If a #3 kg# object moving at #9 m/s# slows down to a halt after moving #27 m#, what is the coefficient of kinetic friction of the surface that the object was moving over?

1 Answer
Apr 1, 2017

Answer:

#mu=0.15#

Explanation:

The work being done on the object is Work due to friction so the following equation is going to be used:

#color(white)(aaaaaaaaaaaa)#Equation (a) #W_f = DeltaKE#

We can rewrite Equation (a) if we break down both sides step-by-step to become:

#color(white)(aaaaaaaaaaaaaaaaaaa)#Equation (b)
#color(white)(aaaaaa)##(mu*mg)*d*costheta = (1/2mv_f^2 - 1/2mv_i^2)#

#mu = "coefficient of kinetic friction"#
#m = "mass (kg)"#
#g = "acceleration due to gravity" (m/s^2)#
#d = "displacement"(m)#
#theta = "angle between friction and displacement"#
#v_f = "velocity final"#
#v_i = "velocity initial"#

Since our object stopped, its final velocity becomes #0# and therefore #"final KE"# becomes #0#. Friction and displacement are opposite one another so #cos(180^@) = -1#. Mass cancels on both sides. Rearrange, plug in, and solve.

#-mu*g*d= - 1/2v_i^2#

#mu=(0.5*9^2)/(9.8*27) = 40.5/264.6 = 0.15#

Answer: 0.15