Two bodies of masses 1 kg and 2 kg are connected by a very light string passed over a clamped light smooth pulley. If the system is released from rest, find the acceleration of the two masses and the tension in the string

Options

A. `3.27 m//s^(2), 13 N`
B. `4.5 m//s^(2), 13 N`
C. `1.27 m//s^(2), 13 N`
D. `0.327 m//s^(2) , 13 N`

Answer

To solve the problem of two bodies connected by a string over a pulley, we will follow these steps:

### Step 1: Identify the Forces Acting on Each Mass
- For the 1 kg mass (m1):
- Weight (W1) = m1 * g = 1 kg * 9.8 m/s² = 9.8 N (acting downwards)
- Tension (T) in the string (acting upwards)

- For the 2 kg mass (m2):
- Weight (W2) = m2 * g = 2 kg * 9.8 m/s² = 19.6 N (acting downwards)
- Tension (T) in the string (acting upwards)

### Step 2: Write the Equations of Motion
- For the 1 kg mass (m1):
\[ T - W1 = m1 \cdot a \implies T - 9.8 = 1 \cdot a \quad (1) \]

- For the 2 kg mass (m2):
\[ W2 - T = m2 \cdot a \implies 19.6 - T = 2 \cdot a \quad (2) \]

### Step 3: Solve the Equations Simultaneously
- From Equation (1):
\[ T = a + 9.8 \quad (3) \]

- Substitute Equation (3) into Equation (2):
\[ 19.6 - (a + 9.8) = 2a \]
\[ 19.6 - 9.8 - a = 2a \]
\[ 9.8 = 3a \]
\[ a = \frac{9.8}{3} \approx 3.27 \, \text{m/s}^2 \]

### Step 4: Calculate the Tension in the String
- Substitute the value of \( a \) back into Equation (3):
\[ T = 3.27 + 9.8 = 13.07 \, \text{N} \approx 13 \, \text{N} \]

### Final Results
- The acceleration of the two masses is approximately \( 3.27 \, \text{m/s}^2 \).
- The tension in the string is approximately \( 13 \, \text{N} \).