Spring Compression problem- PLEASE SOMEONE HELP PLEASE DX<?

[Lo.Lee.Ta.]

Spring Compression problem!- PLEASE SOMEONE HELP! PLEASE! DX<?

Spring Compression problem!

A 2.4kg block is dropped onto a free-standing spring with k=1100N/m from a height of 1.7m above the spring. What is the spring's maximum compression?

Okay, so I drew a picture of the situation.
Frame #1:
Falling Block: kinetic energy, but no potential energy
Spring: No kinetic energy and no potential energy

Frame #2:
Block has now compressed the spring to the maximum it is able.
Block and Spring: potential energy, but no kinetic energy

(If this reasoning is wrong so far, please let me know what is wrong about it and why! Thanks! :D)

So:
Energy in Frame #1 = Energy in Frame #
(Kf + Uf) = (Ki + Ui)
[0 + 1/2(k)(x^2)] = [1/2(m)(v^2) + 0]
([1/2(1100)(x^2)] = [1/2(2.4)(5.77)^2]
x = .2695m

I found the velocity for this equation by: (vf)^2 = (vi)^2 + 2ad
Vf= 5.77m.s

So that was how I found the max. compression of .2695m...

But in class, my teacher got the answer: .29m
He wrote the equation for this problem as: mgh + mgx = 0 + 1/2(k)(x)^2
How does he have 2 potential energies on the left side of the equation?! The block is moving! Wouldn't there be just a kinetic energy?

Please help me! Can the problem be done as I have done it, or is this wrong?
Please explain WHY it is wrong, if that is so.

Thank you SO much! :D You are awesome!

 

[mishek]

Your teacher observed the block before dropping. In that position his PE was max and his KE was zero. His potential energy is calculated with height that is height of 1.7m above the spring, and the compression of the spring. That means, after the spring compresses, the height of the block changes for height h plus x for the spring compression and therefore mgh + mgx.

 

[Lo.Lee.Ta.]

Thank you for replying! :)
I think I understand what you mean. The problem ends up being:
(2.4)(9.8)(1.7) + (2.4)(9.8)(x) = 1/2(1100)(x^2)
Only after using the quatratic equation do I get .29, so I get this method...
But it seems to me that my method should have worked also... Because energy is the same throughout a system, but it just gets converted into different forms (kinetic or potential).
So I chose: (1)- when the block was falling- and (2)- when the block had compressed the spring to the maximum it was able.
But I think I see my error now! The falling object would actually have BOTH kinetic and potential energy, right?
That makes it too difficult to analyze, so we instead choose points where the block is motionless.
So my biggest mistake was saying that the falling block only has kinetic energy, when it also has potential energy, right?
Is this reasoning right? Thanks for responding! :)

 

[mishek]

But I think I see my error now! The falling object would actually have BOTH kinetic and potential energy, right?
That makes it too difficult to analyze, so we instead choose points where the block is motionless.
So my biggest mistake was saying that the falling block only has kinetic energy, when it also has potential energy, right?
Is this reasoning right? Thanks for responding! :)

Exactly. When your block comes to the top of the spring, it has some kinetic energy and potential energy mgx

You could also calculate it the way you started it. You could observe the situation from start to the end of compression. I that situation you would have:

\frac{1}{2}mv^{2}+mgx=\frac{1}{2}kx^{2}