Problem 151 - A Complete Resource Book in Physics for JEE Main

Question

A horizontal force of \(10 \mathrm{~N}\) is necessary to just hold a block stationary against a wall. The co-efficient of friction between the block and wall is \(0.2\). The weight of the block is (A) \(20 \mathrm{~N}\) (B) \(50 \mathrm{~N}\) (C) \(100 \mathrm{~N}\) (D) \(2 \mathrm{~N}\)

Step-by-Step Solution

Answer

The weight of the block is \(2 \mathrm{~N}\) (Option D).

Step 1: Identify the forces acting on the block

Since the block is stationary against the wall, there are three forces acting on it: 1. The gravitational force (weight) acting downward. 2. The horizontal force (provided in the problem) holding the block against the wall. 3. The frictional force acting upward, preventing the block from sliding down the wall.

Step 2: Write down the equations for frictional force and normal force

The frictional force (F_f) is given by the equation: \(F_f = \mu F_n\), where \(\mu\) is the coefficient of friction and \(F_n\) is the normal force. In this problem, the normal force is equal to the applied horizontal force (F_H), since it opposes the horizontal force: \(F_n = F_H\).

Step 3: Determine the weight of the block using the given information

We are given that the horizontal force necessary to hold the block stationary against the wall is \(10 \mathrm{~N}\), and the coefficient of friction is \(0.2\). We can use these values to determine the weight (W) of the block. First, let's find the normal force: \(F_n = F_H = 10 \mathrm{~N}\). Now, we can find the frictional force acting on the block using the formula: \(F_f = \mu F_n = 0.2 * 10 \mathrm{~N} = 2 \mathrm{~N}\). Since the block is stationary against the wall, the frictional force is equal to the weight of the block: \(W = F_f = 2 \mathrm{~N}\). Therefore, the weight of the block is 2 N (Option D).