Gravitational Potential Energy, Elastic Potential Energy, and Kinetic Energyby PeachBanana Tags: elastic, energy, gravitational, kinetic, potential 

#1
Mar112, 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. 



#2
Mar112, 08:59 AM

HW Helper
P: 3,339

you have done it all correctly. Technically speaking, x is the displacement from the equilibrium point. So you can think of it as Δx = x_{2}x_{1}, where the x_{1} 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?




#3
Mar112, 09:37 AM

P: 191

Ok, so Δx and Δh in this case are the same; that makes sense.



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