Problem 33 - A Complete Resource Book in Physics for JEE Main

Question

A block of \(10 \mathrm{~kg}\) is pulled by a constant speed on a rough horizontal surface by a force of \(19.6 \mathrm{~N}\). The co-efficient of friction is (A) \(0.1\) (B) \(0.2\) (C) \(0.3\) (D) \(0.4\)

Step-by-Step Solution

Answer

The coefficient of friction is (B) \(0.2\).

Step 1: Identify the forces acting on the block

The forces acting on the block are the gravitational force (weight) acting vertically downwards, the normal force acting vertically upwards, the applied force acting horizontally, and the frictional force acting horizontally opposite to the applied force.

Step 2: Calculate the weight of the block

The weight of the block is given by the product of its mass and the acceleration due to gravity: W = mg where W is the weight, m is the mass (10 kg), and g is the acceleration due to gravity (9.8 m/s²). Substitute the given values to get: W = 10 × 9.8 W = 98 N

Step 3: Calculate the normal force acting on the block

Since the block is moving horizontally, the normal force is equal to the weight of the block (i.e., N = W = 98 N).

Step 4: Calculate the frictional force acting on the block

As the block is moving at a constant speed, the net force acting on it is zero. Therefore, the frictional force (f) is equal to the applied force (F) but in the opposite direction: f = F = 19.6 N.

Step 5: Calculate the coefficient of friction

The frictional force can be calculated using the formula: f = μN where f is the frictional force, μ is the coefficient of friction, and N is the normal force. We can rearrange this equation to solve for μ: μ = f/N Now, we can substitute the values for f and N to find the coefficient of friction: μ = 19.6/98 μ = 0.2 The correct answer is (B) \(0.2\).