If a #2 kg# object moving at #80 m/s# slows to a halt after moving #9 m#, what is the coefficient of kinetic friction of the surface that the object was moving over?

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Answer 1 · 2017-04-19T08:57:17

The coefficient of kinetic friction is #=36.28#

We apply the equation of motion to find the deceleration

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

The initial velocity is #u=80ms^-1#

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

The distance is #s=9m#

#0=80^2+2*a*9#

#a=-80^2/(2*9)#

#a=-355.6ms^-2#

The frictional force is

#F_r=ma=2*355.6=711.1N#

#N=2g#

The coefficient of kinetic friction is

#mu_k=F_r/N=711.1/(2g)=36.28#