Gravitational Potential Energy, Elastic Potential Energy, and Kinetic Energy


by PeachBanana
Tags: elastic, energy, gravitational, kinetic, potential
PeachBanana
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#1
Mar1-12, 07:25 AM
P: 191
1. The problem statement, all variables and given/known data

A 4.0 kg mass is pressed down on a vertical spring of spring constant 400 N/m, compressing it to 0.250 m. After it is released, the amount of kinetic energy this mass would have when it leaves the spring is ___.

2. Relevant equations

mgy(final) + 1/2 kx^2 (final) + 1/2 mv^2 (final) = mgy (initial) + 1/2kx^2 (initial) + 1/2 mv^2 (initial)

3. The attempt at a solution

Ok so I think what I'm solving for is 1/2 mv^2 (final)

(4.0 kg)(9.8m/s^2)(0m) + 1/2 (400 N/m)(0m)^2 + 1/2(4.0 kg)(v)^2 = (4.0 kg)(9.8 m/s^2)(-0.250 m) + 1/2(400 N/m)(-0.250m)^2 + 1/2(4.0 kg)(0 m/s)^2

That was sort of long so to simplify it a bit:

1/2(4.0 kg)(v)^2 = (4.0 kg)(9.8m/s^2)(-.250m) + 1/2(400 N/m)(-0.250m)^2 + 1/2(4.0 kg)(0 m/s)^2

My main concern: My "x" and "y" are the same. Is that because it's a vertical spring? I'm calling the end of the release x=0 m and y=0 m and the compression -0.250 m.
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BruceW
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#2
Mar1-12, 08:59 AM
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you have done it all correctly. Technically speaking, x is the displacement from the equilibrium point. So you can think of it as Δx = x2-x1, where the x1 is the equilibrium point. We also know that the displacement Δx is purely vertical, so what is the (nice and simple) relationship between Δx and Δh?
PeachBanana
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#3
Mar1-12, 09:37 AM
P: 191
Ok, so Δx and Δh in this case are the same; that makes sense.

BruceW
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Mar1-12, 12:19 PM
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Gravitational Potential Energy, Elastic Potential Energy, and Kinetic Energy


yep, that's right. Often you'll find that its the change in height that is important in questions with uniform gravity.


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