Question

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?

Answer 1

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).