Two bodies of mass 5 kg and 7 kg are suspended at the ends of massless string passing over frictionless pulley. The acceleration of the system.

Options

A. `1.6 m//s^(2)`
B. `10 m//s^(2)`
C. `9.8 m//s^(2)`
D. `7 m//s^(2)`

Explanation

To find the acceleration of the system with two bodies of mass 5 kg and 7 kg suspended over a frictionless pulley, we can follow these steps:

### Step 1: Identify the forces acting on each mass.
- For the 5 kg mass (m1), the force due to gravity is \( F_{g1} = m_1 \cdot g = 5 \cdot g \).
- For the 7 kg mass (m2), the force due to gravity is \( F_{g2} = m_2 \cdot g = 7 \cdot g \).

### 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:
\[ F_{net} = F_{g2} - F_{g1} = 7g - 5g = 2g \]

### Step 3: Calculate the total mass of the system.
- The total mass \( M \) of the system is the sum of the two masses:
\[ M = m_1 + m_2 = 5 + 7 = 12 \, \text{kg} \]

### 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:
\[ a = \frac{F_{net}}{M} \]
- Substituting the values we found:
\[ a = \frac{2g}{12} \]

### Step 5: Substitute the value of \( g \).
- Assuming \( g \approx 10 \, \text{m/s}^2 \):
\[ a = \frac{2 \cdot 10}{12} = \frac{20}{12} = \frac{5}{3} \approx 1.67 \, \text{m/s}^2 \]

### Step 6: Final answer.
- The acceleration of the system is approximately \( 1.67 \, \text{m/s}^2 \).

---