Calculating Max Compression of Massless Hookslaw Spring

AI Thread Summary
The discussion focuses on calculating the maximum compression of a massless Hooke's law spring when a mass is dropped onto it from a height. The energy conservation equation mgh = -mgs + 0.5ks² is central to the problem, where the variables represent gravitational potential energy and spring potential energy. Participants discuss deriving the spring constant k in terms of other parameters to facilitate solving for maximum compression. There is confusion about the relevance of the spring constant and its application in the problem. Understanding the spring constant's role can help clarify the relationship between the mass, height, and spring compression.
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Homework Statement



A massless hooksw law spring has a unstretched length of L, when a mass m is ploaced on it, and slowly lowered until the mass is at rest, the spring is queezed to a length of x. the mass is then dropped on the spring from a height of h, use energ methods

what is the length of the spring at maximum compression as a result of the mass dropping on it

Homework Equations





The Attempt at a Solution



mgh = -mgs + .5ks2

s = -mg + or -(((mg)2 - 4(.5k)(mgh))/k).5

if this is correct
 

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The question is worded aquardly, but I think you're on the right path with your firse equation. Now you need to write the spring constant k in terms of other parameters you have in your problem.

Also, you get two anilytical solutions. Can you think which sollution is correct? What does your non-physical solution correspond do?
 
what would be the benefit from solving for the spring constant

it asks for the length of the spring at compression s
 
joemama69 said:
what would be the benefit from solving for the spring constant

That way the spring constant is expressed in terms of other known parameters.
 
k = (2mgs - 2mgh)s2 = 2mg(s-h)/s2

i still don't understand how that helps, what would i use it for now
 
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