If a #1 kg# object moving at #25 m/s# slows down to a halt after moving #625 m#, what is the coefficient of kinetic friction of the surface that the object was moving over?

1 Answer
Feb 29, 2016

Answer:

#u_k=0,038#

Explanation:

#v_f^2=v_i^2-2*a*x#
#(v/2)^2=v^2-2*a*625#
#v^2/4=v^2-2*a*625#
#v^2=4v^2-8*a*625#
#3v^2=8*a*625#
#v=25 m/s#
#3*25^2=8.a.625#
#3*cancel(625)=8*a*cancel(625)#
#a=3/8 m/s^2" acceleration of the object"#
#F_f=u_k*N" the friction force "#
#N=m*g" N:normal force acting contacting surfaces"#
#F_f=m*a" Newton's the second law"#
#u_k*cancel(m)*g=cancel(m)*3/8#
#u_k=3/(8*9,81)#
#u_k=0,038#