What is the Velocity of a Mass on an Unstretched Spring System?

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SUMMARY

The discussion focuses on calculating the velocity of a 1 kg mass attached to a spring with a spring constant of 200 N/m when the spring and mass are parallel. The conservation of energy principle is applied, leading to the equation: 1/2 m v^{2}_{1} + mgy + 1/2 K S^{2} = 1/2 m v^{2}_{2} + mgy + 1/2 K S^{2}. The initial conditions are established with initial velocity and height both set to zero. The key challenge is determining the correct values for displacement (S) and height (y) at the moment when the spring is parallel to the mass.

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Bingo1915
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Mass m=1kg
Spring constant k=200 N/m
Unstretched spring length 0.1m

When the system as show in the attatachment is realeased from rest the spring contracts. Use conservation of energy to determine the magnitude of velocity of the mass when the string and spring are parallel.


Relevant equations.

1/2 m v^{2}_{1} + mgy + 1/2 K S^{2} = 1/2 m v^{2}_{2} + mgy + 1/2 K S^{2}



The attempt at a solution.

(1/2 * 1 * 0) + (1 * 9.81 * 0) + (1/2 * 200 * ?) = (1/2 * 1 * v^{2}_{2}) + (1 * 9.81 * ?) + (1/2 * 200 * ?)


I know the initial v=0 and the initial y=0. I'm not sure what I should what numbers to use where the "?" are. These are the final y and the S's. Can anyone help?
 

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Still waiting for the picture... but ill elaborate what I can. To begin with, this problem isn't worded well. Rest can be easily mistaken for equilibrium but it is not in this equation. The S of your equation should be the displacement from equilibrium. Although i can't see the pic, hopefully this should be a little more clear.
Good luck.

Chris
 
I hope you can see the pic. I'm not sure what you would consider equilibrium in the pic. I think it would be:\sqrt{0.45^2 + 0.25^2}. Can you advise?
 

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