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]​

Answer 1

Mass =40kg

g=10m/s²

Tmax=600N

T-mg=ma

T=m(g+a)

600=40(10+a)

a=5m/s²

Time taken by the monkey to climb up(length of rope is 10m)

S=ut+½at²

10=0×t+½5×t²

t=2sec

Answer 2

The time taken by monkey to  climb up is 2 seconds.

Explanation:

Given that,

Mass of the monkey, m = 40 kg

Breaking load of the rope, F = 600 N

Length of the rope, d = 10 m

We need to find the time taken by monkey to  climb up. Let a is the acceleration of the monkey. It is given by :

T mg = ma

T = m(g + a)

$a=\dfrac{T-mg}{m}$

$a=\dfrac{600-40\times 10}{40}$

$a=5\ m/s^2$

Let t is the time taken by monkey to  climb up. It can be calculated using second equation of motion as :

$d=ut+\dfrac{1}{2}at^2$

Initial speed of the monkey, u = 0

$d=\dfrac{1}{2}at^2$

$t=\sqrt{\dfrac{2d}{a}}$

$t=\sqrt{\dfrac{2\times 10}{5}}$

t = 2 seconds

So, the time taken by monkey to  climb up is 2 seconds. Hence, this is the required solution.

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Second law of motion

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