A block of mass m is kept on a platform Platform starts moving upwards with an acceleration of `g/2` . Find the work done by the normal force on the block in the first one second.

Options

A. `(3mg^2)/2`
B. zero
C. `(3mg^2)/8`
D. `(3mg^2)/4`

Answer

To solve the problem, we need to find the work done by the normal force on a block of mass \( m \) placed on a platform that accelerates upwards with an acceleration of \( \frac{g}{2} \) in the first one second.

### Step-by-Step Solution:

1. **Identify the Forces Acting on the Block:**
- The gravitational force acting downwards: \( F_g = mg \)
- The normal force acting upwards: \( N \)
- Since the platform is accelerating upwards, we also consider a pseudo force acting downwards on the block due to the acceleration of the platform: \( F_{pseudo} = ma \) where \( a = \frac{g}{2} \).

2. **Calculate the Normal Force:**
- In the accelerating frame of reference (the platform), the normal force must balance the gravitational force and the pseudo force:
\[ N = mg + ma \]
- Substituting the value of \( a \):
\[ N = mg + m\left(\frac{g}{2}\right) = mg + \frac{mg}{2} = mg + \frac{1}{2}mg = \frac{3}{2}mg \]

3. **Determine the Displacement of the Block in One Second:**
- The block starts from rest, so the initial velocity \( u = 0 \).
- The acceleration of the platform is \( a = \frac{g}{2} \).
- Using the second equation of motion:
\[ s = ut + \frac{1}{2}at^2 \]
- Substituting \( u = 0 \), \( a = \frac{g}{2} \), and \( t = 1 \):
\[ s = 0 \cdot 1 + \frac{1}{2} \cdot \frac{g}{2} \cdot (1)^2 = \frac{1}{2} \cdot \frac{g}{2} = \frac{g}{4} \]

4. **Calculate the Work Done by the Normal Force:**
- Work done \( W \) by the normal force is given by:
\[ W = N \cdot s \]
- Substituting \( N = \frac{3}{2}mg \) and \( s = \frac{g}{4} \):
\[ W = \left(\frac{3}{2}mg\right) \cdot \left(\frac{g}{4}\right) = \frac{3}{8}mg^2 \]

5. **Final Result:**
- The work done by the normal force on the block in the first one second is:
\[ W = \frac{3}{8}mg^2 \]