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 kx2.
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.