Problem: A 2 kg block is kept over a 3 kg block, a force of 5 N is applied on 2kg and a force of 10 N in applied on 3 kg. The coefficient of friction between the blocks is 0.2 and between 3 kg block and ground is 0.1. Find individual acceleration of the blocks.

Solution:
To solve the problem, we need to apply Newton's laws of motion. Let's break the solution into the following steps:

Step 1: Draw a free-body diagram for each block.

The free-body diagram for the 2 kg block is as follows:


The free-body diagram for the 3 kg block is as follows:


Step 2: Write down the equations of motion for each block.

The equations of motion for the 2 kg block are as follows:

- F - f = ma
- N = mg

where F is the applied force, f is the frictional force, N is the normal force, m is the mass of the block, and a is the acceleration of the block.

The equations of motion for the 3 kg block are as follows:

- f - F1 = ma
- N1 = mg

where F1 is the applied force, N1 is the normal force, and a is the acceleration of the block.

Step 3: Solve the equations of motion for each block.

Using the equations of motion and the given values, we can solve for the acceleration of each block. The calculations are as follows:

For the 2 kg block:

- F - f = ma
- N = mg

Substituting the values, we get:

- 5 - 0.2N = 2a
- N = 2(9.8)
- N = 19.6 N

Solving for f:

- f = 0.2N
- f = 3.92 N

Substituting f and N in the first equation, we get:

- 5 - 3.92 = 2a
- a = 0.54 m/s^2

Therefore, the acceleration of the 2 kg block is 0.54 m/s^2.

For the 3 kg block:

- f - F1 = ma
- N1 = mg

Substituting the values, we get:

- 0.2N1 - 10 = 3a
- N1 = 3(9.8)
- N1 = 29.4 N

Solving for f:

- f = 0.2N1
- f = 5.88 N

Substituting f and N1 in the first equation, we get:

- 5.88 - 10 = 3a
- a = -1.04 m/s^2

Therefore, the acceleration of the 3 kg block is -1.04 m/s^2, which means it is moving in the opposite direction to the applied force.

Step 4: Check the solution.

We can check the solution by