Question

A 3.7-kg block on a horizontal frictionless surface is attached to an ideal spring whose force constant (spring constant) is 450 n/m. the block is pulled from its equilibrium position at x = 0.000 m to a position x = +0.080 m and is released from rest. the block then executes simple harmonic motion along the horizontal x-axis. the maximum elastic potential energy of the system is closest to

Answer 1

The maximum elastic potential energy of the system occurs when the spring is at its largest displacement from its rest position, i.e. at $\Delta x = 0.08 m-0 m=0.08 m$. In this configuration, the elastic potential energy is given by
$U= \frac{1}{2}k(\Delta x)^2= \frac{1}{2}(450 N/m)(0.08m)^2=144 J$