If a #4 kg# object moving at #5/4 m/s# slows to a halt after moving #8 m#, what is the coefficient of kinetic friction of the surface that the object was moving over?

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Answer 1 · 2016-06-26T05:11:38

#~~0.01#

Given
#m->"Mass of the object"=4kg#

#m->"Velocity of the object"=5/4m/s#
#g->"Acceleration due to gravity"=9.8m/s^2#

#d->"Distance travelled before halt"=8m#

#mu_k->"Coefficient of kinetic friction"=?#

Considering law of conservation of energy we can say,
#"work done against friction =KE of the object"#

#=>mu_kmgxxd=1/2*mv^2#

#=>mu_k=v^2/(2*g*d)=(5/4)^2/(2*9 .8*8)~~0.01#