If a #1/2 kg# object moving at #1/4 m/s# slows to a halt after moving #1/8 m#, what is the coefficient of kinetic friction of the surface that the object was moving over?

1 Answer
Feb 9, 2016

Answer:

#k~=0,408#

Explanation:

#k : "coefficient of friction"#
#V^2=V_i^2-2*a*x#
#V=0 " if object stops."#
#0=(1/4)^2-cancel(2)*a*1/cancel(8)#
#1/16=a/4#
#a=4 m/s^2 " acceleration"#
#k*N=m*a#
#"N: normal force contacting surface"#
#N=m*g#
#k*cancel(m)*g=cancel(m)*a#
#k=a/g#
#k=4/(9,81)#
#k~=0,408#