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?

Question

1581 views around the world

You can reuse this answer
Creative Commons License

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