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 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

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