A 0.50-kg block attached to an ideal spring with a spring constant of 80 N/m oscillates on a horizontal frictionless surface.The total mechanical energy is 0.12 J.The greatest extension of the spring from its equilibrium length is: A)1.5 *10^-3 m B)3.0*10^-3 m C)0.039 m D)0.055 m E)18 m

Question

Quizplus · Accepted Answer

The maximum displacement of the block from equilibrium is given by:
$x_{max}=\sqrt{\frac{2E}{k}}=\sqrt{\frac{2(0.12	ext{ J})}{80	ext{ N/m}}} = 0.055	ext{ m}$

Therefore, the correct choice is (D).

A 050-Kg Block Attached to an Ideal Spring with a Spring