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Energy Concepts

Work done when the force is not along the motion

  • 3 things to know
  • Component 1

Overview

Recall questions on W = Fs cos θ and why the cosine appears, the work done by a force at right angles and at 180° to the motion, resolving the force against thinking of the displacement, and worked problems on a dragged or pushed load.

What you need to know

Answer each one before you reveal it.

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Numbers

A friction force of 45 N acts on a box pushed 8.0 m along a floor. Calculate the work done by friction.

Key term

Explain the origin of the cos θ factor in W = Fx cos θ.

Key term

What is the work done by a force acting at right angles to the motion?

Questions from this subtopic

Each one with the model answer underneath.

  1. 1

    State the equation for the work done by a constant force acting at an angle to the direction of motion, defining every symbol and its unit.

    Model answer

    The equation is W = Fx cos θ, where W is the work done in J, F is the force in N, x is the distance moved in m and θ is the angle between the force and the direction of motion.

  2. 2

    Explain the origin of the cos θ factor in W = Fx cos θ.

    Model answer

    Only the component of the force along the direction of motion, F cos θ, moves through the distance x, and the perpendicular component does no work at all.

  3. 3

    What is the work done by a force acting at right angles to the motion?

    Model answer

    The work done is zero, because cos 90° is zero and the force has no component along the direction of travel.

  4. 4

    A rope pulls a sledge 15.0 m along level ground with a force of 80.0 N at 25.0° above the horizontal. Calculate the work done by the rope.

    Model answer

    The work done is 80.0 × 15.0 × cos 25.0° = 1.09 × 10³ J. The horizontal component is 80.0 cos 25.0° = 72.5 N, and 72.5 × 15.0 = 1088 J confirms the result.

  5. 5

    What value does cos θ take when a force directly opposes the motion, and what does this mean for the work done?

    Model answer

    The angle is 180° so cos θ is −1, and the work done by the force is negative, meaning energy is taken from the moving body.

  6. 6

    A friction force of 45 N acts on a box pushed 8.0 m along a floor. Calculate the work done by friction.

    Model answer

    Friction acts at 180° to the motion, so the work done by it is 45 × 8.0 × cos 180° = −360 J. The negative sign shows that 360 J is removed from the box.

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