Two blocks with masses `m_(1) = 0.2` kg and `m_(2) = 0.3` kg hang one under other as shown in figure. Find the tensions in the strings (massless) in the following situations :
`(g = 10 m//s^(2))`
they accelerate downward at `2 m//s^(2)`

Explanation

To find the tensions in the strings when the two blocks are accelerating downward at \(2 \, \text{m/s}^2\), we can follow these steps:

### Step 1: Identify the forces acting on each block
For Block 1 (mass \(m_1 = 0.2 \, \text{kg}\)):
- Weight of Block 1: \(W_1 = m_1 \cdot g = 0.2 \cdot 10 = 2 \, \text{N}\)
- Tension in the string above Block 1: \(T_1\)

For Block 2 (mass \(m_2 = 0.3 \, \text{kg}\)):
- Weight of Block 2: \(W_2 = m_2 \cdot g = 0.3 \cdot 10 = 3 \, \text{N}\)
- Tension in the string above Block 2: \(T_2\)

### Step 2: Write the equations of motion for each block
For Block 1 (accelerating downward):
\[ W_1 - T_1 = m_1 \cdot a \]
Substituting the known values:
\[ 2 - T_1 = 0.2 \cdot 2 \]
\[ 2 - T_1 = 0.4 \]
\[ T_1 = 2 - 0.4 = 1.6 \, \text{N} \]

For Block 2 (also accelerating downward):
\[ W_2 - T_2 = m_2 \cdot a \]
Substituting the known values:
\[ 3 - T_2 = 0.3 \cdot 2 \]
\[ 3 - T_2 = 0.6 \]
\[ T_2 = 3 - 0.6 = 2.4 \, \text{N} \]

### Step 3: Final Results
The tensions in the strings are:
- \(T_1 = 1.6 \, \text{N}\)
- \(T_2 = 2.4 \, \text{N}\)