`Mr. X` of mass `80 kg` enters a lift and select the floor he wants .The lift now acceleration upwards at `2 ms^(-2)` for `2 s` and then moves with constant velocity .As the lift approaches his floor ,it decelerates at the same rate as it previously accelerates .If the lift cabes can safely withstant a tension of `2 xx 10^(4) N` and the lift itself has a mass of `500 kg` ,how many `Mr.X' s` could it safely carry at one time?

Options

A. `22`
B. `14`
C. `18`
D. `12`

Explanation

To solve the problem, we need to analyze the forces acting on the lift and Mr. X when the lift is accelerating upwards and when it is decelerating. We will use Newton's second law of motion to derive the number of Mr. X's that the lift can safely carry.

### Step-by-Step Solution:

1. **Identify the Given Data:**
- Mass of Mr. X, \( m = 80 \, \text{kg} \)
- Mass of the lift, \( M = 500 \, \text{kg} \)
- Acceleration of the lift, \( a = 2 \, \text{m/s}^2 \)
- Maximum tension in the cable, \( T_{\text{max}} = 2 \times 10^4 \, \text{N} \)
- Acceleration due to gravity, \( g = 10 \, \text{m/s}^2 \)

2. **Calculate the Weight of Mr. X and the Lift:**
- Weight of Mr. X, \( W_x = m \cdot g = 80 \cdot 10 = 800 \, \text{N} \)
- Weight of the lift, \( W_L = M \cdot g = 500 \cdot 10 = 5000 \, \text{N} \)

3. **Set Up the Equation for Tension:**
When the lift is accelerating upwards, the net force acting on the system can be expressed as:
\[ T - W_L - n \cdot W_x = (M + n \cdot m) \cdot a \]
where \( n \) is the number of Mr. X's in the lift.

4. **Substituting Known Values:**
Rearranging the equation gives:
\[ T = W_L + n \cdot W_x + (M + n \cdot m) \cdot a \]
Substituting the values:
\[ T = 5000 + n \cdot 800 + (500 + n \cdot 80) \cdot 2 \]

5. **Simplifying the Equation:**
Expanding the equation:
\[ T = 5000 + 800n + 1000 + 160n = 6000 + 960n \]
Therefore, we have:
\[ T = 6000 + 960n \]

6. **Setting the Maximum Tension:**
Since the maximum tension the cable can withstand is \( 2 \times 10^4 \, \text{N} \):
\[ 6000 + 960n \leq 20000 \]

7. **Solving for \( n \):**
Rearranging gives:
\[ 960n \leq 20000 - 6000 \]
\[ 960n \leq 14000 \]
\[ n \leq \frac{14000}{960} \approx 14.58 \]

8. **Final Calculation:**
Since \( n \) must be a whole number, the maximum number of Mr. X's that can safely travel in the lift is:
\[ n = 14 \]

### Conclusion:
The lift can safely carry **14 Mr. X's** at one time.