With what minimum acceleration can a fireman slides down a rope while breaking strength of the rope is 2/3 his weight

The correct answer is g/3.

Let's carefully understand the physical setup and solve it step by step.

Given Information:
- Breaking strength of the rope = $\frac{2}{3}$ × weight of fireman
- Let the mass of the fireman = m
- So, maximum tension the rope can withstand: $T=\frac{2}{3}mg$

Step 1: Understand the forces acting on the fireman

When the fireman slides down the rope, two forces act on him:

1. Weight (downward): $W=mg$
2. Tension in rope (upward): $T$

Step 2: Apply Newton's Second Law

Since the fireman is sliding downward, taking downward as positive, the net force equation is:

$mg-T=ma$

where $a$ is the downward acceleration of the fireman.

Step 3: Find the condition for minimum acceleration

The fireman wants to slide down as slowly (safely) as possible, meaning he wants minimum acceleration downward. To minimize acceleration, the tension in the rope must be as large as possible.

The maximum tension the rope can provide without breaking is:

$T=\frac{2}{3}mg$

Step 4: Substitute and solve for minimum acceleration

Substituting the maximum tension into Newton's second law:

$mg-\frac{2}{3}mg=ma\text{(min)}$

$mg\left(1-\frac{2}{3}\right)=ma\text{(min)}$

$mg\times \frac{1}{3}=ma\text{(min)}$

Dividing both sides by $m$:

$a\text{(min)}=\frac{g}{3}$

Step 5: Physical Interpretation

- If the fireman were in free fall, he would accelerate at $g$ downward (rope has zero tension).
- If the fireman were stationary or moving at constant speed, the rope tension would equal his full weight $mg$ — but this would break the rope since the rope can only hold $\frac{2}{3}mg$.
- Therefore, the fireman must have some downward acceleration. The minimum safe downward acceleration while keeping the rope intact is $\frac{g}{3}$.

The minimum acceleration with which the fireman must slide down so the rope does not break is:

$a\text{(min)}=\frac{g}{3}$