Answer:
c. The coefficient of kinetic friction is less than the coefficient of static friction
Explanation:
When the box finally does break loose. Then the component of the box weight which is parallel to the board weight parallel component, is equal to the [tex]\mu_gn[/tex].
[tex]w_{box}=\mu_gn[/tex]
For the box to acce;erate thee must be non-zero net force acting on the box parallel to the board. Or we can say,
[tex]w_1>f_g\\w_1>\mu_gn[/tex]
Therefore the force of kinetic friction must be less than the force of static friction. Thus,
[tex]\mu_g<\mu_s[/tex]