If a #3/2 kg# object moving at #4/3 m/s# slows to a halt after moving #1/3 m#, what is the coefficient of kinetic friction of the surface that the object was moving over?

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Answer 1 · 2018-06-16T18:14:19

#mu_k = 0.272#

The original kinetic energy was #KE = 1/2*3/2 kg*(4/3 m/s)^2 = 4/3 J#.

Work of 4/3 J needs to be done by the friction to stop it.

#"work" = 4/3 J = F*1/3 m#

#F = (4/cancel(3) J)/(1/cancel(3) m) = 4 J/m = 4 N#

To find the value of #mu_k#, we solve

#F_"fk" = 4 N = mu_k*m*g = mu_k*3/2 kg*9.8 m/s^2 = mu_k*14.7 N#

#mu_k = (4 N)/(14.7 N) = 0.272#

I hope this helps,
Steve