If a #2# #kg# object moving at #10# #ms^-1# slows to a halt after moving #25# #m#, what is the coefficient of kinetic friction of the surface that the object was moving over?

1 Answer
Jun 17, 2017

Answer:

#\mu=F_"frict"/(mg)=4/(2xx9.8)=0.20#

Explanation:

The object's final velocity will be #v=0# #ms^-1#, the initial velocity #u=10# #ms^-1# and the distance #25# #m#. We can use this to find the object's acceleration (deceleration):

#v^2=u^2+2as#

#a=(v^2-u^2)/(2s)=(0^1-10^2)/(2xx25)=(-100)/50=-2# #ms^-2#

The frictional force will be given by #F_"frict"=ma=2xx-2=-4# #N#

That is, the force is #4# #N# in the direction opposite to the initial motion.

The relationship between the frictional force and the object's weight force is #F_"frict"=\muF_w=\mumg#.

Rearranging:

#\mu=F_"frict"/(mg)=4/(2xx9.8)=0.20#

(frictional coefficients have no units)