Answer: 0.6
Explanation:
If we draw a free body diagram of the box we will have the following:
Net force in the x-axis:
[tex]N-W-Fsin \theta=0[/tex] (1)
Net force in the y-axis:
[tex]-F_{r}+Fcos \theta=0[/tex] (2)
Where:
[tex]N[/tex] is the normal force
[tex]W=325 N[/tex] is the weight of the box
[tex]F=425 N[/tex] is the force exerted on the box
[tex]\theta=35.2\°[/tex] is the angle below the horizontal
[tex]F_{r}=\mu_{k}N[/tex] is the friction force, being [tex]\mu_{k}[/tex] the coefficient of kinetic friction
Isolating [tex]N[/tex] from (1):
[tex]N=W+Fsin \theta[/tex] (3)
Substituting (3) in (2):
[tex]-\mu_{k}(W+Fsin \theta)+Fcos \theta=0[/tex] (4)
Finding [tex]\mu_{k}[/tex]:
[tex]\mu_{k}=\frac{Fcos \theta}{W+Fsin \theta}[/tex] (5)
[tex]\mu_{k}=\frac{425 N cos (35.2\°)}{325 N+425 N sin (35.2\°)}[/tex] (6)
Finally:
[tex]\mu_{k}=0.6[/tex] (5)
