# Monkey and Banana problem

I need someone to clarify something I don't understand.

There's a monkey climbing a rope that passes over a frictionless pulley with bananas higher up on the other side of the pulley. The bananas and the monkey weigh the same.

Since the monkey accelerates upwards, he must have a net force upwards equal to the bananas upwards net force. Since their mass is the same, they both accelerate at the same rate upwards so the distance between them stays constant. When the monkey lets go of the rope, there is no more tension in the rope so both the monkey and banana fall at $$g$$, and the distance between them is still constant.

I've solved up to here but I don't understand the last part of the problem. The monkey grabs the rope to stop his fall before he reaches the ground, what do the bananas do?

My thinking is, the monkey exerts a force equal to his weight and so the rope pulls him up with a tension equal to his weight. Therefore, it should continue downwards at constant velocity. But the bananas also experience the same tension which balances the forces on them, so they should also continue moving down at constant velocity, but they can't both move down if the rope is inextensible, so which is it?

• Hint: Think about what the bananas did when the monkey let go of the rope. – zeta-band Nov 6 '19 at 22:59
• They were accelerating down at g since the rope had no tension on it, right? So they should continue down at their final speed $v_f$ when the monkey grabs a hold of the rope again. – user552217 Nov 6 '19 at 23:01
• Since the monkey accelerates upwards - Where does his come from? What actually is the question? Have you abbreviated it or missed something out? As it is written, there is no question. – Oscar Bravo Nov 7 '19 at 7:11
• The question was ambiguous and just said the monkey climbs up, i assumed it was accelerated and not at a constant speed. Either way, it doesn't affect the final part of the problem that I was concerned with. – user552217 Nov 7 '19 at 12:03