Use Hooke's Law to determine the variable force in the spring problem. A force of 5 inches compresses a 15-inch spring a total of 3 inches. How much work is done in compressing the spring 7 inches?

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. 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= 5  to compress a 15 -inch spring a total of 3 inches, we get:
F=kx
5=k*3
k=5/3
Plug-in k =5/3 on Hooke's law, we get:
F = (5/3)x
Works is done when a force is applied to move an object to a new position, It can be defined with formula: W = F*Deltax  where:
 F = force or ability to do work.
Deltax = displacement as 
With  variable force function: F (x)= (5/3)x , we set-up the integral application for work as:
W = int_a^b F(x) dx
W = int_0^7 (5/3)xdx
Apply basic integration property: int c*f(x)dx= c int f(x)dx.
W = (5/3)int_0^7 xdx
 
Apply Power rule for integration: int x^n(dx) = x^(n+1)/(n+1).
W = (5/3) * x^(1+1)/(1+1)|_0^7
W = (5/3) * x^2/2|_0^7
W = (5x^2)/6|_0^7
Apply definite integral formula: F(x)|_a^b = F(b)-F(a) .
W = (5(7)^2)/6 -(5(0)^2)/6
W =245/6 - 0
W=245/6 or 40.83 inch-lbs