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

[southernbelle]

Homework Statement

A block of mass 1 kg is forced against a horizontal spring of negligible mass, compressing the spring an amount x1= 0.2m. When released, the block moves on a horizontal tabletop a distance x2 = 1.0m before coming to rest. The spring constant k is 100 N. What is the coefficient of friction, μ, between the block and the table?

Homework Equations

K₁ + U₁ = K₂ + U₂
fs= -kx
Kinetic energy = 1/2kv²
Potential energy = mgh

The Attempt at a Solution

At x1 kinetic energy is 0 and potential energy is 1/2kΔx²
So the total energy would be = 40.
However, I do not know how to tie this in with the coefficients of friction.

 

[Rake-MC]

potential energy of a linear spring = \frac{1}{2} kx^2

therefore potential energy of this spring at this compression = \frac{1}{2} (100) (.04) = 2J

conservation of momentum E_k = E_p

so kinetic energy also = 2J at the moment that the spring reaches full length. \frac{1}{2}mv^2 = 2

Now solve for velocity at the instant it leaves the spring. Calculate the force due to gravity on the block.
Let me know how you go from there, it should be clear how to finish the question.

 

[southernbelle]

Okay I understand now, thank you!

 

[Rake-MC]

Okay I understand now, thank you!

No worries, I just realized that there was an error in the equation I wrote out on the second line of my post, however the result is still the same. I have corrected it now. apologies.