# If the work required to stretch a spring 1 foot beyond its natural length is 12 foot-pounds, how much work is needed to stretch it 9 inches beyond its natural length?

Let’s look at the integral for work (for springs):

##W=int_a^b kx dx = k int_a^b x dx ##

Here’s what we know:

##W=12## ##a=0## ##b=1##

So, let’s substitute these in:

##12=k[(x^2)/2]_0^1## ##12=k(1/2-0)## ##24=k##

Now:

9 inches = 3/4 foot = ##b##

So, let’s substitute again with ##k##:

##W=int_0^(3/4) 24xdx## ##=(24x^2)/2|_0^(3/4)## ##=12(3/4)^2## ##=(27)/4## ft-lbs

Always set up the problem with what you know, in this case, the integral formula for work and springs. Generally, you will need to solve for ##k##, that’s why ##2## different lengths are provided. In the case where you are given a single length, you’re probably just asked to solve for ##k##.

If you are given a problem in metric, be careful if you are given mass to stretch or compress the spring vertically because mass is not force. You will have to multiply by 9.8 ##ms^(-2)## to compute the force (in newtons).