We will have the following:
1. In the first case, we will calculate the normal force:
[tex]N=(15\operatorname{kg}\cdot9.8m/s^2)\Rightarrow N=147N[/tex]Then, we will have that the fricion coefficient will be:
[tex]\mu_f=\frac{30.0N}{147N}\Rightarrow\mu_f=\frac{10}{49}\Rightarrow\mu_f\approx0.20[/tex]So the magnitude of the friccional force when applied horizontally will be of 30.0N.
2. And the magnitude when applied horizontal but the block is on an inclined surface of 60°, we will have that:
[tex]F=(10/49)(15.0\operatorname{kg}\cdot9.8m/s^2)\cos (60)\Rightarrow F=15N[/tex]So, the friction force when the force is applied, is 15N.
