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

Options

A. `(2)/(3) g`
B. g
C. `(1)/(3) g`
D. zero

Explanation

To solve the problem of determining the minimum acceleration with which a fireman can slide down a rope without breaking it, we can follow these steps:

### Step 1: Identify the forces acting on the fireman
The fireman experiences two main forces:
- The gravitational force acting downwards, which is equal to his weight \( W = mg \) (where \( m \) is the mass of the fireman and \( g \) is the acceleration due to gravity).
- The tension \( T \) in the rope acting upwards.

### Step 2: Set up the equation of motion
Since the fireman is sliding down the rope with acceleration \( a \), we can apply Newton's second law. The net force acting on the fireman can be expressed as:
\[ F_{\text{net}} = mg - T \]
According to Newton's second law, this net force is also equal to the mass times acceleration:
\[ F_{\text{net}} = ma \]
Thus, we can write:
\[ ma = mg - T \]

### Step 3: Determine the maximum tension in the rope
The problem states that the breaking strength of the rope is \( \frac{2}{3} \) of the fireman's weight. Therefore, we can express the maximum tension \( T \) as:
\[ T = \frac{2}{3} mg \]

### Step 4: Substitute the tension into the equation of motion
Now, we substitute the expression for \( T \) into the equation of motion:
\[ ma = mg - \frac{2}{3}mg \]

### Step 5: Simplify the equation
We can simplify the right side:
\[ ma = mg - \frac{2}{3}mg = \frac{1}{3}mg \]

### Step 6: Solve for acceleration \( a \)
Now, we can solve for \( a \) by dividing both sides by \( m \):
\[ a = \frac{1}{3}g \]

### Conclusion
The minimum acceleration with which the fireman can slide down the rope without breaking it is:
\[ \boxed{\frac{1}{3}g} \]