# What is the work required to stretch the spring from a length of 21 inches to a length of 26 inches if a spring has a natural length of 18 inches and a force of 20 lbs is required to stretch and hold the spring to a length of 24 inches?

Mar 27, 2015

This is quite a good one!
It involves the elastic force which is given by Hooke's Law as ${F}_{e l} = - k x$ where $x$ is the elongation and $k$ is the spring constant (and the negative sign indicates that is a force always opposite to the displacement from equilibrium!).
But this is a VARIABLE force (depends on $x$) so for the work you have to integrate and get:
${W}_{e l} = {\int}_{i}^{f} F \mathrm{dx} = {\int}_{i}^{f} - k x \mathrm{dx} = {\left[- k {x}^{2} / 2\right]}_{i}^{f} =$
$- \frac{k}{2} \left({f}^{2} - {i}^{2}\right) = - \frac{k}{2} \left({26}^{2} - {21}^{2}\right)$ (I)

Here $i = 21$ inches and $f = 26$ inches

You can find $k$ from the second part of your problem where you have:
$- k \cdot \left(24 - 18\right) = - 20 = {F}_{e l}$ It is $- 20$ because you immagine the stretching (from a force) in the positive $x$ direction so that your ${F}_{e l}$ is in the negative $x$ direction.
So $k = \frac{20}{6} = \frac{10}{3}$ lbs/inches that substiituted in the expression for work (I) gives:
${W}_{e l} = - 391.67$ lbs.inches
This is the work done by the elastic force. The work done by you to stretch the spring is the same but with opposite sign.

Please check my maths and my units as I am...metric!!!
:-)