A 231 kg block is released at height h = 3.9 m. The track is frictionless except for a portion of length 6.6 m. The block travels down the track, hits a spring of force constant k = 1624 N/m, and compresses it 1.7 m from its equilibrium position before coming to rest momentarily.

Question

A 231 kg block is released at height h = 3.9 m. The track is frictionless except for a portion of length 6.6 m. The block travels down the track, hits a spring of force constant k = 1624 N/m, and compresses it 1.7 m from its equilibrium position before coming to rest momentarily. 
The acceleration of gravity is 9.8 m/s^2.
Determine the coefficient of kinetic friction between surface and block over the 6.6 m track length.

bobpursley · Answer

finalenergy=initialenergy-frictionwork

1/2 kx^2=mgh-frictionwork
1/2 1624*1.7^2=231*9.8*3.9-friction work.
friction work= 6482.14 J

but frictionwork=mg*cosTheta*mu*6.6
I dont see how it can proceed further without knowing more about the angle of the track.

Dj · Answer

I figured it out, thank you! Using 6482.14 J, I was able to find the coefficient of kinetic friction over the 6.6m segment by using the equation:
W = mu*F*x.
6482.14 J = mu(231kg*9.8m/s^2*6.6m)
mu = 6482.14/14,941.08 = 0.4338.