A wooden block of mass 2 kg rests on a soft horizontal floor . When aniron cylinder of mass 25 kg is placed on top of the block , the floor yields steadily , and the block and the cylinder go down with an acceleration of `0.1 ms^(-2)` What is the action of the block on the floor (a) before and (b) after the floor yields ? Take `g = 10 ms^(-2)` . Identify the action reaction pairs in the problem .

Explanation

To solve the problem, we need to analyze the forces acting on the wooden block before and after the iron cylinder is placed on it.

### Step-by-Step Solution:

**(a) Action of the block on the floor before the floor yields:**

1. **Identify the forces acting on the block:**
- The weight of the block (downward force) is given by \( W = mg \), where \( m = 2 \, \text{kg} \) and \( g = 10 \, \text{m/s}^2 \).
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