If a 3 kg3kg object moving at 8 m/s8ms slows to a halt after moving 180 m180m, what is the coefficient of kinetic friction of the surface that the object was moving over?

1 Answer
Mar 28, 2017

k~=0.0136k0.0136

Explanation:

enter image source here

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

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

180=((8+4))/2*t" , "180=12/2*t" , "t=30 "sec"180=(8+4)2t , 180=122t , t=30sec

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

tan alpha=(4-8)/30=-4/30=-2/15" "m/s^2tanα=4830=430=215 ms2

"We can calculate the friction force between contacted surface "We can calculate the friction force between contacted surface "using formula " F_f=k*Nusing formula Ff=kN

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

"and N: Normal Force."and N: Normal Force.

"Note that N="m*gNote that N=mg

F_f=k*m*gFf=kmg

"according to Newton's second law of Dynamics "F=m*aaccording to Newton's second law of Dynamics F=ma

"Thus , we can write as "m*a=k*m*gThus , we can write as ma=kmg

"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