Lectures on Physics has been derived from Benjamin Crowell's Light and Matter series of free introductory textbooks on physics. See the editorial for more information....

# Varying Force

Up until now we have done no actual calculations of work in cases where the force was not constant. The question of how to treat such cases is mathematically analogous to the issue of how to generalize the equation (distance) = (velocity)(time) to cases where the velocity was not constant. There, we found that the correct generalization was to find the area under the graph of velocity versus time. The equivalent thing can be done with work:

#### general rule for calculating work

The work done by a force F equals the area under the curve on a graph of FII versus x. (Some ambiguities are encountered in cases such as kinetic friction.)

The examples in this section are ones in which the force is varying, but is always along the same line as the motion, so F is the same as FII.

 Self-Check In which of the following examples would it be OK to calculate work using Fd, and in which ones would you have to use the area under the F - x graph? (a) A fishing boat cruises with a net dragging behind it. (b) A magnet leaps onto a refrigerator from a distance. (c) Earth's gravity does work on an outward-bound space probe. Answer, p. 158

An important and straightforward example is the calculation of the work done by a spring that obeys Hooke's law,

The minus sign is because this is the force being exerted by the spring, not the force that would have to act on the spring to keep it at this position. That is, if the position of the cart in the figure below is to the right of equilibrium, the spring pulls back to the left, and vice-versa.

 The spring does work on the cart. (Unlike the ball in section 3.1, the cart is attached to the spring.)

We calculate the work done when the spring is initially at equilibrium and then decelerates the car as the car moves to the right. The work done by the spring on the cart equals the minus area of the shaded triangle, because the triangle hangs below the x axis. The area of a triangle is half its base multiplied by its height, so

This is the amount of kinetic energy lost by the cart as the spring decelerates it.

 The area of the shaded triangle gives the work done by the spring as the cart moves from the equilibrium position to position x.

It was straightforward to calculate the work done by the spring in this case because the graph of F versus x was a straight line, giving a triangular area. But if the curve had not been so geometrically simple, it might not have been possible to find a simple equation for the work done, or an equation might have been derivable only using calculus. Optional section 3.4 gives an important example of such an application of calculus.

 Energy production in the sun.

Last Update: 2009-06-21