Work done by a constant force

The work \(W\) done on a system by another object, system or field exerting a constant force on the system is the product of the magnitude \(F\) of the force, the magnitude \(\Delta r\) of the displacement of the point of application of the force, and \(\cos\theta\), where \(\theta\) is the angle between the force and displacement vectors:

\[W := F \Delta r\cos\theta.\]

Using a more concise mathematical definition, we can say that work is the scalar product of the force vector and the displacement vector:

\[W := \mathbf{F} \cdot \Delta\mathbf{r}.\]

Work is always associated with a force.

Work is a scalar quantity.

The SI unit of work is the joule:
\[\left[W\right] = 1\,\textrm{N} \cdot \textrm{m} = 1\,\mathrm{kg} \cdot \frac{\mathrm{m}}{\mathrm{s}^2} = 1\,\textrm{J}.\]

The above definition of work can be interpreted in two ways:

the product of the magnitude of the force (\(F\)) and the displacement component in the direction of the force (\(\Delta r \cos\theta\));
the work of a force can also be defined as the product of the displacement (\(\Delta r\)) and the force component acting in the direction of the displacement (\(F \cos\theta\)).

Direction of force and displacement

When the force is perpendicular to displacement (\(\theta = 90°\)) – see \(\mathbf{F}_{\mathrm{N}}\) and \(m\mathbf{g}\) in the figure below – no work is done:
\[W = F \Delta r \cos90° = F \Delta r \cdot 0 = 0.\]

The normal force \(\mathbf{F}_{\mathrm{N}}\) and the weight \(m\mathbf{g}\) do no work on the object. In the situation shown here, \(\mathbf{F}\) is the only force doing work on the object.

When the force component parallel to the displacement is in the same direction as the displacement (\(0 \leq \theta < 90°\)) – see the figure below – the force contributes to the displacement and the work is positive:

\[\textrm{if } 0 \leq \theta < 90°:\ \cos\theta > 0, \Rightarrow W > 0.\]

When the force component parallel to the displacement is in the direction opposite the displacement (\(90° < \theta \leq 180°\)) – see the figure below – the force hinders the displacement and the work is negative:

\[\textrm{if } 90° < \theta \leq 180°:\ \cos\theta < 0, \Rightarrow W < 0.\]