speed at the bottom is 18 m s⁻¹. Find the average resistive force.

  1. loss in Ep = 60 × 9.81 × 25 = 14 715 J; gain in Ek = ½ × 60 × 18² = 9720 J C1
  2. work done against resistance = 14 715 − 9720 = 4995 J C1
  3. F = 4995/120 = 42 N A1
Common mistakeAssuming the vertical drop equals the distance moved. Work against resistance uses the distance along the slope (120 m); the change in Ep uses the vertical height (25 m). The 2024 Paper 2 report flagged a similar slip with a spring's compression.

2Power and efficiency

Definitions

Power — work done per unit time (rate of doing work or transferring energy).

Efficiency — the ratio of useful energy output (or power output) to total energy input (or power input).

P = W/t ; P = Fv
W = J s⁻¹; F is the force in the direction of motion
  • Deriving P = Fv: P = W/t = Fs/t = F × (s/t) = Fv, for a constant force and speed v.
  • At maximum speed a vehicle's driving force equals the total resistive force (no acceleration), so P = (resistive force) × v_max.
Worked example

A car of mass 1200 kg has an engine power of 40 kW. At 20 m s⁻¹ the resistive force is 1200 N. Find its acceleration at this speed.

  1. driving force F = P/v = 40 000/20 = 2000 N C1
  2. a = (2000 − 1200)/1200 C1
  3. = 0.67 m s⁻² A1
Worked example

A motor lifts a 50 kg load through 8.0 m in 12 s. The input power is 600 W. Find the efficiency.

  1. useful power = mgh/t = 50 × 9.81 × 8.0/12 = 327 W C1
  2. efficiency = 327/600 = 0.545 = 55% A1
Common mistakeMixing up useful and total. State which power is input and which is useful output; in 2024 candidates who wrote "power increases" without saying which one lost the mark.
Worked example

A cyclist of total mass 80 kg rides up a slope inclined at 5.0° at a steady 6.0 m s⁻¹ against a resistive force of 20 N. Find the power output.

  1. force needed = mg sin θ + resistance = 80 × 9.81 × sin 5.0° + 20 = 88.4 N C1
  2. P = Fv = 88.4 × 6.0 = 530 W A1
Linking questionWork done by a gas and internal energy extend these ideas in Thermodynamics; the work done stretching a wire is the area under a force–extension graph in Deformation of solids; and gravitational potential energy far from Earth is −GMm/r in Gravitational fields.

Test yourself — cover the answers first

Quick check

  1. Define work done.
  2. Define power.
  3. Derive P = Fv.
  4. What is the efficiency if 400 J of 1000 J input is useful?
  5. Why does a vertical force do no work on an object moving horizontally?

Exam-style

  1. Derive Ek = ½mv² from the equations of uniformly accelerated motion.
  2. A car's engine delivers a constant 60 kW. The resistive force is 0.80v² N, where v is the speed in m s⁻¹. Find the maximum speed on a level road.
  3. A 2.0 kg block slides down a rough slope from rest, dropping 1.5 m vertically, and reaches 4.0 m s⁻¹. Calculate the energy transferred to thermal energy.
Show answers
  1. force × displacement in the direction of the force
  2. work done per unit time
  3. P = W/t = Fs/t = Fv
  4. 40%
  5. there is no displacement in the direction of the force
  6. from rest: v² = 2as; W = Fs = ma × v²/(2a) = ½mv²
  7. at max speed P = Fv = 0.80v³ ⇒ v³ = 75 000 ⇒ v = 42 m s⁻¹
  8. Ep lost = 2.0 × 9.81 × 1.5 = 29.4 J; Ek = 16 J; thermal = 13 J
Practise this chapter
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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.