We will have the following:
1. In the first case, we will calculate the normal force:
![N=(15\operatorname{kg}\cdot9.8m/s^2)\Rightarrow N=147N]()
Then, we will have that the fricion coefficient will be:
![\mu_f=(30.0N)/(147N)\Rightarrow\mu_f=(10)/(49)\Rightarrow\mu_f\approx0.20](https://img.qammunity.org/2023/formulas/physics/high-school/4ss8lzd5nqfoh25i0t75mcwk0ckpdjvvb6.png)
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:
![F=(10/49)(15.0\operatorname{kg}\cdot9.8m/s^2)\cos (60)\Rightarrow F=15N]()
So, the friction force when the force is applied, is 15N.