Two blocks with masses `m_(1) = 0.2` kg and `m_(2) = 0.3` kghang 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 upward at `2 m//s^(2)`

Explanation

To find the tensions in the strings when two blocks of masses \( m_1 = 0.2 \, \text{kg} \) and \( m_2 = 0.3 \, \text{kg} \) are accelerating upward at \( 2 \, \text{m/s}^2 \), we can follow these steps:

### Step 1: Identify the forces acting on the blocks
- For block \( m_1 \):
- The tension \( T_1 \) acts upward.
- The weight \( m_1 g \) acts downward.
- For block \( m_2 \):
- The tension \( T_2 \) acts upward.
- The weight \( m_2 g \) acts downward.

### Step 2: Write the equations of motion
- For block \( m_1 \):
\[ T_1 - m_1 g = m_1 a \]
- For block \( m_2 \):
\[ T_2 - m_2 g = m_2 a \]

### Step 3: Substitute the values
Given:
- \( g = 10 \, \text{m/s}^2 \)
- \( a = 2 \, \text{m/s}^2 \)
- \( m_1 = 0.2 \, \text{kg} \)
- \( m_2 = 0.3 \, \text{kg} \)

Substituting these values into the equations:

1. For block \( m_1 \):
\[ T_1 - (0.2)(10) = (0.2)(2) \]
\[ T_1 - 2 = 0.4 \]
\[ T_1 = 2 + 0.4 = 2.4 \, \text{N} \]

2. For block \( m_2 \):
\[ T_2 - (0.3)(10) = (0.3)(2) \]
\[ T_2 - 3 = 0.6 \]
\[ T_2 = 3 + 0.6 = 3.6 \, \text{N} \]

### Step 4: Final results
The tensions in the strings are:
- \( T_1 = 2.4 \, \text{N} \)
- \( T_2 = 3.6 \, \text{N} \)

### Summary
- Tension in the string above block \( m_1 \) (T1): \( 2.4 \, \text{N} \)
- Tension in the string above block \( m_2 \) (T2): \( 3.6 \, \text{N} \)

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