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

Answer

To solve the problem, we need to determine the reading on the weighing scale when the man is in a lift that is accelerating upwards.

### Step-by-Step Solution:

1. **Identify the Forces Acting on the Man:**
- The weight of the man (W) acts downwards, which can be calculated using the formula:
\[ W = mg \]
- The normal force (N) exerted by the weighing scale acts upwards.

2. **Calculate the Weight of the Man:**
- Given that the mass (m) of the man is 80 kg and taking the acceleration due to gravity (g) as approximately \(10 \, \text{m/s}^2\):
\[ W = 80 \, \text{kg} \times 10 \, \text{m/s}^2 = 800 \, \text{N} \]

3. **Apply Newton's Second Law:**
- Since the lift is moving upwards with an acceleration (a) of \(5 \, \text{m/s}^2\), we can apply Newton's second law. The net force (F_net) acting on the man can be expressed as:
\[ F_{\text{net}} = N - W \]
- According to Newton's second law, this net force is also equal to the mass times the acceleration:
\[ F_{\text{net}} = ma \]

4. **Set Up the Equation:**
- Combining the two expressions for net force, we have:
\[ ma = N - W \]
- Rearranging gives:
\[ N = ma + W \]

5. **Substitute the Known Values:**
- Now substitute the values into the equation:
- Mass (m) = 80 kg
- Acceleration (a) = \(5 \, \text{m/s}^2\)
- Weight (W) = \(800 \, \text{N}\)
\[ N = (80 \, \text{kg} \times 5 \, \text{m/s}^2) + 800 \, \text{N} \]
- Calculate:
\[ N = 400 \, \text{N} + 800 \, \text{N} = 1200 \, \text{N} \]

6. **Conclusion:**
- The reading on the scale, which is the normal force (N), is \(1200 \, \text{N}\).

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