Work with variable forces

Now what if the force is variable, that is it depends on position? For example, for a spring $F(x) ~=~ -kx$, so the value of the force is not constant but depends on position. Can we figure out what the work was in this case? Of course, I wouldn't have mentioned otherwise.

So look at the diagram below. We're applying a force to the spring in order to move it from position $x_i$ to $x_f$.

The red curve is the force as a function of position $x$. Let's calculate the total work in going from $x_i$ to $x_f$. Well we know how to get the answer if the force is constant, but here its clearly varying. So we'll try to get an approximate answer by replacing the smooth curve in red by a the staircase shown in blue. We're replacing it with this because we know how to calculate the work in going along one of the steps. In the diagram we see the work done in going along between $x_j$ and $x_{j+1}$ can be calculated because the force is constant in that interval. It's just $F_j \Delta x$. So to get the total work we have to sum up over all these intervals:

$$W ~ \approx ~ \sum_j \Delta x F_j$$ (1.5)

Now we'll do the old calculus trick of taking the limit as $\Delta x \rightarrow 0$. In this limit we get an integral so the work becomes

$$W ~=~ \int_{x_i}^{x_f} F(x) dx$$ (1.6)

This can be interpreted, as usual, as the area under the curve $F(x)$.

In the case of the spring, the force $F = -kx$ so the work done by the spring in moving an object from $x_i$ to $x_f$ is

$$W ~=~ - {1\over 2}k (x_f^2-x_i^2)$$ (1.7)