Two masses of 10 kg and 6 kg connected at the two ends of an inextensible string pass over a smooth frictionless pulley . Calculate acceleration of the system and tension in the string .

Options

A. 71.5 N
B. 72.5 N
C. 73.5 N
D. 74.5 N

Explanation

To solve the problem of two masses connected by a string over a frictionless pulley, we can follow these steps:

### Step 1: Identify the forces acting on the masses
- For the 10 kg mass (let's call it \( m_1 \)), the force acting downward due to gravity is \( F_{g1} = m_1 \cdot g = 10 \, \text{kg} \cdot 9.81 \, \text{m/s}^2 = 98.1 \, \text{N} \).
- For the 6 kg mass (let's call it \( m_2 \)), the force acting downward due to gravity is \( F_{g2} = m_2 \cdot g = 6 \, \text{kg} \cdot 9.81 \, \text{m/s}^2 = 58.86 \, \text{N} \).

### Step 2: Set up the equations of motion
- Since the system is connected by an inextensible string, the tension \( T \) in the string will be the same on both sides.
- For mass \( m_1 \) (10 kg), the net force acting on it can be expressed as:
\[ F_{net1} = m_1 \cdot a = F_{g1} - T \]
Therefore, we have:
\[ 10a = 98.1 - T \quad \text{(1)} \]

- For mass \( m_2 \) (6 kg), the net force acting on it can be expressed as:
\[ F_{net2} = m_2 \cdot a = T - F_{g2} \]
Therefore, we have:
\[ 6a = T - 58.86 \quad \text{(2)} \]

### Step 3: Solve the equations simultaneously
- From equation (1):
\[ T = 98.1 - 10a \quad \text{(3)} \]

- Substitute equation (3) into equation (2):
\[ 6a = (98.1 - 10a) - 58.86 \]
\[ 6a = 39.24 - 10a \]
\[ 6a + 10a = 39.24 \]
\[ 16a = 39.24 \]
\[ a = \frac{39.24}{16} = 2.465 \, \text{m/s}^2 \]

### Step 4: Calculate the tension in the string
- Now, substitute the value of \( a \) back into equation (3) to find \( T \):
\[ T = 98.1 - 10(2.465) \]
\[ T = 98.1 - 24.65 = 73.45 \, \text{N} \]

### Final Answer:
- The acceleration of the system is \( a = 2.465 \, \text{m/s}^2 \).
- The tension in the string is \( T = 73.45 \, \text{N \).