Work done is the energy transferred when a force moves an object

Work, Energy & Power — AQA A-Level Physics

Key Definition

Work done — The amount of energy transferred when an external force causes an object to move over a certain distance. W = Fs. Unit: joule (J). 1 J = 1 N m.

$$W = QV$$

$$W = Fs$$

$W$: work doneEnergy transferred when a force moves an object. In electrical circuits, W = QV (chargeA property of matter that causes it to experience a force in an electromagnetic field. Measured in coulombs (C). times potential difference). (J)
$F$: force applied parallel to displacementThe distance moved in a particular direction from a starting point. A vector quantity. Measured in metres (m). (N)
$s$: displacementThe distance moved in a particular direction from a starting point. A vector quantity. Measured in metres (m). (m)

The force must be in the direction of the displacementThe distance moved in a particular direction from a starting point. A vector quantity. Measured in metres (m). for $W = Fs$ to apply directly.
If the force is at angle theta to the displacement, use $W = Fs \cos(\theta)$.
The component of force perpendicular to the displacement does no work.
Work doneEnergy transferred when a force moves an object. In electrical circuits, W = QV (chargeA property of matter that causes it to experience a force in an electromagnetic field. Measured in coulombs (C). times potential difference). against friction is transferred to thermal energyThe capacity to do work. Measured in joules (J). and sound.

W = Fs \cos \theta

$W$: work doneEnergy transferred when a force moves an object. In electrical circuits, W = QV (chargeA property of matter that causes it to experience a force in an electromagnetic field. Measured in coulombs (C). times potential difference). (J)
$F$: applied force (N)
$s$: displacement (m)
$θ$: angle between force and displacement

Worked Example

A barrel of weight $2.5 \times 10^{3}$ N sits on a frictionless slope at 40 degrees. A force parallel to the slope pushes it 6.0 m up the slope at constant speed. Calculate the work done.

Show Solution

Find the force parallel to the slope

The force must overcome the component of weight parallel to the slope:

$$F = W \sin 40° = 2.5 \times 10^3 \times \sin 40° = 1607 \text{ N}$$

Calculate work done

$$W = F \times d = 1607 \times 6.0 = 9.6 \times 10^3 \text{ J}$$

Answer

$W = 9.6 \times 10^3$ J

Common Mistake MEDIUM

Students often: Using the total force rather than the component parallel to the displacement.
Instead: Only the component of force in the direction of displacement does work. Resolve the force first if it is at an angle.