If a #2 kg# object moving at #80 m/s# slows to a halt after moving #4 m#, what is the coefficient of kinetic friction of the surface that the object was moving over?

1 Answer
Nov 5, 2017

Answer:

The coefficient of kinetic friction is #=81.6#

Explanation:

Apply the equation of motion

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

to determine the acceleration #a#

The initial speed is #u=80ms^-1#

The final speed is #v=0ms^-1#

The distance is #s=4m#

The acceleration is

#a=(v^2-u^2)/(2s)=(0^2-80^2)/(2 xx 4)=-80*80/8=-800ms^-2#

The frictional force is #F_r=ma=2xx800=1600N#

Let the acceleration due to gravity be #g=9.8ms^-2#

The normal force is #N=mg=2*9.8=19.6N#

The coefficient of kinetic friction is

#mu _k=F_r/N=1600/19.6=81.6#