# 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?

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The answer is ##(27)/4## ft-lbs.

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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?
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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).

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