Conservation of energy involving spring and friction

  • #1
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Homework Statement



A 0.62 kg wood block is firmly attached to a very light horizontal spring (k = 180 N/m) . This block-spring system, when compressed .05 m and released, stretches out 0.023 m beyond the equilibrium position before stopping and turning back.

What is the coefficient of kinetic friction between the block and the table?

GIANCOLI.ch08.p037.jpg


The Attempt at a Solution



This is how far I've gotten:

At x = -.05, there is 0 kinetic energy, and potential energy = 1/2 kx2.

So:

1/2 kx2 = 1/2 mv22 + [tex]\mu[/tex]mg(energy lost due to friction)

But this has 2 unknown variables.

Any help is greatly appreciated.
 

Answers and Replies

  • #2
190
0
Nevermind, figured it out.

1/2 kxinitial2 = 1/2 kx2final + [tex]\mu[/tex]mg(abs((xfinal - xinitial))
 

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