### Problem 8:A block that weighs 60 N is initially at rest on a horizontal surface. A horizontal force \( F_1 \) of magnitude 22 N and a vertical force \( F_2 \) are then applied to the block. The coefficients of friction for the block and the surface are \( \mu_s = 0.50 \) and \( \mu_k = 0.25 \). Determine the magnitude of the frictional force acting on the block if the magnitude of \( F_2 \) is:(a) 10 N (b) 20 N#### Diagram:The diagram shows a block resting on a horizontal surface. Two forces are acting on the block:- \( \mathbf{F_1} \) is a horizontal force acting to the right.- \( \mathbf{F_2} \) is a vertical force acting upwards from the center of the block.#### Explanation of Forces:1. **Block Weight (W):** The force due to gravity on the block, which is 60 N, acts vertically downward.2. **Applied Horizontal Force (\( F_1 \)):** A force of 22 N acts horizontally to the right.3. **Applied Vertical Force (\( F_2 \)):** - Case (a): \( F_2 = 10 \) N - Case (b): \( F_2 = 20 \) N4. **Frictional Force:** The force opposing the motion of the block due to contact with the surface. The frictional force can be calculated using the normal reaction and the coefficients of friction.#### Coefficients of Friction:- Static Friction Coefficient (\( \mu_s \)): 0.50- Kinetic Friction Coefficient (\( \mu_k \)): 0.25

Solution: Given Values, Weight of the block(W)=60 N Horizontal force(F1)=22 N Vertical force=F2…

Answered: -8. A block that weighs 60 N is…

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