If a #4 kg# object moving at #5/2 m/s# slows to a halt after moving #12 m#, what is the coefficient of kinetic friction of the surface that the object was moving over?

1 Answer
Apr 15, 2018

Answer:

The coefficient of kinetic friction is #=0.027#

Explanation:

Apply the equation of motion

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

To calculate the acceleration

The initial velocity is #u=5/2ms^-1#

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

The distance is #s=12m#

The acceleration is

#a=(v^2-u^2)/(2s)=(0-(5/2)^2)/(2*12)=-0.26ms^-2#

According to Newton's Second Law, the force of friction is

#F_r=ma=4*0.26=1.042N#

The normal force is

#N=mg=4*9.8=39.2N#

The coefficient of kinetic friction is

#mu_k=F_r/N=1.042/39.2=0.027#