Question

A monkey of mass 20kg is holding a vertical rope, which breaks under the weight of mass $25kg$. The maximum acceleration with which the monkey can climb up the rope is:
(A) $7.5m/{s^2}$
(B) $5m/{s^2}$
(C) $2m/{s^2}$
(D) $2.5m/{s^2}$

Answer 1

Hint: The maximum tension in the rope is fixed. So it can hold a mass of $25kg$with no acceleration or a mass less than it with some amount of acceleration. When the monkey moves up through the rope, the force on the rope is a sum of the monkey’s weight and the force due to its acceleration.

Complete step by step solution:
Mass of the monkey,$M = 20kg$
It is given that the rope breaks at the mass of, $M' = 25kg$
The maximum tension that the rope can handle is given by,$F = m'g$
If g is taken as $10m/{s^2}$, the maximum tension is given by-
$F = 25 \times 10 = 250N$
The free body diagram of the monkey and the rope is given by-

From the diagram it is evident that the tension (T) in the rope is equal to the normal contact force (N) between the monkey and the rope. The rope is stationary while the monkey moves up with an acceleration a.
Thus the acceleration of the monkey can be given by,
(assumed direction, upwards positive)
$Ma = N - Mg$
From the free body diagram of the rope we know that,
$N = T$
By substituting this value,
$Ma = T - Mg$
Rewriting it as-
$T = Mg + Ma$
$T = M(g + a)$
Substituting values of $M = 20kg$,$T = 250N$ and $g = 10m/{s^2}$
We have,
$250 = 20(10 + a)$
$a = \dfrac{{250}}{{20}} - 10$
$a = 12.5 - 10$
$a = 2.5m/{s^2}$

Therefore, option (D) is correct.

Note The force experienced by the rope when the monkey climbs is given by the tension produced in it. Since the rope is fixed, it does not move, hence only the monkey has a net acceleration. The mass of the rope is neglected (it is considered massless) as it is not provided in the question.