How Is the Coefficient of Kinetic Friction Calculated in a Block-Spring System?

[Quincy]

Homework Statement

A 0.620 kg wood block is firmly attached to a spring (k = 180 N/m). It is noted that the block-spring system, when compressed 0.050 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?

Homework Equations

-(Wnc) = PEf - PEo + KEf - KEo

The Attempt at a Solution

1/2 kx² = 1/2 mv² + umgd

This equation has two unknowns (u and v). I don't know any other equation to solve for the unknowns, so I'm stuck here... What other equation do I use?

 

[marmoset]

I would look at the block at the two times it has zero velocity (i.e. before it is released, and just as it changes direction, call these points A and B), and compare the potential energies stored in the spring. Then work done against friction in going from A to B is the difference in potential energies:

$$W = \frac{k}{2}(0.05^2 - 0.023^2)$$

You know the distance traveled is the distance AB, so using the work calculated in the last line:

$$W = F(0.023 + 0.05)$$

Then you need to work out the coefficient of friction. First, the reaction force R of the table on the block:

$$R = mg$$

Then,

$$F = \mu R$$

I'm not sure whether this is correct.