Problem 42 - A Complete Resource Book in Physics for JEE Main

Question

A block of mass \(0.1 \mathrm{~kg}\) is held against a wall by applying a horizontal force of \(5 \mathrm{~N}\) on the block. If the co-efficient of friction between the block and the wall is \(0.5\), the magnitude of the frictional force acting on the block is (A) \(2.5 \mathrm{~N}\) (B) \(0.98 \mathrm{~N}\) (C) \(4.9 \mathrm{~N}\) (D) \(0.49 \mathrm{~N}\)

Step-by-Step Solution

Answer

The magnitude of the frictional force acting on the block is \(0.49 \mathrm{~N}\) (option D).

Step 1: Identifying the forces

In this problem, we have three forces acting on the block: the gravitational force pulling the block downwards, the horizontal force keeping the block against the wall, and the normal force acting perpendicular to the wall.

Step 2: Calculate the gravitational force

The gravitational force acting on the block can be calculated as follows: \(F_g = mg\) where: - \(F_g\) = gravitational force - \(m\) = mass of the block (0.1 kg) - \(g\) = acceleration due to gravity (approximately 9.8 m/s²) \(F_g = (0.1 \mathrm{~kg})(9.8 \mathrm{~m/s^2})\) \(F_g = 0.98 \mathrm{~N}\)

Step 3: Calculate the normal force

Since there is no vertical acceleration of the block, the normal force must be equal in magnitude and opposite in direction to the gravitational force, resulting in no net vertical force. Therefore, the normal force is: \(F_n = F_g\) \(F_n = 0.98 \mathrm{~N}\)

Step 4: Calculate the frictional force

The frictional force can be calculated using the following formula: \(F_f = \mu F_n\) where: - \(F_f\) = frictional force - \(\mu\) = coefficient of friction (0.5) - \(F_n\) = normal force (0.98 N) \(F_f = (0.5)(0.98 \mathrm{~N})\) \(F_f = 0.49 \mathrm{~N}\) Thus, the magnitude of the frictional force acting on the block is 0.49 N, which corresponds to option (D).