- #1

cupcaked

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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...?