Block Against Vertical Wall – Rankers Physics

Friction: Practice Problem & Solution

Assertion (A): A block of weight \(10 \text{ N}\) is pushed against a vertical wall by a horizontal force of \(15 \text{ N}\). The coefficient of friction between the wall and the block is \(0.6\). Then the magnitude of maximum frictional force is \(9 \text{ N}\). Reason (R): For given system block will remain stationary.  
(1) Both (A) & (R) are true and the (R) is the correct explanation of the (A)
(2) Both (A) & (R) are true but the (R) is not the correct explanation of the (A)
(3) (A) is true but (R) is false
(4) Both (A) and (R) are false

Short Solution (Best for NEET):

To solve this problem, we apply the core principles of Friction. Understanding the underlying formula is key to arriving at the correct answer below:

Normal force \(N = 15 \text{ N}\). Maximum static friction \(f_{\text{max}} = \mu_{\text{s}} N = 0.6 \times 15 = 9 \text{ N}\). So (A) is true. Weight of block is \(W = 10 \text{ N}\). Since \(W > f_{\text{max}}\), the block will slide down. So (R) is false.

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