## SAT Physics Subject Test

**Chapter 4 ****Work, Energy, and Power**

**WORK DONE BY A VARIABLE FORCE**

If a force remains constant over the distance through which it acts, then the work done by the force is just the product of force and distance. However, if the force doesn’t remain constant, then the work done by the force isn’t just a simple product. Focusing only on displacements that are along a straight line (say the *x*-axis), let **F** be a force whose component in the *x* direction varies with position according to the equation **F** = **F**(*x*). If we have a graph of **F** *versus* *x*, then the work done by **F** as it acts from *x* = *x*_{1} to *x* = *x*_{2} is equal to the area bounded by the graph of *F*, the *x*-axis, and the vertical lines *x* = *x*_{1} and *x* = *x*_{2}.

8. The force exerted by a spring when it’s displaced by *x* from its natural length is given by the equation **F**(*x*) = –*kx*, whered *k* is a positive constant. What is the work done by a spring as it pushes out from *x* = –*x*_{2} to *x* = –*x*_{1} (where *x*_{2} > *x*_{1}) ?

Here’s How to Crack It

We sketch the graph of **F**(*x*) = –*kx* and calculate the area under the graph from *x* = –*x*_{2} to *x* = –*x*_{1}.

Here, the region is a trapezoid with area *A* = (base_{1} + base_{2}) × height, so