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AS & A-Level Physics 05 — Work, Energy and Power

AS & A-Level Physics 05 — Work, Energy and Power

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Independent Deckloop revision aligned with the Cambridge International AS & A Level Physics (9702) syllabus, 2025–2027. Not affiliated with or endorsed by Cambridge International Education. Chapter 5 of 26: Work, energy and power. Concepts, worked applications and misconception checks.

Physics EN AS & A-Level
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Defining and Calculating Work Done

In physics, 'work' has a very specific meaning. It is the energy transferred when a force causes an object to move, or be displaced. For work to be done, two conditions must be met: a force must be applied, and the object must move in the direction of that force. The amount of work done is the product of the force and the displacement. If the force acts at an angle to the direction of motion, only the component of the force parallel to the displacement does work. Work is a scalar quantity, meaning it has magnitude but no direction, and its SI unit is the joule (J). One joule of work is done when a force of one newton moves its point of application one metre in the direction of the force.

Key points

  • Work is done when a force causes displacement: W=FsW = Fs, where ss is the displacement in the direction of the force FF.
  • If a force FF acts at an angle θ\theta to the displacement ss, the work done is calculated using the component of the force in the direction of displacement: W=(Fcos⁡θ)sW = (F \cos\theta)s.
  • A force acting perpendicular to the direction of motion does no work.
  • Work done is a scalar quantity measured in joules (J), where 1 J = 1 N m.

Worked example

Question

A traveller pulls a suitcase with a force of 50 N along a horizontal floor for 12 m. The strap makes an angle of 30° with the horizontal. Calculate the work done by the traveller on the suitcase.

Solution

1. First, find the component of the force that acts in the horizontal direction of motion. This is given by Fh=Fcos⁡θF_h = F \cos\theta.
2. Fh=50 N×cos⁡(30∘)=43.30... NF_h = 50 \text{ N} \times \cos(30^\circ) = 43.30... \text{ N}.
3. Now, calculate the work done using W=Fh×sW = F_h \times s, where ss is the horizontal displacement.
4. W=43.30... N×12 m=519.6... JW = 43.30... \text{ N} \times 12 \text{ m} = 519.6... \text{ J}.

The work done is 520 J (to 2 significant figures).

Common pitfalls

  • Confusing the physics definition of work with everyday effort. Holding a heavy object stationary requires muscular effort but does no work because there is no displacement (s=0s=0).
  • Forgetting that only the force component parallel to the displacement does work. A common mistake is to use the full magnitude of the force when it acts at an angle to the motion.

Prerequisites

  • Resolving vectors into components is essential for calculating work done when the force is not parallel to the displacement.
  • Basic calculations involving multiplication and trigonometry are required.
Further resources