Answer:
0.58
Explanation:
Using the frictional force formula expressed as;
[tex]F_f = \mu R\\[/tex]
[tex]\mu[/tex] is the coefficient of kinetic friction
R is the normal reaction
[tex]\mu = \frac{F_f}{R}[/tex]
[tex]F_f = Wsin \theta\\R = W cos\theta\\\mu = \frac{Wsin \theta}{W cos \theta}\\\mu = \frac{sin\theta}{cos \theta} \\\mu = tan \theta[/tex]
Given
[tex]\theta = 30^0\\\mu = tan 30^0\\\mu = 0.58[/tex]
hence the coefficient of kinetic friction is 0.58
