Work done
Work, Energy & Power - OCR A-Level Physics
Key Definition
Work done
The energy transferred when a force causes an object to move through a displacement in the direction of the force. Measured in joules (J), where $1 \text{ J} = 1 \text{ N m}$.
The energy transferred when a force causes an object to move through a displacement in the direction of the force. Measured in joules (J), where $1 \text{ J} = 1 \text{ N m}$.
$$W = Fx\cos\theta$$
Variables
$W$ is the work done in J, $F$ is the constant force in N, $x$ is the distance moved in m, and $\theta$ is the angle between the force vector and the displacementThe distance moved in a particular direction from a starting point. A vector quantity. Measured in metres (m). vector. The cosine resolves the force into the component along the direction of motion.
- Force parallel to displacement ($\theta = 0^\circ$): $W = Fx$. Pushing a trolley 4 m with a horizontal 20 N force does $80 \text{ J}$ of work.
- Force perpendicular to displacement ($\theta = 90^\circ$): $W = 0$. A satellite in a circular orbit has gravity acting perpendicular to its motion, so gravity does no work and the orbital speed is constant.
- Force opposing displacement ($\theta = 180^\circ$): $W$ is negative. Friction acting backwards on a sliding box removes energy from the box.
- The equation assumes $F$ is constant. For a varying force, use the area under a force-displacement graph.
- Work against friction or air resistance is transferred to the thermal energyThe energy stored in a system due to the random motion of its particles; it is related to temperature. store of the object and its surroundings.
Diagram pending
A sledge on flat ground pulled by a rope at angle $\theta$ above the horizontal. Arrow labelled $F$ along the rope. Horizontal arrow labelled $F\cos\theta$ shows the component along the displacement $x$. The vertical component $F\sin\theta$ is shown but labelled as doing no work.
Will be replaced with a GeoGebra SVG in stream 2.
Common Mistake
MEDIUM
Wrong: Using the full magnitude of the force when it acts at an angle to the displacement, writing $W = Fx$ instead of $W = Fx\cos\theta$.
Right: Only the component of force parallel to the displacement does work. Identify $\theta$ as the angle between the force and the direction of motion, then multiply by $\cos\theta$.
Right: Only the component of force parallel to the displacement does work. Identify $\theta$ as the angle between the force and the direction of motion, then multiply by $\cos\theta$.
Worked Example [3 marks]
A child pulls a sledge $50 \text{ m}$ across flat ground using a rope at $30^\circ$ to the horizontal. The tension in the rope is $40 \text{ N}$. Calculate the work done by the tension.
Show Solution
1
Identify the values
$F = 40 \text{ N}$, $x = 50 \text{ m}$, $\theta = 30^\circ$.
[1]2
$W = Fx\cos\theta = 40 \times 50 \times \cos 30^\circ$.
[1]3
$W = 40 \times 50 \times 0.866 = 1732 \text{ J} \approx 1700 \text{ J}$ (2 s.f.).
[1]Answer
$1700 \text{ J}$
Examiner Tips and Tricks
- The angle $\theta$ in the formula is the angle between the force and displacement vectors, not the angle to the ground. Always draw a quick sketch.
- If asked for the work done by a specific named force (gravity, friction, tension), use only that force in $F$, not the resultant.