If a #2 kg# object moving at #13 m/s# slows to a halt after moving #19 m#, what is the coefficient of kinetic friction of the surface that the object was moving over?

1 Answer
May 24, 2016

Answer:

#u_k~=0.4#

Explanation:

#v_f^2=v_i^2-2*a*Delta x#

#v_f=13/2" "m/s" final velocity"#

#v_i=13" "m/s " initial velocity"#

#Delta x=19" " m" displacement of object"#

#(13/2)^2=13^2-2*a* 19#

#2*a*19=169-169/4#

#38*a=507/4#

#a=597/(38*4)#

#a=597/152 =3.93 m/s^2#

#F_f=u_k*m*g#

#a=F_f/m#

#3.93=(u_k*cancel(m)*g)/cancel(m)#

#u_k=(3.93)/(9.81)#

#g=9.81 N/(kg)#

#u_k~=0.4#