If a #3 kg# object moving at #8 m/s# slows to a halt after moving #180 m#, what is the coefficient of kinetic friction of the surface that the object was moving over?

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Answer 1 · 2017-03-28T17:21:39

#k~=0.0136#

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#"First,let us draw a velocity-time graph."#

#"Area under velocity-time graph gives object's displacement."#

#180=((8+4))/2*t" , "180=12/2*t" , "t=30 "sec"#

#"The slope of velocity-time graph gives deceleration of object."#

#tan alpha=(4-8)/30=-4/30=-2/15" "m/s^2#

#"We can calculate the friction force between contacted surface "##"using formula " F_f=k*N#

#"Where "F_f ":The friction force , k:coefficient of friction force ,"#

#"and N: Normal Force."#

#"Note that N="m*g#

#F_f=k*m*g#

#"according to Newton's second law of Dynamics "F=m*a#

#"Thus , we can write as "m*a=k*m*g#

#"Let us simplify "cancel(m)*a=k*cancel(m)*g#

#"We get "a=k*g#

#k=a/g#

#k=2/(15*9.81)#

#k=0,013591573#

#k~=0.0136#