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?

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Answer 1 · 2017-04-15T08:05:55

The coefficient of kinetic friction is #=0.015#

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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#