If a #5 kg# object moving at #5 m/s# slows to a halt after moving #25 m#, what is the coefficient of kinetic friction of the surface that the object was moving over?

1 Answer
Jul 7, 2016

Answer:

#mu_k=0.038" , has no unit"#

Explanation:

#v_f^2=v_i^2-2*a*Delta x#
#v_f:"represents the final velocity of object"#
#v_i:"represents the initial velocity of object"#
#a:"represents deceleration of object"#
#Delta x:"represents the displacement of object" #

#"Where; "v_i=5 " "m/s " ; " v_f=5/2 " "m/s" ; "Delta x=25" "m#

#(5/2)^2=5^2-2*a*25#

#25/4=25-2*a*25#

#50*a=25-25/4" ; "50a=75/4" ; "a=75/200#

#a=3/8" "m/s^2#

#F_f=m*a#
#F_f:"represents the friction force between contacting surface"#
#N:"represents the normal force"#
#mu_k:"represents the coefficient of kinetic friction"#

#N=m*g#

#F_f=mu_k*m*g#

#mu_k*cancel(m)*g=cancel(m)*a#

#mu_k=a/(g)#

#mu_k=(3/8)/(9.81)#

#mu_k=3/(8*9.81)#

#mu_k=3/(78,48)#

#mu_k=0.038" , has no unit"#