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
Aug 21, 2014

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

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).