A block of mass 2 kg slides along a horizontal tabletop. A horizontal applied force of 12 N and a vertical...

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A block of mass 2 kg slides along a horizontal tabletop. A horizontal applied force of 12 N and a vertical applied force of 15 N act on the block, as shown above. If the coefficient of kinetic friction between the block and the table is 0.2, the frictional force exerted on the block is most nearly

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To determine the frictional force exerted on the block, we first need to calculate the normal force acting on the block and then use that to find the frictional force. ### Step 1: Identify Forces 1. **Weight of the Block (W)**: $$ W = m \cdot g $$ Where: - $ m = 2 \, \text{kg} $ (mass of the block) - $ g = 9.8 \, \text{m/s}^2 $ (acceleration due to gravity) $$ W = 2 \, \text{kg} \cdot 9.8 \, \text{m/s}^2 = 19.6 \, \text{N} $$ 2. **Vertical Applied Force (Fₙ)**: The vertical applied force is given as 15 N. ### Step 2: Calculate Normal Force (N) The normal force is affected by both the weight of the block and the vertical applied force. Since the applied vertical force acts upwards, the normal force is calculated as follows: $$ N = W - F_v $$ Where: - $ F_v = 15 \, \text{N} $ (vertical applied force) $$ N = 19.6 \, \text{N} - 15 \, \text{N} = 4.6 \, \text{N} $$ ### Step 3: Calculate the Frictional Force The frictional force (f_friction) can be calculated using the equation: $$ f_{\text{friction}} = \mu_k \cdot N $$ Where: - $ \mu_k = 0.2 $ (coefficient of kinetic friction) Substituting the normal force into the equation: $$ f_{\text{friction}} = 0.2 \cdot 4.6 \, \text{N} $$ $$ f_{\text{friction}} = 0.92 \, \text{N} $$ ### Conclusion The frictional force exerted on the block is most nearly: The frictional force exerted on the block is most nearly 0.92 N.

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A block of mass 2 kg slides along a horizontal tabletop. A horizontal applied force of 12 N and a vertical...

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