Answer to Question #44914 in Mechanics | Relativity for Roshni dhar

Question

A monkey of 20kg mass is holding a light rope that passes over a frictionless pulley.A bunch of bananas of same mass is tied to the other end of rope.In order to get access to the bunch the monkey starts climbing .The distance between the monkey and the bananas is
a)decreasing
b)increasing
c)unchanged
d)nothing can be stated

Answer 1

Since the monkey and bananas have the same mass, they have the same weight, so the tension in the string is the same on the monkey’s end and the banana’s end. Thus, the same forces are acting on both objects (the force of gravity and the tension force in the string,) so the net force on both the monkey and the bananas is the same, which means the acceleration (direction and magnitude) are the same as well.
As the monkey climbs up the rope, the bananas move up as well. This is because the acceleration is the same, so both objects move in the same in the same direction at the same rate.
The distance between the monkey and the bananas stays constant because they have the same velocity (in magnitude and direction) so they are always moving up or down at the same rate.
If the monkey lets go of the rope, the distance between the monkey and the bananas again stays the same. Because both him and the bananas will be in freefall and they have the same mass and the same initial velocity, they will move at the same rate and the distance between them will not change.
If the monkey grabs back on to the rope, the bananas will slow down at the same rate as the bananas because they have the same forces acting upon them, and the same net force. Thus, if the monkey comes to a stop, so will the bananas.
The distance between the monkey and the bananas stays the same - equal and opposite forces act on the monkey and the bananas and since they have the same mass they both accelerate and at the same rate - the distance between them therefore doesn't alter.
Answer: c) unchanged.