two blocks of masses 4 kg and 6 kgs connected by a massless string are kept on a rough surface having a coefficient of sliding friction is 0.2 the two horizontal forces 30N and 10 N applied on 4kg and 6 kg blocks.find the friction force on both the blocks and tension in the string

To solve this problem, we need to analyze the forces acting on both blocks and determine the frictional forces and the tension in the string connecting them. Let's break it down step by step.

Understanding the Forces

We have two blocks: Block A (4 kg) and Block B (6 kg). They are connected by a massless string, and both are on a rough surface with a coefficient of friction (μ) of 0.2. The forces applied are 30 N on Block A and 10 N on Block B. First, we need to calculate the maximum static friction force for each block.

Calculating the Normal Forces

The normal force (N) for each block is equal to its weight, which can be calculated using the formula:

• N = mass × gravity (g = 9.81 m/s²)

For Block A (4 kg):

• N₁ = 4 kg × 9.81 m/s² = 39.24 N

For Block B (6 kg):

• N₂ = 6 kg × 9.81 m/s² = 58.86 N

Determining the Maximum Frictional Forces

The maximum frictional force (f_max) can be calculated using the formula:

• f_max = μ × N

For Block A:

• f_max₁ = 0.2 × 39.24 N = 7.848 N

For Block B:

• f_max₂ = 0.2 × 58.86 N = 11.772 N

Net Forces and Acceleration

Next, we need to determine the net forces acting on each block. The net force (F_net) on each block can be calculated by subtracting the frictional force from the applied force.

For Block A (4 kg)

The applied force is 30 N, and the frictional force is 7.848 N. Thus:

• F_net₁ = Applied Force - Frictional Force = 30 N - 7.848 N = 22.152 N

For Block B (6 kg)

The applied force is 10 N, and the frictional force is 11.772 N. Thus:

• F_net₂ = Applied Force - Frictional Force = 10 N - 11.772 N = -1.772 N

Since the net force on Block B is negative, it indicates that the frictional force exceeds the applied force, meaning Block B does not move, and the frictional force acting on it is equal to the applied force (10 N).

Calculating Tension in the String

Now, we can find the tension (T) in the string connecting the two blocks. Since Block B is not moving, the tension in the string must equal the frictional force acting on Block A minus the net force acting on Block A:

• T = F_net₁ + f_max₁ = 22.152 N + 7.848 N = 30 N

Summary of Results

To summarize:

• The frictional force on Block A is 7.848 N.
• The frictional force on Block B is 10 N (equal to the applied force).
• The tension in the string is 30 N.

This analysis shows how the forces interact and how friction plays a crucial role in determining the motion of connected objects. If you have any further questions or need clarification on any part of this explanation, feel free to ask!