Q.A block of mass $2\ \text{kg}$, initially at rest on a frictionless horizontal surface, is acted on by a variable force $F(x) = 3x^2\ \text{N}$ (with $x$ in metres) directed along the positive $x$-axis, from $x = 0$ to $x = 2\ \text{m}$. Find the work done by the force and the speed of the block at $x = 2\ \text{m}$.

When either the magnitude of the force or the angle it makes with the displacement changes as a body moves -- a spring that pulls harder the more it stretches is the standard example -- the work done can no longer be found by simple multiplication. The small work done over an infinitesimal displacement is still $dW=(F\cos\theta)\,dr$, but now $F\cos\theta$ is itself a function of position, so the total work must be obtained by integrating this small element over the whole path:

Graphically, this is the area under a (generally curved, not rectangular) plot of $F\cos\theta$ against $r$. A representative worked case is a force that grows with position as $F=kx^2$: integrating $kx^2\,dx$ from an initial to a final position gives the work done, exactly analogous to how the elastic potential energy of a stretched spring, $U=\tfrac12kx^2$, is obtained by integrating the variable spring force $F=kx$ from zero extension up to $x$. Recognising whether a given force is constant or variable -- and hence whether a straightforward multiplication or an integral is required -- is the first and most important step in any work calculation.