The potential energy of a spring is given by the equation \(PE=\dfrac{kx^2}2\) where \(k\) is a constant and \(x\) is the distance the spring is stretched. If \(K =16\) and the spring is stretched to \(2\) feet and then to \(3\) feet, how much potential energy is gained by the spring from the moment it is stretched \(2\) feet to the moment it is stretched \(3\) feet?

A. 200

B. 128

C. 72

D. 40

E. 32

Answer: D

Source: Veritas Prep