# Use Hooke's Law to determine the variable force in the spring problem. A force of 20 pounds stretches a spring 9 inches in an exercise machine. Find the work done in stretching the spring 1 foot from its natural position. Hooke's law states that a force is needed to stretch or compress a spring by a distance of x. The force is proportional to the distance x. It is written as `F = kx`

where:

`F `  = force

`k` = proportionality constant or spring constant

`x` = length displacement from its natural length

Applying the given variable force: `F= 20` pounds  to stretch a spring a total of `9` inches, we get:

`F=kx`

`20=k*9`

`k=20/9`

Plug-in `k =20/9` on Hooke's law , we get:

`F = (20/9)x`

Work can be define with formula: `W = F*Deltax`  where:

`F ` = force or ability to do work.

`Deltax` = displacement of the object’s position

With force function: `F(x)= (20/9)x ` and condition to stretch the spring by 1 foot (or 12 inches) from its natural position, we set-up the integral application for work as:

`W = int_a^b F(x) dx`

`W = int_0^12 (20/9)xdx`

Apply basic integration property: `int c*f(x)dx= c int f(x)dx` .

`W = (20/9)int_0^12 xdx`

Apply Power rule for integration: `int x^n(dx) = x^(n+1)/(n+1)` .

`W = (20/9) * x^(1+1)/(1+1)|_0^12`

`W = (20/9) * x^2/2|_0^12`

`W = (20x^2)/18|_0^12`

`W= (10x^2)/9|_0^12`

Apply definite integral formula: `F(x)|_a^b = F(b)-F(a)` .

`W = (10(12)^2)/9-(10(0)^2)/9`

`W =160 - 0`

`W=160` inch-lbs