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A monkey is climbing on a rope that goes over a smooth light pulley and supports a block of equal mass at the other end. Show that whatever force the monkey exerts on the rope, the monkey and the block move on the same direction with equal acc:n. If initially both were at rest, their separation will not change as time passes. A monkey is climbing on a rope that goes over a smooth light pulley and supports a block of equal mass at the other end. Show that whatever force the monkey exerts on the rope, the monkey and the block move on the same direction with equal acc:n. If initially both were at rest, their separation will not change as time passes.
Solution:Suppose the monkey accelerate upward with acceleration `a` and the block accelerates downward with acceleration a`.Let the force exerted by monkey is equal to T.Then T-mg-ma=0T=mg+ma-----(i)Again,for the blockT+ma`-mg=0From (i),mg+ma+ma`-mg=0ma=-ma`a`=-aAcceleration `-a` downwards i.e,`a` upward.The block and the monkey move in the direction with equal acceleration.If initially they are at rest (no force is exerted by monkey)motion of block occurs as they have same weight. Their separation will not change as the time passes.
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