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 step by step, we will calculate the work done by the normal reaction of the elevator floor on the man as the elevator moves upward with a given acceleration.

### Step 1: Identify the parameters
- Mass of the man (m) = 50 kg
- Acceleration due to gravity (g) = 10 m/s²
- Acceleration of the elevator (a) = g/3 = 10/3 m/s²
- Distance moved by the elevator (h) = 12 m

### Step 2: Calculate the effective acceleration
When the elevator is moving upward with an acceleration of \( g/3 \), the effective acceleration acting on the man is the sum of gravitational acceleration and the elevator's acceleration.

\[ \text{Effective acceleration} = g + a = g + \frac{g}{3} = g \left(1 + \frac{1}{3}\right) = g \cdot \frac{4}{3} = \frac{40}{3} \text{ m/s}^2 \]

### Step 3: Calculate the work done by the normal force
The work done (W) by the normal force can be calculated using the formula:

\[ W = F \cdot h \]

Where \( F \) is the normal force acting on the man. The normal force can be calculated as:

\[ F = m \cdot \text{Effective acceleration} = m \cdot \frac{40}{3} \]

Substituting the values:

\[ F = 50 \cdot \frac{40}{3} = \frac{2000}{3} \text{ N} \]

Now, substituting \( F \) into the work done formula:

\[ W = \left(\frac{2000}{3}\right) \cdot 12 \]

### Step 4: Simplify the expression for work done
Calculating the work done:

\[ W = \frac{2000 \cdot 12}{3} = \frac{24000}{3} = 8000 \text{ J} \]

### Conclusion
The work done by the normal reaction of the elevator floor on the man when the elevator moves a distance of 12 m is **8000 Joules**.

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