The person o( mass 50 kg slands on a weighing scale on a lift. If the lift is ascending upwards with a uniform acceleration of `9ms^(-2)`, what would be the reading of the weighting scale? `("Take g"=10ms^(-2))`

Options

A. 50kg
B. 60kg
C. 96kg
D. 176kg

Explanation

To solve the problem, we need to determine the reading on the weighing scale when a person of mass 50 kg is in a lift that is accelerating upwards at 9 m/s². We will use Newton's laws of motion to find the normal force acting on the person, which corresponds to the reading on the scale.

### Step-by-Step Solution:

1. **Identify the Forces Acting on the Person**:
- The weight of the person (W) acting downwards is given by:
\[ W = m \cdot g \]
where \( m = 50 \, \text{kg} \) and \( g = 10 \, \text{m/s}^2 \).
- Therefore, the weight \( W \) is:
\[ W = 50 \, \text{kg} \cdot 10 \, \text{m/s}^2 = 500 \, \text{N} \]

2. **Determine the Net Force Acting on the Person**:
- Since the lift is accelerating upwards, we need to consider the effective acceleration. The effective acceleration \( a \) is the sum of gravitational acceleration and the lift's acceleration:
\[ a_{\text{effective}} = g + a = 10 \, \text{m/s}^2 + 9 \, \text{m/s}^2 = 19 \, \text{m/s}^2 \]

3. **Calculate the Normal Force (N)**:
- The normal force \( N \) is what the scale reads, and it can be calculated using the formula:
\[ N = m \cdot a_{\text{effective}} = m \cdot (g + a) \]
- Substituting the values:
\[ N = 50 \, \text{kg} \cdot 19 \, \text{m/s}^2 = 950 \, \text{N} \]

4. **Convert the Normal Force to kg Reading on the Scale**:
- The reading on the scale in kg can be found by dividing the normal force by \( g \):
\[ \text{Reading on scale} = \frac{N}{g} = \frac{950 \, \text{N}}{10 \, \text{m/s}^2} = 95 \, \text{kg} \]

### Final Answer:
The reading of the weighing scale will be **95 kg**.