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A block of weight 5 N is pushed against ...

A block of weight 5 N is pushed against a vertical wall by a force 12 N . The coefficient of friction between the wall and block is `0.6` . Find the magnitude of the force exerted by the wall on the block :

A

12 N

B

5 N

C

7.2 N

D

13 N

Text Solution

AI Generated Solution

The correct Answer is:
To solve the problem, we need to find the total force exerted by the wall on the block. Let's break it down step by step. ### Step 1: Identify the Forces Acting on the Block - The weight of the block (W) acting downwards = 5 N. - The applied force (F) pushing the block against the wall = 12 N. - The normal force (N) exerted by the wall on the block, which acts horizontally and is equal in magnitude to the applied force = 12 N. - The frictional force (f) acting upwards, which opposes the weight of the block. ### Step 2: Calculate the Maximum Frictional Force The maximum frictional force can be calculated using the formula: \[ f_{\text{max}} = \mu \cdot N \] Where: - \(\mu\) = coefficient of friction = 0.6 - \(N\) = normal force = 12 N Substituting the values: \[ f_{\text{max}} = 0.6 \cdot 12 = 7.2 \, \text{N} \] ### Step 3: Compare the Weight and Frictional Force The weight of the block is 5 N, and the maximum frictional force is 7.2 N. Since the maximum frictional force is greater than the weight of the block, the frictional force will be equal to the weight of the block in this case to maintain equilibrium. Thus, the frictional force (f) acting upwards is: \[ f = W = 5 \, \text{N} \] ### Step 4: Calculate the Total Force Exerted by the Wall The total force exerted by the wall on the block consists of two components: 1. The normal force (N) = 12 N (horizontal) 2. The frictional force (f) = 5 N (vertical) To find the resultant force (F_net) exerted by the wall on the block, we can use the Pythagorean theorem: \[ F_{\text{net}} = \sqrt{N^2 + f^2} \] Substituting the values: \[ F_{\text{net}} = \sqrt{(12)^2 + (5)^2} \] \[ F_{\text{net}} = \sqrt{144 + 25} \] \[ F_{\text{net}} = \sqrt{169} \] \[ F_{\text{net}} = 13 \, \text{N} \] ### Final Answer The magnitude of the force exerted by the wall on the block is **13 N**. ---

To solve the problem, we need to find the total force exerted by the wall on the block. Let's break it down step by step. ### Step 1: Identify the Forces Acting on the Block - The weight of the block (W) acting downwards = 5 N. - The applied force (F) pushing the block against the wall = 12 N. - The normal force (N) exerted by the wall on the block, which acts horizontally and is equal in magnitude to the applied force = 12 N. - The frictional force (f) acting upwards, which opposes the weight of the block. ...
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Knowledge Check

  • A 20 N block is being pushed across a horizontal table by an 18 N force. If the coefficient of kinetic friction between the block and the table is 0.4, find the acceleration of the block.

    A
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    B
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    C
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    D
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