A 500 g block is released from rest and slides down a frictionless track that begins 2.00 m above the horizontal. At the bottom of the track, where the surface is horizontal, the block strikes and sticks to a light spring with a spring stiffness constant 20.0 N/m. Find the maximum distance the spring is compressed.
The Attempt at a Solution
1/2(20) A^2= 1/2(.5)(6.26)^2
A= .99 m
I did this because I figured that if I found the velocity when the box first hit the spring, it would be considered the maximum velocity since the spring is at equilibrium at that point in time. Then, since it asked for the maximum distance it would be compressed, I just found the amplitude. Is this right? I really hope so because that is what I did on my test...Thanks for the help.