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