If a #4 kg# object moving at #16 m/s# slows down to a halt after moving #800 m#, what is the friction coefficient of the surface that the object was moving over? Physics Forces and Newton's Laws Frictional Forces 1 Answer ali ergin Jun 2, 2016 #mu_k=0.013" has no unit"# Explanation: #v_f^2=v_i^2-2*a*Delta x# #v_f=8" "m/s# #v_i=16" "m/s# #Delta x=800" "m# #a:"acceleration of the object"# #8^2=16^2-2*a*800# #64=256-a*1600# #1600*a=256-64# #1600*a=192# #a=192/1600" "m/s^2# #F_f=mu_k*cancel(m)*g=cancel(m)*a# #mu_k="coefficient of friction"# #mu_k*g=a# #mu_k=a/g# #mu_k=(192/1600)/(9.18)# #mu_k=192/(1600*9.18)# #mu_k=192/14688# #mu_k=0.013# 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 1298 views around the world You can reuse this answer Creative Commons License