With what minimum acceleration can a fireman slide down a rope whose breaking strength is `(2//3)` of his weight?

Options

A. `(g)/(2)`
B. `(2)/(3)g`
C. `(3)/(2)g`
D. `(g)/(3)`

Explanation

To solve the problem, we need to determine the minimum acceleration at which a fireman can slide down a rope without breaking it. The breaking strength of the rope is given as \( \frac{2}{3} \) of the fireman's weight.

### Step-by-Step Solution:

1. **Identify Forces Acting on the Fireman:**
- The weight of the fireman, \( W = mg \), acts downward.
- The tension in the rope, \( T \), acts upward.
- The net force acting on the fireman when he slides down is given by \( F_{net} = mg - T \).

2. **Apply Newton's Second Law:**
- According to Newton's second law, the net force is also equal to mass times acceleration:
\[ F_{net} = ma \]
- Therefore, we can write:
\[ mg - T = ma \]

3. **Express Tension in Terms of Weight and Acceleration:**
- Rearranging the equation gives us:
\[ T = mg - ma \]

4. **Set Up the Breaking Strength Condition:**
- The rope will break if the tension exceeds its breaking strength. The breaking strength is given as:
\[ T_{breaking} = \frac{2}{3} mg \]
- Thus, we set up the inequality:
\[ mg - ma \leq \frac{2}{3} mg \]

5. **Rearranging the Inequality:**
- Rearranging the inequality gives:
\[ mg - \frac{2}{3}mg \leq ma \]
- Simplifying the left side:
\[ \frac{1}{3}mg \leq ma \]

6. **Dividing by Mass:**
- Assuming \( m \neq 0 \), we can divide both sides by \( m \):
\[ \frac{1}{3}g \leq a \]
- This indicates that the minimum acceleration \( a \) must be greater than or equal to \( \frac{1}{3}g \).

7. **Conclusion:**
- Therefore, the minimum acceleration with which the fireman can slide down the rope without breaking it is:
\[ a = \frac{g}{3} \]

### Final Answer:
The minimum acceleration with which a fireman can slide down the rope is \( \frac{g}{3} \).

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