I am having some trouble with conservation of energy

  • Thread starter kyin01
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Homework Statement


2-5.png



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:

Answers and Replies

  • #2
PhanthomJay
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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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