Block dropped onto vertical spring, finding work/speed/height

AI Thread Summary
The discussion focuses on a physics problem involving a 0.25 kg block dropped onto a vertical spring with a spring constant of 250 N/m, which compresses 0.12 m before stopping. The work done by gravity during compression is calculated as 0.294 J, while the work done by the spring is found to be -1.8 J. The speed of the block just before impact is derived from the conservation of mechanical energy, leading to the equation v = √(19.6h). The final part of the problem involves equating the potential energy lost by the block to the energy stored in the spring to find the height from which the block was released. The discussion highlights the need for clarity in solving the last part of the problem.
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I need some help with this problem, to check if my answers and formulas are right/how to solve the last parts...

A 0.25kg block is dropped onto a vertical spring with spring constant k= 250 N/m. The block becomes attached to the spring and the spring compresses 0.12m before coming momentarily to rest.

a) While the spring is being compressed what is the work done by the force of gravity?
W = Fdcosθ
W = (9.8 * 0.25)(0.12)(cos0)
W = 0.294 J

b) What is the work done by the spring?
W = -1/2kx^2
W = -1/2(250)(0.12)^2
W = 1.8 J

c) What was the speed of the block just before it hits the spring?
ME = KE + PE
MEi = (1/2)(0.25)(0 + (0.25)(9.8)(h)
MEi = 2.45h

MEf = (1/2)(0.25)(v^2) + (0.25)(9.8)(0)
MEf = 0.125v^2

MEi = MEf
2.45h = 0.125v^2
v^2 = 19.6h
v = \sqrt{19.6h} ...??

d) From what height was the block released relative to the top of the spring?
I'm not sure how to do this, maybe I did the last part wrong or I keep overlooking something stupid...?
 
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Fall in PE of the block = mg(h+x)
Energy stored in the spring = 1/2*k*x^2
Equate them and solve for h.
 
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