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? Physics Forces and Newton's Laws Frictional Forces 1 Answer ali ergin May 24, 2016 #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# Answer link Related questions Question #d6539 Question #242b7 Question #6bde4 Question #50c79 Question #a2018 Question #f7a62 Question #27931 Question #0b375 Question #d70af Question #dab6f See all questions in Frictional Forces Impact of this question 1309 views around the world You can reuse this answer Creative Commons License