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

  • Thread starter cupcaked
  • Start date
  • #1
5
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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 = [itex]\sqrt{19.6h}[/itex] ...??

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...?
 

Answers and Replies

  • #2
rl.bhat
Homework Helper
4,433
9
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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