1. **State the problem:** A 5 kg block is pushed 3 m across a horizontal floor by a 25 N force applied at an angle of 30° above the horizontal. The coefficient of kinetic friction is 0.2. We need to find the normal force acting on the block. 2. **Identify forces and formula:** The normal force $N$ is the force exerted by the surface perpendicular to the block. It balances the vertical forces. The forces acting vertically are: - The weight of the block: $W = mg$ - The vertical component of the applied force: $F_y = F \sin(30^\circ)$ The normal force balances the weight minus the vertical component of the applied force: $$N = mg - F \sin(30^\circ)$$ 3. **Calculate the weight:** $$m = 5, \quad g = 9.8$$ $$W = 5 \times 9.8 = 49 \text{ N}$$ 4. **Calculate the vertical component of the applied force:** $$F = 25, \quad \sin(30^\circ) = 0.5$$ $$F_y = 25 \times 0.5 = 12.5 \text{ N}$$ 5. **Calculate the normal force:** $$N = 49 - 12.5 = 36.5 \text{ N}$$ **Final answer:** The normal force acting on the block is $36.5$ N.