If a #4 kg# object moving at #1 m/s# slows down to a halt after moving #90 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 Sidharth Feb 16, 2016 #5 * 10 ^-4# Explanation: #2as = v^2 - u^2# #a =( Deltav^2)/(2s) = 1 / 180# #f = mu N# #N = mg = 40 N# #D_f = mu N/m = mu g# #mu g = 1/180# #mu = 1/1800 = 0.00055555555 approx 5 * 10 ^-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 1497 views around the world You can reuse this answer Creative Commons License