A 1.00 kg object is attached to a horizontal spring. The spring is initially stretched by 0.400 m, and the obj

Question

A 1.00 kg object is attached to a horizontal spring. The spring is initially stretched by 0.400 m, and the object is released from rest there. It proceeds to move without friction. The next time the speed of the object is zero is 0.400 s later. What is the maximum speed of the object?

Answer 1

The energy put into the system is

W = Fd = mgd = 1 kg * 9.81 m/s^2 * 0.4 m = 39.24 J.

The stored energy becomes kinetic energy when the object is released.

Et = total energy in the system.

Et = m(vo)²/2  (vo = maximum velocity).

vo = √2Ek/m

vo = √(2 * 39.24 J)/1 kg

vo = 8.86 m/s