A horizantal force `''F''` produces an acceleration of `6m//s^(2)` on a block resting on a smooth horizantal surface. The same force produces an acceleration of `3m//s^(2)` on a second block resting on a smooth horizantal surface. If the two blocks are tied together and the same force acts, the acceleration produced will be

Options

A. `9 m//s^(2)`
B. `2 m//s^(2)`
C. `4 m//s^(2)`
D. `1//2 m//s^(2)`

Explanation

To solve the problem, we need to find the acceleration produced when two blocks are tied together and the same force \( F \) is applied to them. Let's break it down step by step.

### Step 1: Determine the mass of each block
1. For the first block, the force \( F \) produces an acceleration of \( 6 \, \text{m/s}^2 \).
- Using Newton's second law, \( F = m_1 \cdot a_1 \), we can express the mass of the first block (\( m_1 \)):
\[ m_1 = \frac{F}{a_1} = \frac{F}{6} \]

2. For the second block, the same force \( F \) produces an acceleration of \( 3 \, \text{m/s}^2 \).
- Similarly, we can express the mass of the second block (\( m_2 \)):
\[ m_2 = \frac{F}{a_2} = \frac{F}{3} \]

### Step 2: Find the total mass when both blocks are tied together
- The total mass \( m \) of the system when both blocks are tied together is the sum of their individual masses:
\[ m = m_1 + m_2 = \frac{F}{6} + \frac{F}{3} \]

### Step 3: Simplify the total mass
- To add the two fractions, we need a common denominator:
\[ m = \frac{F}{6} + \frac{2F}{6} = \frac{3F}{6} = \frac{F}{2} \]

### Step 4: Calculate the acceleration of the combined system
- Now, we apply the same force \( F \) to the combined mass \( m \):
\[ F = m \cdot a \implies a = \frac{F}{m} \]
Substituting \( m \) from the previous step:
\[ a = \frac{F}{\frac{F}{2}} = 2 \, \text{m/s}^2 \]

### Final Answer
The acceleration produced when the two blocks are tied together and the same force \( F \) acts on them is:
\[ \boxed{2 \, \text{m/s}^2} \]