If a #1 kg# object moving at #10 m/s# slows down to a halt after moving #50 m#, what is the coefficient of kinetic friction of the surface that the object was moving over?

1 Answer
Dec 8, 2015

Answer:

#mu=1/9.8#

Explanation:

First we must find the acceleration experienced by the object.
Here we will use the kinematic equation:
#(vf)^2=(vi)^2+2ad#
plug in our given data...
#0=100+100a#
#a=-1m/s^2#
We can now use this acceleration to calculate force using #F=ma#
#F=1kg*-1m/s^2#
#F=-1N# this must be the force of friction because no other resistive forces are given.

Kinetic friction is #fk=muN#
The normal force in this case is the weight of the object. #N=mg#
#N=1kg*9.8m/s^2#
#N=9.8#

Plugging in our values for #fk# and #N#....
#1=mu*9.8#
#mu=1/9.8#