If a #2 kg# object moving at #1 m/s# slows to a halt after moving #1/4 m#, what is the coefficient of kinetic friction of the surface that the object was moving over?

1 Answer
Apr 12, 2016

Answer:

#0.20#, rounded to second decimal place.

Explanation:

To find out the retardation due to kinetic friction let us use the equation
#v^2-u^2=2as#,

where #v,u,a,s# are the final velocity, initial velocity, acceleration and distance moved respectively.
#0^2-1^2=2xxaxx1/4#
or #a=-2ms^-2#

Now Force #F=m.a#
or Retarding force which is the force #|F_f|=2xx2=4N#
We know that ths #F_f=muxx"Normal Reaction"#, where #mu# is the coefficient of kinetic friction.

Assuming that the surface is a level surface, #:. "Normal Reaction"=mg#
Taking #g=9.81ms^-2#, we obtain
#4=muxx2xx9.81#, solving for #mu#
#mu=4/(2xx9.81)=0.20#, rounded to second decimal place.