With what minimum acceleration can a fireman slide down a rope whose breaking strength is two third of his weight?

Question

With what minimum acceleration can a fireman slide down a rope whose breaking strength is two third of his weight? (A) (g/2) (B) (2/3) g (C) (3/2) g (D) (g/3)

Tardigrade Admin · Accepted Answer

Net downward force in the string, ma=mg-T But Tmin =(2/3)mg ∴ mamin =mg-(2/3)mg amin=g-(2/3)g=(g/3)

Q. With what minimum acceleration can a fireman slide down a rope whose breaking strength is two third of his weight?

$\frac{g}{2}$

$\frac{2}{3}$ $g$

$\frac{3}{2}$ $g$

$\frac{g}{3}$

Net downward force in the string, $ma = m g - T$
But $T_{min}$ $= \frac{2}{3} m g$ $∴$ $m a_{min}$ $= m g - \frac{2}{3} m g$
$a_{min} = g - \frac{2}{3} g = \frac{g}{3}$

Q. With what minimum acceleration can a fireman slide down a rope whose breaking strength is two third of his weight?

2g​

32​ g

23​ g

3g​

Net downward force in the string, ma=mg−T But Tmin​ =32​mg ∴ mamin​ =mg−32​mg amin​=g−32​g=3g​

$\frac{g}{2}$

$\frac{2}{3}$ $g$

$\frac{3}{2}$ $g$

$\frac{g}{3}$

Net downward force in the string, $ma = m g - T$
But $T_{min}$ $= \frac{2}{3} m g$ $∴$ $m a_{min}$ $= m g - \frac{2}{3} m g$
$a_{min} = g - \frac{2}{3} g = \frac{g}{3}$