A man weighs `80kg` . He stands on a weighing scale in a lift which is moving upwords with a uniform acceleration of `5m//s^(2)` . What would be the reading on the scale?

Options

A. `400N`
B. `800N`
C. `1200N`
D. Zero

Explanation

To solve the problem, we need to determine the reading on the weighing scale when a man weighing 80 kg is standing on it in a lift that is accelerating upwards at 5 m/s².

### Step-by-Step Solution:

1. **Identify the Forces Acting on the Man:**
- The weight of the man (W) acting downwards is given by:
\[ W = m \cdot g \]
where \( m = 80 \, \text{kg} \) and \( g = 10 \, \text{m/s}^2 \) (approximate value of gravitational acceleration).
- Therefore,
\[ W = 80 \, \text{kg} \cdot 10 \, \text{m/s}^2 = 800 \, \text{N} \]

2. **Determine the Pseudo Force:**
- When the lift accelerates upwards, a pseudo force (F_p) acts downwards on the man due to the acceleration of the lift. This force can be calculated using:
\[ F_p = m \cdot a \]
where \( a = 5 \, \text{m/s}^2 \) (the upward acceleration of the lift).
- Therefore,
\[ F_p = 80 \, \text{kg} \cdot 5 \, \text{m/s}^2 = 400 \, \text{N} \]

3. **Calculate the Normal Force (Reading on the Scale):**
- The normal force (N), which is the reading on the scale, must balance both the weight of the man and the pseudo force acting downwards. Thus, we can write:
\[ N = W + F_p \]
- Substituting the values we calculated:
\[ N = 800 \, \text{N} + 400 \, \text{N} = 1200 \, \text{N} \]

4. **Conclusion:**
- The reading on the scale when the lift is moving upwards with an acceleration of \( 5 \, \text{m/s}^2 \) is:
\[ \text{Reading on the scale} = 1200 \, \text{N} \]

### Final Answer:
The reading on the scale is **1200 N**.

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