If a #1/24 kg# object moving at #1/3 m/s# slows to a halt after moving #10/27 m#, what is the coefficient of kinetic friction of the surface that the object was moving over?

Question

1503 views around the world

You can reuse this answer
Creative Commons License

Answer 1 · 2017-04-15T08:05:55

The coefficient of kinetic friction is #=0.015#

enter image source here

The coefficient of kinetic friction is

#mu_k=F_r/N#

#F_f=ma#

We apply the equation of motion

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

#0=(1/3)^2+2*a*10/27#

#20/27a=-1/9#

#a=-27/9*1/20=-3/20ms^-2#

The negative sign indicates that it is a deceleration

Therefore,

#F_r=1/24*-3/20=-1/160N#

Therefore,

#mu_k=1/160*1/(1/24g)#

#=1/65.3=0.015#