I am having some trouble with conservation of energy

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SUMMARY

The discussion centers on the application of the conservation of energy principle in a mechanics problem involving kinetic energy, potential energy, and work done by friction. The equation used is K1 + U1 + Wnc = K2 + U2, where the user attempts to solve for kinetic energy after a mass compresses a spring. The key mistake identified is the incorrect assumption regarding the work done by friction and the neglect of the condition that the object's speed is zero when the spring returns to its uncompressed state. This insight clarifies the dependency of the solution on the variable 'x' and corrects the user's approach.

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kyin01
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Homework Statement


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Homework Equations


[tex]K_{1}[/tex]+[tex]U_{1}[/tex]+[tex]W_{nc}[/tex]=[tex]K_{2}[/tex]+[tex]U_{2}[/tex]


The Attempt at a Solution


Using the above equation

1/2m[tex]v^{2}[/tex] + 0 + xmg[tex]\mu[/tex] = 0 + 1/2k[tex]x^{2}[/tex]

On the left side
When it hits the spring there is 0 potential because it has not yet compress the spring and it still remains on a horizontal surface. It has a kinetic energy and kinetic friction work
On the right side
I have 0 kinetic because it just finished compressing the spring and about to decompress so instead I get maximum potential energy.

So solving for K i get

[tex]\frac{mv^{2}+xmg\mu}{x^{2}}[/tex]

But according to the answer it does not depend on the variable X (distance compressed by spring)
So I'm not entirely sure I know where I made my mistake
 
Last edited:
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Aside from your algebra error and your incorrect assumption that the work done by friction is positive, you have neglected to incorporate the given condition that the object has no speed when the spring returns to its uncompressed length (that is, its speed is zero both when the spring is fully compressed, and when it returns to its uncompressed state). That will give you the info you need to eliminate the 'x' term. Remember that the work done by friction is negative.
 

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