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

AI Thread 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 is unstretched and the system is released from rest. The conservation of energy equation is provided, but the user is uncertain about how to apply the variables for displacement and height in the equation. Clarification is sought regarding the correct values to use for the final position of the mass and the spring's displacement from equilibrium. The user also notes confusion regarding the wording of the problem and the definition of equilibrium in this context. Overall, the thread emphasizes the need for guidance on applying conservation of energy principles to this spring-mass system.
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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