# A More-Mathematical View

We can generalize the method shown above to derive an equation that gives the work done in stretching a spring a certain distance.

The diagram at right shows a graph of force vs. stretch for a spring. The blue line represents the graph of F = kx (Hooke's Law). We know that the area shaded in red represents the work you would need to do to stretch the spring a distance x from its rest position.

Since this area is a triangle, the shaded area = (1/2)(base)(height) = (1/2)(x)(kx) = (1/2)kx2.

Therefore, the work done in stretching a spring a distance x from its rest position is:

## Examples:

Example 1: If the force to stretch a spring is given by F = (100 N/m)x, how much work does it take to stretch the spring 4 meters from rest?

Solution: We have k = 100 N/m and x = 4 m. Therefore, Work = (1/2)kx2 = (1/2)(100 N/m)(4 m)2 = 800 Joules

Example 2: A spring is stretched 60 cm, and it takes a force of 40 Newtons to hold it there. How much work did it take to stretch the spring to this point?

Solution: We know that F = 40 N when x = 60 cm = 0.60 m. Since F = kx, k = F/x = (40 N)/(0.60 m) = 67 N/m. Now, work = (1/2)kx2 = (1/2)(67 N/m)(0.60 m)2 = 12 Joules

last update December 11, 2003 by JL Stanbrough