If an object with a mass of #10 kg # is moving on a surface at #18 m/s# and slows to a halt after # 4 s#, what is the friction coefficient of the surface?

1 Answer
Mar 3, 2016

Answer:

#u_k~=0,459#

Explanation:

#v_l=v_i-a*t#
#v/2=v-a*t" "v-v/2=a*t" "v/2=a*t#
#18=a*4" "a=9/2 m/s^2#
#F=m*a" "u_k*N=m*a" "u_k*cancel(m)*g=cancel(m)*a#
#u_k=a/g" "u_k=9/(2*9,81) " "u_k~=0,459#