A .235 kg piece of putty is slowly placed on the top of a massless..?

Question

vertical spring of constant, 19.6 N/m - (A) determine the maximum compression of the spring. (B) what is the answer to part A, if the putty is dropped onto the spring from a height of .500 m from the top of the spring?

I need step by step instructions on how the answer for B is done. I have part A just need help step wise to resolve B.

Answer 1

The answer to A is max compression = static compression xs = w/k = mg/k = 0.1175 m (assuming g=9.8). w = 2.303 N. Energy is not conserved in this process because the PE of the spring = w*xs/2 whereas PE of the height change is w*xs. "Slowly" adding the mass prevents any accumulation of KE.
For B, energy is conserved. The putty will hit the spring and deflect it beyond xs to a maximum point xd (d for dynamic) where its KE = 0. Then it will oscillate about xs with amplitude xd-xs.
At xd the PE of the spring = the PE of the entire Δx (xd+0.5) of the mass at xd.
PEmass = w*(xd+0.5) = PEspring = k*xd^2/2
2w(xd+0.5) = k*xd^2
k(xd^2) - 2w(xd) - w = 0 (quadratic)
The positive root is 0.4799 m.
Ans. xd = 0.4799 m
Checking,
PEmass = w*(xd+0.5) = 2.257 Nm
PEspring = k*xd^2/2 = 2.257 Nm
The solution is good.