Question

With what minimum acceleration (in $m/s{sec}^2$) can a fireman slide down a rope whose breaking strength is 50% of his weight? $(g=10m/s^2)$

Answer 1

Let the weight of man be $x.N$

Breaking point of rope is $\frac{x}{g}N$

$\Rightarrow$ mass of man $=\frac{x}{g}kg$

$\Rightarrow$ Net acceleration $=(g-a)$ downward.

Force on rope $=ma$ net

$=\frac{x}{g}(g-a)$

$\Rightarrow ATQ\frac{x}{2}=\frac{x}{g}(g-a)$

$g=2g-2a$

$2a=-g$

$a=\frac{g}{2}$

Hence, the answer is $\frac{g}{2}.$