A block `A` of mass `m_1` rests on a horizontal table. A light string connected to it passes over a frictionless pulley at the edge of table and from its other end another block `B` of mass `m_(2)` is suspended. The coefficient of knetic friction between the block and table is `mu_(k)` . When the block `A` is sliding on the table, the tension in the string is.

Options

A. `((m_(2)+mu_(k)m_(1))g)/(m_(1)+m_(2))`
B. `((m_(2)-mu_(k)m_(1))g)/(m_(1)+m_(2))`
C. `(m_(1)m_(2)(1+mu_(k))g)/(m_(1)+m_(2))`
D. `(m_(1)m_(2)(1-mu_(k))g)/(m_(1)+m_(2))`

Answer

To solve the problem, we will analyze the forces acting on both blocks A and B and derive the expression for the tension in the string.

### Step 1: Identify the forces acting on Block A (mass \( m_1 \))
- The forces acting on block A are:
- The tension \( T \) in the string acting horizontally towards the pulley.
- The kinetic friction force \( F_k \) acting opposite to the direction of motion.
- The normal force \( N \) acting vertically upwards.
- The weight \( m_1 g \) acting vertically downwards.

### Step 2: Write the equations of motion for Block A
Using Newton's second law for block A, we have:
\[ T - F_k = m_1 a \]
Where \( F_k = \mu_k N \) and since \( N = m_1 g \), we can substitute:
\[ F_k = \mu_k m_1 g \]
Thus, the equation becomes:
\[ T - \mu_k m_1 g = m_1 a \tag{1} \]

### Step 3: Identify the forces acting on Block B (mass \( m_2 \))
- The forces acting on block B are:
- The weight \( m_2 g \) acting downwards.
- The tension \( T \) acting upwards.

### Step 4: Write the equations of motion for Block B
Using Newton's second law for block B, we have:
\[ m_2 g - T = m_2 a \tag{2} \]

### Step 5: Solve the equations simultaneously
From equation (1), we can express \( T \):
\[ T = m_1 a + \mu_k m_1 g \]
Substituting this expression for \( T \) into equation (2):
\[ m_2 g - (m_1 a + \mu_k m_1 g) = m_2 a \]
Rearranging gives:
\[ m_2 g - \mu_k m_1 g = m_1 a + m_2 a \]
Factoring out \( a \) on the right side:
\[ m_2 g - \mu_k m_1 g = (m_1 + m_2) a \]
Now, solving for \( a \):
\[ a = \frac{m_2 g - \mu_k m_1 g}{m_1 + m_2} \]

### Step 6: Substitute \( a \) back to find \( T \)
Now substituting \( a \) back into the expression for \( T \):
\[ T = m_1 \left(\frac{m_2 g - \mu_k m_1 g}{m_1 + m_2}\right) + \mu_k m_1 g \]
This simplifies to:
\[ T = \frac{m_1 m_2 g - \mu_k m_1^2 g}{m_1 + m_2} + \mu_k m_1 g \]
Combining the terms gives:
\[ T = \frac{m_1 m_2 g + \mu_k m_1 (m_1 + m_2) g}{m_1 + m_2} \]
Thus, the final expression for the tension \( T \) in the string is:
\[ T = \frac{m_1 m_2 g + \mu_k m_1 g (m_1 + m_2)}{m_1 + m_2} \]

### Final Answer
The tension in the string is:
\[ T = \frac{m_1 m_2 g + \mu_k m_1 g (m_1 + m_2)}{m_1 + m_2} \]