2.5 Physical Applications 

[STARTS AT 13:31]

PROFESSOR: Suppose it takes a force of 8 pounds to stretch a spring 6 inches from the equilibrium position. How much work is done to stretch the spring 1 foot from of the equilibrium position? So let's see what we've got here. So, first, we need to find our force function. So our force is kx. So the force is 8 pounds. 

And I'm going to go ahead and convert this to-- from 6 inches into feet because that is the standard that we're going to use. So this is 0.5 times k. Solving that, our spring constant should be 16. So if our spring constant is 16, our function is f of x equals 16x. 

Now, to find our work, that is the integral from 0-- and we are stretching 1 foot. So the integral from 0 to 1 of our force 16x-- that's our force function-- times our distance, which is dx. 

Finding the antiderivative there, we get 8x squared evaluated from 0 to 1, which is going to be 8. So because we were dealing with feet and pounds, our answer is in foot pounds. 

[ENDS AT 15:03]