What is the coefficient of kinetic friction between the block and the incline?

[alevis]

Homework Statement

A 2.00-kg block situated on a rough incline is connected to a spring of negligible mass having a spring constant of 100 N/m. The block is released from rest when the spring is unstretched, and the pulley is frictionless. The block moves 20.0 cm down the incline before coming to rest.
Find the coefficient of kinetic friction between block and incline.

Homework Equations

PE = KE
F= ma
Ff = \muF_n
\mu = Ff/F_n

PE = KE

The Attempt at a Solution

Ff = mgsin\theta
Fn = mgcos\theta
\mu = (mgsin\theta)/mgcos\theta
stuck!

mgh = 1/2mv²

 

[hage567]

You know, that's not really an attempt at a solution. You need to show more than some equations slapped together. Explain what you have done and why. We can't read your mind!

Did you draw a digram of all the forces on the block? Always start with that.

 

[AEM]

Homework Statement

A 2.00-kg block situated on a rough incline is connected to a spring of negligible mass having a spring constant of 100 N/m. The block is released from rest when the spring is unstretched, and the pulley is frictionless. The block moves 20.0 cm down the incline before coming to rest.
Find the coefficient of kinetic friction between block and incline.

Homework Equations

PE = KE
F= ma
Ff = \muF_n
\mu = Ff/F_n

PE = KE

The Attempt at a Solution

Ff = mgsin\theta
Fn = mgcos\theta
mgh = 1/2mv²

Because of friction your two energy equations are not quite correct. You could try thinking of it this way: The potential energy at the start goes into the work to overcome friction and the energy stored in the spring. That's assuming that you put your reference level for the potential energy where the mass stops.