Answer to Question #260069 in Optics for haze

Question

A block attached to a spring of the spring constant k = 400 N/m is pulled to stretch the spring by 0.40 m from the equilibrium point. Next moment, the block is released. When the block passes the spring’s equilibrium point, what is the block’s kinetic energy in J?

Answer 1

$k = 400N/m$

$x = 0.40m$

$KE = \frac{1}{2}kx^{2}$

$KE = \frac{1}{2}(400N/m)(0.40m)^{2} \\ KE = 32.00J$