A 0.50-kg block attached to an ideal spring with a spring constant of 80 N/m oscillates on a horizontal frictionless surface. When the spring is 4.0 cm longer than its equilibrium length, the speed of the block is 0.50 m/s. The greatest speed of the block is:

Question

Quizplus · Accepted Answer

The greatest speed of the block occurs when the spring is at its equilibrium length. The potential energy stored in the spring at this point is equal to the kinetic energy of the block. Therefore, the greatest speed of the block is:$$v = \sqrt{\frac{2k}{m}} = \sqrt{\frac{2(80 N/m)}{0.50 kg}} = 0.78 m/s$$

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