If an object with a mass of #10 kg # is moving on a surface at #15 m/s# and slows to a halt after # 4 s#, what is the friction coefficient of the surface?

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Answer 1 · 2017-04-27T09:37:34

The coefficient of friction is #=0.38#

We start by calculating the deceleration.

We apply the equation of motion

#v=u+at#

The initial velocity is #u=15ms^-1#

The time is #t=4s#

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

Therefore,

#0=15+4a#

#a=-15/4ms^-2#

According to Newton's Second Law, the frictional force is #F_r=ma#

So,

#F_r=10*15/4=75/2N#

The normal force is

#N=mg=10*9.8=98N#

The coefficient of friction is

#mu_k=F_r/N=37.5/98=0.38#