The elastic limit of an elavator cable is `2xx10^(9)N//m^(2)`. The maximum upward acceleration that an elavator of mass `2xx10^(3)kg` can have when supported by a cable whose cross sectional area is `10^(-4)m^(2)`, provided the stres in cable would not exceed half to the elastic limit would be

Options

A. `10ms^(-2)`
B. `50 ms^(-2)`
C. `40 ms^(-2)`
D. Not possible to move up

Explanation

To find the maximum upward acceleration that an elevator can have without exceeding half of the elastic limit of the cable, we can follow these steps:

### Step 1: Understand the Problem
We have an elevator of mass \( m = 2 \times 10^3 \, \text{kg} \) and a cable with a cross-sectional area \( A = 10^{-4} \, \text{m}^2 \). The elastic limit of the cable is \( \sigma_{\text{max}} = 2 \times 10^9 \, \text{N/m}^2 \). We need to find the maximum upward acceleration \( a \) such that the stress in the cable does not exceed half of the elastic limit.

### Step 2: Calculate the Maximum Allowable Stress
Since the stress in the cable should not exceed half of the elastic limit, we calculate:
\[ \sigma_{\text{max allowable}} = \frac{\sigma_{\text{max}}}{2} = \frac{2 \times 10^9}{2} = 1 \times 10^9 \, \text{N/m}^2 \]

### Step 3: Write the Equation for Tension
The tension \( T \) in the cable can be expressed using Newton's second law. The forces acting on the elevator are the tension \( T \) upward and the weight \( mg \) downward. Thus, we have:
\[ T - mg = ma \]
Rearranging gives:
\[ T = mg + ma \]

### Step 4: Relate Tension to Stress
Stress in the cable is defined as:
\[ \sigma = \frac{T}{A} \]
Substituting the expression for \( T \):
\[ \sigma = \frac{mg + ma}{A} \]

### Step 5: Set Up the Inequality
We need to ensure that the stress does not exceed the maximum allowable stress:
\[ \frac{mg + ma}{A} \leq \sigma_{\text{max allowable}} \]
Substituting the values:
\[ \frac{(2 \times 10^3 \, \text{kg})(9.81 \, \text{m/s}^2) + (2 \times 10^3 \, \text{kg})a}{10^{-4} \, \text{m}^2} \leq 1 \times 10^9 \, \text{N/m}^2 \]

### Step 6: Calculate the Weight of the Elevator
Calculating the weight:
\[ mg = 2 \times 10^3 \times 9.81 = 19620 \, \text{N} \]

### Step 7: Substitute and Solve for \( a \)
Substituting \( mg \) into the inequality:
\[ \frac{19620 + 2000a}{10^{-4}} \leq 1 \times 10^9 \]
Multiplying both sides by \( 10^{-4} \):
\[ 19620 + 2000a \leq 1 \times 10^5 \]
Rearranging gives:
\[ 2000a \leq 1 \times 10^5 - 19620 \]
Calculating the right side:
\[ 2000a \leq 80380 \]
Dividing by 2000:
\[ a \leq \frac{80380}{2000} = 40.19 \, \text{m/s}^2 \]

### Step 8: Conclusion
Thus, the maximum upward acceleration \( a \) that the elevator can have is approximately:
\[ a \approx 40 \, \text{m/s}^2 \]