Work done by a varying force

If the force varies with the displacement, we cannot apply the definition above directly because that is only valid when the force is constant in magnitude and direction.

Consider a particle being displaced along the \(x\) axis under the action of a force that varies with position. In this case, we can break down the total displacement into intervals so small that within such an interval (\(\Delta x\)) the force changes so little that it can be approximated as constant. Within such a small interval \(\Delta x\), the work of the force can be approximated as

\[\Delta W(x) \approx F_x(x)\Delta x.\]

The total work over the full displacement can be obtained as the sum of the work values within the small intervals:
\[W \approx \sum\limits_{x = x_0}^{x_1}\Delta W(x) = \sum\limits_{x = x_0}^{x_1} F_x(x)\Delta x.\]

If we refine the resolution \(\Delta x\) of the displacement, what we get is that the work of a force \(F_x\) that varies with the position \(x\) is the area under the \(F_x(x)\) curve from the initial position \(x_0\) to the final position \(x_1\):

The rigorous mathematical operation corresponding to the area under the curve is integration. If we make the interval length \(\Delta x\) approach zero:

\[W = \lim\limits_{\Delta x \to 0} \sum\limits_{x = x_0}^{x_1}{F_x(x)\Delta x} = \int_{x_0}^{x_1} F_x(x)\mathrm{d}x.\]

The work done by a varying force

The work done by a one-dimensional force \(F_x\) that varies with the coordinate \(x\) is the integral of the force with respect to the position between the initial and the final coordinate:

\[W = \int_{x_0}^{x_1} F_x(x) \mathrm{d}x.\]

In three dimensions, we have to integrate the force vector as a scalar product:

\[W = \int_{\mathbf{r}_0}^{\mathbf{r}_1} \mathbf{F}(\mathbf{r}) \mathrm{d}\mathbf{r}.\]

Remember, the intuitive understanding of the integral is the ‘area under the curve’.