# 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?

##### 1 Answer

The answer is

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

#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

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