Conservation of energy involving spring and friction

[IniquiTrance]

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 kx².

So:

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

But this has 2 unknown variables.

Any help is greatly appreciated.

 

[IniquiTrance]

Nevermind, figured it out.

1/2 kx_initial² = 1/2 kx²_final + $$\mu$$mg(abs((x_final - x_initial))

 

[thomate1]

Your approach is correct so far. In this problem, conservation of energy can be applied to find the unknown coefficient of kinetic friction. As you have correctly identified, at the compressed position of -0.05 m, the block has no kinetic energy but has potential energy stored in the spring. At the equilibrium position, the block has both kinetic and potential energy. However, as the block moves beyond the equilibrium position, it will lose some of its energy due to friction with the table. This energy loss can be accounted for in the equation you have set up.

To solve for the coefficient of kinetic friction, you will need to use the fact that at the maximum displacement (0.023 m), the block has come to a stop and has no kinetic energy. This means that all of the energy it had at the equilibrium position has been lost due to friction. Therefore, you can equate the potential energy at the equilibrium position to the energy lost due to friction at the maximum displacement:

1/2 k(0.023)^2 = \mu mg(0.023)

Solving for the coefficient of kinetic friction, we get:

\mu = (1/2 k(0.023)^2)/(mg(0.023))

Substituting in the given values, we get:

\mu = (1/2 * 180 * (0.023)^2)/(0.62*9.8*(0.023))

\mu = 0.071

Therefore, the coefficient of kinetic friction between the block and the table is approximately 0.071. This means that the friction between the two surfaces is relatively low, allowing the block to travel a considerable distance beyond the equilibrium position before coming to a stop.