Two bodies of mass 3 kg and 4 kg are suspended at the ends of massless string passing over a frictionless pulley. The acceleration of the system is `(g - 9.8 m//s^(2))`.

Options

A. `4.9 m//s^(2)`
B. `2.45 m//s^(2)`
C. `1.4 m//s^(2)`
D. `9.5 m//s^(2)`

Explanation

To solve the problem of two bodies of mass 3 kg and 4 kg suspended at the ends of a massless string over a frictionless pulley, we will follow these steps:

### Step 1: Identify the forces acting on each mass
- For the 4 kg mass (let's call it m1), the force acting downward due to gravity is \( F_{m1} = m_1 \cdot g = 4 \, \text{kg} \cdot 9.8 \, \text{m/s}^2 = 39.2 \, \text{N} \).
- For the 3 kg mass (let's call it m2), the force acting downward due to gravity is \( F_{m2} = m_2 \cdot g = 3 \, \text{kg} \cdot 9.8 \, \text{m/s}^2 = 29.4 \, \text{N} \).

### Step 2: Determine the net force acting on the system
- The net force acting on the system is the difference between the gravitational forces acting on the two masses:
\[ F_{\text{net}} = F_{m1} - F_{m2} = 39.2 \, \text{N} - 29.4 \, \text{N} = 9.8 \, \text{N} \]

### Step 3: Calculate the total mass of the system
- The total mass of the system is the sum of the two masses:
\[ m_{\text{total}} = m_1 + m_2 = 4 \, \text{kg} + 3 \, \text{kg} = 7 \, \text{kg} \]

### Step 4: Apply Newton's second law to find the acceleration
- According to Newton's second law, the acceleration \( a \) of the system can be calculated using the formula:
\[ F_{\text{net}} = m_{\text{total}} \cdot a \]
Rearranging gives:
\[ a = \frac{F_{\text{net}}}{m_{\text{total}}} = \frac{9.8 \, \text{N}}{7 \, \text{kg}} \approx 1.4 \, \text{m/s}^2 \]

### Final Answer
The acceleration of the system is approximately \( 1.4 \, \text{m/s}^2 \).

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