Acceleration of the Blocks in a Pulley Mass System

The system shown in the figure consists of two blocks connected by a massless string that passes over a massless and frictionless pulley. The objective is to find the acceleration of the blocks when the system is released from rest.

Free Body Diagrams:
To analyze the system, we need to draw free body diagrams for each block involved.

1. Block A: The weight of block A acts vertically downward, and the tension in the string pulls it upwards. Therefore, we have the following forces acting on block A:
- Weight (W_A) acting downwards
- Tension (T) acting upwards

2. Block B: The weight of block B acts vertically downward, and the tension in the string pulls it upwards. Therefore, we have the following forces acting on block B:
- Weight (W_B) acting downwards
- Tension (T) acting upwards

Equations of Motion:
Using Newton's second law (F = ma), we can write the equations of motion for both blocks.

1. Block A:
- T - W_A = m_A * a (equation 1)

2. Block B:
- T - W_B = m_B * a (equation 2)

Acceleration Calculation:
Since the blocks are connected by a string, their accelerations must be equal in magnitude and opposite in direction. Let's assume the acceleration of the system is 'a'. Therefore, the acceleration of block A is 'a' upwards, and the acceleration of block B is 'a' downwards.

Solving the Equations:
To find the acceleration 'a', we can add equations 1 and 2:

(T - W_A) + (T - W_B) = (m_A + m_B) * a

Simplifying the equation:

2T - (W_A + W_B) = (m_A + m_B) * a

W_A and W_B can be replaced with their respective weights:

2T - (m_A * g + m_B * g) = (m_A + m_B) * a

Simplifying further:

2T - (m_A + m_B) * g = (m_A + m_B) * a

Dividing both sides by (m_A + m_B):

2T / (m_A + m_B) - g = a

Therefore, the acceleration of the system is given by:

a = 2T / (m_A + m_B) - g

Conclusion:
By using the equations of motion and considering the forces acting on the blocks, we derived the expression for the acceleration of the system