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Lo.Lee.Ta.
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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!
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!