Q172 / 180 · COMEDK 2022 · Physics
With what minimum acceleration can a fireman slide down a rope while breaking strength of the rope is $\dfrac{2}{3}$ of the weight?
- A$\dfrac{2}{3} g$
- B$g$
- Correct answerC$\dfrac{1}{3} g$
- Dzero
Solution
Let $m$ be the mass of the fireman and $g$ be the acceleration due to gravity. The weight of the fireman is $W = mg$.
The breaking strength of the rope is given as $T_{max} = \dfrac{2}{3} mg$.
When the fireman slides down with an acceleration $a$, the equation of motion is $mg - T = ma$, where $T$ is the tension in the rope.
To prevent the rope from breaking, the tension $T$ must not exceed $T_{max}$. Thus, $T \le \dfrac{2}{3} mg$.
Substituting the maximum tension into the equation of motion: $mg - \dfrac{2}{3} mg = ma$.
Solving for $a$: $\dfrac{1}{3} mg = ma$, which gives $a = \dfrac{1}{3} g$.
Since the fireman must slide with at least this acceleration to keep the tension within the breaking limit, the minimum acceleration required is $\dfrac{1}{3} g$.
Answer: $\dfrac{1}{3} g$
The breaking strength of the rope is given as $T_{max} = \dfrac{2}{3} mg$.
When the fireman slides down with an acceleration $a$, the equation of motion is $mg - T = ma$, where $T$ is the tension in the rope.
To prevent the rope from breaking, the tension $T$ must not exceed $T_{max}$. Thus, $T \le \dfrac{2}{3} mg$.
Substituting the maximum tension into the equation of motion: $mg - \dfrac{2}{3} mg = ma$.
Solving for $a$: $\dfrac{1}{3} mg = ma$, which gives $a = \dfrac{1}{3} g$.
Since the fireman must slide with at least this acceleration to keep the tension within the breaking limit, the minimum acceleration required is $\dfrac{1}{3} g$.
Answer: $\dfrac{1}{3} g$
Hence, the correct answer is option: 'C'.