If a #13 kg# object moving at #4 m/s# slows down to a halt after moving #80 m#, what is the friction coefficient of the surface that the object was moving over?

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Answer 1 · 2018-03-29T17:05:38

The coefficient of friction is #=0.01#

Apply the equation of motion to find the acceleration

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

The initial velocity is #u=4ms^-1#

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

The distance is #s=80m#

The acceleration is

#a=(v^2-u^2)/(2s)=(0-4^2)/(2*80)=-16/160=-0.1ms^-2#

The mass of the object is #m=13kg#

According to Newton's Second Law of Motion

#F=ma#

The force of friction is #F_r=13*0.1=1.3N#

The acceleration due to gravity is #g=9.8ms^-2#

The normal reaction is #N=mg=13*9.8=127.4N#

The coefficient of kinetic friction is

#mu_k=F_r/N=1.3/127.4=0.01#