The mass of man when standing on the lift is 60 kg. The weight when the lift is moving upwards with acceration `4.9 ms^(-2)` is

Options

A. 882 N
B. 600 N
C. 306 N
D. zero

Answer

To solve the problem, we need to determine the weight of a man when he is standing in a lift that is moving upwards with an acceleration of \(4.9 \, \text{m/s}^2\). The mass of the man is given as \(60 \, \text{kg}\).

### Step-by-Step Solution:

1. **Identify the Forces Acting on the Man**:
- The weight of the man acting downwards is given by \(W = mg\), where \(m\) is the mass and \(g\) is the acceleration due to gravity.
- There is also an additional force due to the upward acceleration of the lift, which can be represented as \(F = ma\), where \(a\) is the acceleration of the lift.

2. **Calculate the Weight of the Man**:
- The weight of the man when the lift is stationary (or not accelerating) is:
\[ W = mg \]
- Given \(m = 60 \, \text{kg}\) and taking \(g = 9.8 \, \text{m/s}^2\):
\[ W = 60 \, \text{kg} \times 9.8 \, \text{m/s}^2 = 588 \, \text{N} \]

3. **Calculate the Pseudo Force**:
- The pseudo force acting on the man due to the lift's upward acceleration is:
\[ F = ma \]
- Here, \(a = 4.9 \, \text{m/s}^2\):
\[ F = 60 \, \text{kg} \times 4.9 \, \text{m/s}^2 = 294 \, \text{N} \]

4. **Calculate the Effective Weight in the Lift**:
- When the lift is accelerating upwards, the effective weight \(W'\) of the man is the sum of his weight and the pseudo force:
\[ W' = W + F \]
- Substituting the values we calculated:
\[ W' = 588 \, \text{N} + 294 \, \text{N} = 882 \, \text{N} \]

### Final Answer:
The weight of the man when the lift is moving upwards with an acceleration of \(4.9 \, \text{m/s}^2\) is \(882 \, \text{N}\).

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