If a #5 kg# object moving at #6 m/s# slows to a halt after moving #2 m#, what is the coefficient of kinetic friction of the surface that the object was moving over?

1 Answer
Jan 25, 2018

Answer:

The coefficient of kinetic friction is #=0.92#

Explanation:

The mass of the object is #m=5kg#

The initial speed of the object is #u=6ms^-1#

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

The distance is #s=2m#

Apply the equation of motion

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

to calculate the acceleration

The acceleration is

#a=(v^2-u^2)/(2s)=(0^2-6^2)/(2*2)=-36/4=-9ms^-2#

According to Newton's Second Law

#F=ma#

The frictional force is #F_r=5*9=45N#

The acceleration due to gravity is #g=9.8ms^-2#

The normal force is #N=mg=5*9.8=49N#

The coefficient of kinetic friction is

#mu_k=F_r/N=45/49=0.92#