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 force exerted by the elevator floor on the man as the elevator moves upward.

### Step-by-Step Solution:

1. **Identify the Given Values:**
- Mass of the man, \( m = 50 \, \text{kg} \)
- Acceleration due to gravity, \( g = 10 \, \text{m/s}^2 \)
- Acceleration of the elevator, \( a = \frac{g}{3} = \frac{10}{3} \, \text{m/s}^2 \)
- Distance moved by the elevator, \( d = 12 \, \text{m} \)

2. **Calculate the Total Acceleration:**
The total effective acceleration acting on the man while the elevator is moving upward is the sum of the gravitational acceleration and the elevator's acceleration:
\[ a_{\text{effective}} = g - a = 10 - \frac{10}{3} = \frac{30}{3} - \frac{10}{3} = \frac{20}{3} \, \text{m/s}^2 \]

3. **Calculate the Normal Force:**
The normal force \( N \) exerted by the elevator floor on the man can be calculated using Newton's second law:
\[ N = m \cdot a_{\text{effective}} = 50 \cdot \frac{20}{3} = \frac{1000}{3} \, \text{N} \]

4. **Calculate the Work Done by the Normal Force:**
Work done \( W \) by the normal force when the elevator moves a distance \( d \) is given by:
\[ W = N \cdot d = \left(\frac{1000}{3}\right) \cdot 12 \]
\[ W = \frac{1000 \cdot 12}{3} = \frac{12000}{3} = 4000 \, \text{J} \]

### 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 \( 4000 \, \text{J} \).