If a #11 kg# object moving at #15 ms^-1# slows down to a halt after moving #225 m,# what is the coefficient of kinetic friction of the surface that the object was moving over?

1 Answer
Jan 19, 2016

Answer:

The coefficient of friction is related to the force acting on the object to accelerate (decelerate) it and the normal force: #mu =F/F_N = 5.5/107.8=0.05#.

Explanation:

First we need to calculate the acceleration (deceleration) of the object:

#v^2 = u^2 + 2 ad#

#0^2 = 15^2 -2*a*225#

#a = 0.5 ms^-2#

Now we have an acceleration and a mass and want to find a force... hmm, that sounds a lot like Newton's Second Law to me!

#F=ma=11*0.5=5.5 N#

This is the actual magnitude of the frictional force. The coefficient of friction is defined as:

#mu =F/F_N# where #mu# is the coefficient and #F_N# is the normal force acting. In this case that is the weight force of the object, given by #F_N = mg = 11*9.8 = 107.8 N#

Then #mu =F/F_N = 5.5/107.8=0.05#

Note that a frictional coefficient is a 'dimensionless' number which does not have units (because it is calculated as a force in #N# divided by another force in #N#, so the units cancel).