A block of mass m is resting one a smooth horizontal surface. One end of a uniform rope of mass m/3 is fixed to the block, which is pulled in the horizontal direction by applying force F at the other end. The tension in the middle of the rope is

Options

A. `(8)/(6) F`
B. `(1)/(7) F`
C. `(1)/(8)F`
D. `(7)/(8)F`

Explanation

To find the tension in the middle of the rope when a block of mass \( m \) is pulled with a force \( F \) and the rope has a mass of \( \frac{m}{3} \), we can follow these steps:

### Step 1: Determine the total mass of the system
The total mass of the system consists of the mass of the block and the mass of the rope. Therefore, the total mass \( M \) is given by:
\[ M = m + \frac{m}{3} = \frac{3m}{3} + \frac{m}{3} = \frac{4m}{3} \]

### Step 2: Calculate the acceleration of the system
Using Newton's second law, the net force \( F \) acting on the system is equal to the total mass times the acceleration \( a \):
\[ F = M \cdot a \]
Substituting the total mass \( M \):
\[ F = \frac{4m}{3} \cdot a \]
Solving for acceleration \( a \):
\[ a = \frac{3F}{4m} \]

### Step 3: Analyze the left half of the rope
To find the tension at the middle of the rope, we can consider the left half of the rope and the block. The mass of the left half of the rope is \( \frac{m}{6} \) (since the total mass of the rope is \( \frac{m}{3} \), half of that is \( \frac{m}{6} \)). The total mass being pulled by the tension \( T_O \) at the middle point includes the mass of the block \( m \) and the mass of the left half of the rope \( \frac{m}{6} \):
\[ \text{Total mass on the left} = m + \frac{m}{6} = \frac{6m}{6} + \frac{m}{6} = \frac{7m}{6} \]

### Step 4: Apply Newton's second law to the left side
The tension \( T_O \) at the middle of the rope is responsible for accelerating the total mass on the left side. Therefore, we can write:
\[ T_O = \left(\frac{7m}{6}\right) \cdot a \]
Substituting the expression for acceleration \( a \):
\[ T_O = \left(\frac{7m}{6}\right) \cdot \left(\frac{3F}{4m}\right) \]

### Step 5: Simplify the expression for tension
Now, we can simplify the expression for \( T_O \):
\[ T_O = \frac{7m}{6} \cdot \frac{3F}{4m} = \frac{7 \cdot 3F}{6 \cdot 4} = \frac{21F}{24} = \frac{7F}{8} \]

### Conclusion
Thus, the tension in the middle of the rope is:
\[ T_O = \frac{7F}{8} \]