A man of mass 50 kg is standing in an elevator. If elevator is moving up with an acceleration `(g)/(3)` then work done by normal reaction of elevator floor on man when elevator moves by a distance 12 m is `(g=10 m//s^(2))`:-

Options

A. 2000 J
B. 4000 J
C. 6000 J
D. 8000 J

Answer

To solve the problem, we need to calculate the work done by the normal reaction of the elevator floor on the man when the elevator moves upwards by a distance of 12 m. Here’s the step-by-step solution:

### Step 1: Identify the Given Data
- Mass of the man, \( m = 50 \, \text{kg} \)
- Acceleration of the elevator, \( a = \frac{g}{3} \)
- Distance moved by the elevator, \( d = 12 \, \text{m} \)
- Gravitational acceleration, \( g = 10 \, \text{m/s}^2 \)

### Step 2: Calculate the Weight of the Man
The weight of the man, \( W \), can be calculated using the formula:
\[ W = mg \]
Substituting the values:
\[ W = 50 \, \text{kg} \times 10 \, \text{m/s}^2 = 500 \, \text{N} \]

### Step 3: Calculate the Pseudo Force
The pseudo force acting on the man due to the upward acceleration of the elevator is given by:
\[ F_{\text{pseudo}} = ma = m \left(\frac{g}{3}\right) \]
Substituting the values:
\[ F_{\text{pseudo}} = 50 \, \text{kg} \times \left(\frac{10 \, \text{m/s}^2}{3}\right) = \frac{500}{3} \, \text{N} \]

### Step 4: Calculate the Normal Reaction Force
The normal reaction force \( N \) exerted by the elevator floor on the man can be calculated using:
\[ N = W + F_{\text{pseudo}} = mg + ma \]
Substituting the values we calculated:
\[ N = 500 \, \text{N} + \frac{500}{3} \, \text{N} \]
To combine these, convert \( 500 \, \text{N} \) into a fraction:
\[ N = \frac{1500}{3} \, \text{N} + \frac{500}{3} \, \text{N} = \frac{2000}{3} \, \text{N} \]

### Step 5: Calculate the Work Done by the Normal Force
The work done \( W_d \) by the normal force when the elevator moves upwards by a distance \( d \) is given by:
\[ W_d = N \cdot d \cdot \cos(\theta) \]
Since the normal force and the displacement are in the same direction, \( \theta = 0 \) degrees and \( \cos(0) = 1 \):
\[ W_d = N \cdot d \]
Substituting the values:
\[ W_d = \left(\frac{2000}{3} \, \text{N}\right) \times 12 \, \text{m} = \frac{24000}{3} \, \text{J} = 8000 \, \text{J} \]

### Final Answer
The work done by the normal reaction of the elevator floor on the man when the elevator moves by a distance of 12 m is:
\[ \boxed{8000 \, \text{J}} \]