A 60 kg man stands on an elevator floor The elevator is going up with constant acceleration of 1.96 m/`s^2`. Percentage change in the apparent weight of the person is

Options

A. 10
B. 15
C. 20
D. 25

Answer

To solve the problem, we will follow these steps:

### Step 1: Identify the given values
- Mass of the man (m) = 60 kg
- Acceleration of the elevator (a) = 1.96 m/s²
- Acceleration due to gravity (g) = 9.8 m/s²

### Step 2: Calculate the original weight of the man
The original weight (W) of the man can be calculated using the formula:
\[ W = mg \]
Substituting the values:
\[ W = 60 \, \text{kg} \times 9.8 \, \text{m/s}^2 = 588 \, \text{N} \]

### Step 3: Calculate the apparent weight of the man in the accelerating elevator
When the elevator is accelerating upwards, the apparent weight (N) of the man is given by:
\[ N = mg + ma \]
This can be simplified to:
\[ N = m(g + a) \]
Substituting the values:
\[ N = 60 \, \text{kg} \times (9.8 \, \text{m/s}^2 + 1.96 \, \text{m/s}^2) \]
\[ N = 60 \, \text{kg} \times 11.76 \, \text{m/s}^2 = 705.6 \, \text{N} \]

### Step 4: Calculate the change in weight
The change in weight (ΔW) is given by:
\[ \Delta W = N - W \]
Substituting the values:
\[ \Delta W = 705.6 \, \text{N} - 588 \, \text{N} = 117.6 \, \text{N} \]

### Step 5: Calculate the percentage change in weight
The percentage change in weight can be calculated using the formula:
\[ \text{Percentage Change} = \left( \frac{\Delta W}{W} \right) \times 100\% \]
Substituting the values:
\[ \text{Percentage Change} = \left( \frac{117.6 \, \text{N}}{588 \, \text{N}} \right) \times 100\% \]
\[ \text{Percentage Change} = 0.2 \times 100\% = 20\% \]

### Final Answer
The percentage change in the apparent weight of the person is **20%**.

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