Two bodies whose masses are `m_1 = 50 kg and m_2 = 150kg` are tied by a light string and are placed on a frictionless horizontal table . When m 1 is pulled by a horizontal force F = 1000 N, find the acceleration produced in the bodies and the tension in the string .

Options

A. `5m//s^(2),750N`
B. `20m//s^(2),650N`
C. `10m//s^(2),550N`
D. `15m//s^(2),450N`

Answer

To solve the problem, we need to determine the acceleration of the two bodies and the tension in the string connecting them. Here’s a step-by-step breakdown of the solution:

### Step 1: Understand the System
We have two masses, \( m_1 = 50 \, \text{kg} \) and \( m_2 = 150 \, \text{kg} \), connected by a light string on a frictionless table. A force \( F = 1000 \, \text{N} \) is applied to \( m_1 \).

### Step 2: Define the Variables
Let:
- \( a \) = acceleration of the system
- \( T \) = tension in the string

### Step 3: Write the Equations of Motion
For \( m_1 \):
The net force acting on \( m_1 \) is the applied force minus the tension in the string:
\[ F - T = m_1 \cdot a \tag{1} \]

For \( m_2 \):
The only force acting on \( m_2 \) is the tension in the string:
\[ T = m_2 \cdot a \tag{2} \]

### Step 4: Combine the Equations
From equation (1):
\[ F - T = m_1 \cdot a \]
Substituting equation (2) into equation (1):
\[ F - m_2 \cdot a = m_1 \cdot a \]
Rearranging gives:
\[ F = m_1 \cdot a + m_2 \cdot a \]
Factoring out \( a \):
\[ F = (m_1 + m_2) \cdot a \]

### Step 5: Solve for Acceleration
Substituting the known values:
\[ 1000 \, \text{N} = (50 \, \text{kg} + 150 \, \text{kg}) \cdot a \]
\[ 1000 = 200 \cdot a \]
\[ a = \frac{1000}{200} = 5 \, \text{m/s}^2 \]

### Step 6: Find the Tension
Now, we can find the tension using equation (2):
\[ T = m_2 \cdot a \]
Substituting the known values:
\[ T = 150 \, \text{kg} \cdot 5 \, \text{m/s}^2 = 750 \, \text{N} \]

### Final Answers
- The acceleration \( a = 5 \, \text{m/s}^2 \)
- The tension \( T = 750 \, \text{N} \)