When the block moves up the inclined plane at constant speed, the net force is zero. Forces acting along the slope:

• Applied Force $F_{up} = 20$ N (Upwards)
• Gravity component $mg \sin \theta$ (Downwards)
• Kinetic Friction $f_k$ (Downwards, opposing motion)

Equation 1: $$F_{up} = mg \sin \theta + f_k$$ $$20 = mg \sin \theta + f_k$$

The problem states a force of 8 N is required to slide the block downwards. This phrasing usually implies that the block would not slide down on its own (i.e., friction is strong, or gravity is weak). Therefore, we must push it down.

Forces acting along the slope:

• Applied Force $F_{down} = 8$ N (Downwards)
• Gravity component $mg \sin \theta$ (Downwards)
• Kinetic Friction $f_k$ (Upwards, opposing motion)

Equation 2: $$F_{down} + mg \sin \theta = f_k$$ $$8 + mg \sin \theta = f_k$$

We now have a system of two linear equations: 1. $mg \sin \theta + f_k = 20$ 2. $-mg \sin \theta + f_k = 8$

Add the two equations: $$(mg \sin \theta + f_k) + (-mg \sin \theta + f_k) = 20 + 8$$ $$2 f_k = 28$$ $$f_k = 14 \text{ N}$$

Answer: The force of kinetic friction is 14 N.