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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?

## 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 kx

^{2}.

So:

1/2 kx

^{2}= 1/2 mv

_{2}

^{2}+ [tex]\mu[/tex]mg(energy lost due to friction)

But this has 2 unknown variables.

Any help is greatly appreciated.