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A Level · Cambridge (CIE) · 9702

Calculate power and derive P = Fv

Calculate power as work or energy transferred per unit time and connect it to force and velocity.

Work, energy and power · Cambridge International AS & A Level Physics 9702 · 5.1

AS Level; Cambridge International Physics 9702, 2025–2027 syllabus, Topic 5. These original explanations and calculations support theory study; full timed assessment and human subject-teacher review remain outstanding.

Before you begin

  • Use force, displacement, speed, mass and gravitational field strength with consistent SI units.

Work on paper. Try each question before revealing help, and explain why your method works.

Quick readiness check

Which unit measures power?

What you will learn

  • Define power as work done per unit time
  • Use P = W/t
  • Derive and apply P = Fv when force and velocity are in the same direction

Power describes the rate of energy transfer

Power is the work done per unit time: P = W/t. Since work is an energy transfer, power is also energy transferred per unit time. Its SI unit is the watt, where 1 W = 1 J s⁻¹.

For the same work, completing it in less time requires greater average power. Use the duration of the transfer, not the total time the system has existed.

Derive the force-speed form

For a constant force acting along the direction of motion, work done W = Fs. Substitute into P = W/t to obtain P = Fs/t. Since speed v = s/t, P = Fv.

The force and velocity must be in the same direction for this simple magnitude form. If a force is not parallel to velocity, only its component along the velocity contributes to the power transferred by that force.

P = W/t · when force is parallel to velocity: P = Fv

Pause and explain

How does P = Fv follow for a force parallel to motion?

Put the idea to work

Worked example

A motor does 2400 J of work in 20 s. Find its average power. Then find the power from a 300 N driving force at 4.0 m s⁻¹ in the same direction.

Show the worked solution
  1. P = W/t = 2400/20 = 120 W.
  2. For parallel force and velocity, P = Fv.
  3. P = 300 × 4.0 = 1200 W for the second situation.

Answer Average power in the first situation is 120 W; power in the second is 1.2 kW.

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From guided practice to a new situation

Write your answer and reasoning first. Use a hint only if you are stuck. The model solution is for self-checking; your written work is not automatically marked.

1 · Guided

A crane transfers 18 kJ of energy in 30 s. Calculate its average power.

Give me a hint

Convert kilojoules to joules before dividing by time.

Compare my reasoning
  1. 18 kJ = 18 000 J.
  2. P = W/t = 18 000/30.
  3. Give power in watts.

600 W.

Look for these in your work

  • Energy converted to joules
  • Work divided by time
  • W unit stated
2 · Independent

A car moves at 12 m s⁻¹ while its engine exerts a forward force of 1500 N. Calculate the power delivered by this force.

Give me a hint

The force is parallel to velocity, so use P = Fv.

Compare my reasoning
  1. P = Fv.
  2. P = 1500 × 12.
  3. Convert to kilowatts if useful.

18 000 W = 18 kW.

Look for these in your work

  • Force-speed relation applied
  • Product correct
  • Power unit included
3 · Transfer

An object moves upward at constant speed 2.0 m s⁻¹ while a 500 N lifting force acts upward. Find the lifting power and explain the role of gravity if the object has constant speed.

Give me a hint

Use the force component along motion; constant speed means zero resultant force.

Compare my reasoning
  1. P = Fv = 500 × 2.0 = 1000 W.
  2. At constant speed, the resultant force is zero.
  3. For a vertical lift with no other vertical forces, weight balances the lifting force; gravity does negative work on the upward displacement.

The lifting force transfers energy at 1.0 kW. At constant speed the net force is zero; the lifting force balances weight, and gravity transfers energy oppositely to the motion.

Look for these in your work

  • Power calculated
  • Constant-speed force balance used
  • Gravity's opposite energy transfer noted

Common mistakes

  • Using P = Fv when force is not along velocity without resolving components
  • Confusing work in joules with power in watts
  • Multiplying force by time instead of displacement or velocity
Recall without your notes

What can you explain now?

Which unit measures power?

Compare with the explanation

W

A watt is one joule transferred per second.

After trying it yourself, choose your next review. This is your self-assessment.

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Make it stick

Which motion direction must be used when applying P = Fv to one force?

If you needed the solution, close it and solve the problem again from a blank page. A correct answer is stronger when you can explain the reason for each step.

Keep learning Physics

Course scope and assessment

One hundred and sixty-two original lessons cover all 12 published subject-content outcomes in Topic 1 (sections 1.1–1.4), all nine outcomes in Topic 2 (section 2.1), all 13 outcomes in Topic 3 (sections 3.1–3.3), all 13 outcomes in Topic 4 (sections 4.1–4.3), all 11 outcomes in Topic 5 (sections 5.1–5.2), all 10 outcomes in Topic 6 (sections 6.1–6.2), all 16 outcomes in Topic 7 (sections 7.1–7.5), all 12 outcomes in Topic 8 (sections 8.1–8.4), all 15 outcomes in Topic 9 (sections 9.1–9.3), all 16 outcomes in Topic 10 (sections 10.1–10.3), all 18 outcomes in Topic 11 (sections 11.1–11.2), all seven outcomes in Topic 12 (sections 12.1–12.2), all 12 outcomes in Topic 13 (sections 13.1–13.4), all eight outcomes in Topic 14 (sections 14.1–14.3), and all nine outcomes in Topic 15 (sections 15.1–15.3), plus all four outcomes in Topic 16 (sections 16.1–16.2), and all ten outcomes in Topic 17 (sections 17.1–17.3), plus all 12 outcomes in Topic 18 (sections 18.1–18.5), all nine outcomes in Topic 19 (sections 19.1–19.3), all 19 outcomes in Topic 20 (sections 20.1–20.5), and all eight outcomes in Topic 21 (sections 21.1–21.2), and all 17 outcomes in Topic 22 (sections 22.1–22.4), and all 13 outcomes in Topic 23 (sections 23.1–23.2), and all 16 outcomes in Topic 24 (sections 24.1–24.3), and all 11 outcomes in Topic 25 (sections 25.1–25.3), of the 2025–2027 syllabus. Each lesson has guided, independent and application practice with explained answers. All 25 syllabus topics now have mapped teaching and practice; the 2028–2030 changes outside Topics 23–25 still need reconciliation. Online study cannot demonstrate supervised Paper 3 laboratory competence. Paper 5 planning and analysis, full timed cumulative assessment and human subject-teacher review remain outstanding.

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