A monkey of mass 40 kg climbs up a rope, of breaking load 600 N hanging from a ceiling. If it climbs up the rope with the maximum possible acceleration, then the time taken by monkey to climb up is [Length of rope is 10 m ]

Question

A monkey of mass 40 kg climbs up a rope, of breaking load 600 N hanging from a ceiling. If it climbs up the rope with the maximum possible acceleration, then the time taken by monkey to climb up is [Length of rope is 10 m ] (A) 2 s (B) 1 s (C) 4 s (D) 3 s

Tardigrade Admin · Accepted Answer

600-400 =40 a a =(200/40)=5 m / s 2 S =u t+(1/2) a t2 10 =(1/2)(5) t2 ⇒ t =2 second

Q. A monkey of mass $40 k g$ climbs up a rope, of breaking load $600 N$ hanging from a ceiling. If it climbs up the rope with the maximum possible acceleration, then the time taken by monkey to climb up is [Length of rope is $10 m]$

2 s

1 s

4 s

3 s

$600 - 400 = 40 a$
$a = \frac{200}{40} = 5 m / s^{2}$
$S = u t + \frac{1}{2} a t^{2}$
$10 = \frac{1}{2} (5) t^{2}$
$\Rightarrow t = 2$ second

Q. A monkey of mass 40kg climbs up a rope, of breaking load 600N hanging from a ceiling. If it climbs up the rope with the maximum possible acceleration, then the time taken by monkey to climb up is [Length of rope is 10m]

2 s

1 s

4 s

3 s

600−400=40a a=40200​=5m/s2 S=ut+21​at2 10=21​(5)t2 ⇒t=2 second

Q. A monkey of mass $40 k g$ climbs up a rope, of breaking load $600 N$ hanging from a ceiling. If it climbs up the rope with the maximum possible acceleration, then the time taken by monkey to climb up is [Length of rope is $10 m]$

$600 - 400 = 40 a$
$a = \frac{200}{40} = 5 m / s^{2}$
$S = u t + \frac{1}{2} a t^{2}$
$10 = \frac{1}{2} (5) t^{2}$
$\Rightarrow t = 2$ second