The potential energy of a spring is given by the equation$PE=\dfrac{kx^2}{2}$ where k is a constant and...

Source: Veritas Prep

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

The OA is D

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