calculus/physics
posted by deb .
A force of 10 pounds stretches a spring 2 inches. Find the work done in stretching this spring 3 inches beyond its natural length

Assuming an ideal spring (linear) and using Hooke's law, the force generated by the spring is:
F = kx (1)
where x is the distance the spring is stretched from the rest position. So, substituting the info into (1) you get:
10 = k(2)
or
k = 5 lb/in
work is the 1st integral of force with respect to x, which would be:
W = (1/2)kx^2
= (1/2)5(3)^2
= 22.5 inlb