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Question: A man and a block, both of mass M, are suspended from two ends of a rope, passing over a frictionles...

A man and a block, both of mass M, are suspended from two ends of a rope, passing over a frictionless pulley. If the man starts moving in such a way that acceleration of the block becomes 2 m/s2 in upward direction then acceleration of the man with respect to the rope is (Pulley is massless and

neglect friction everywhere):

A

1 m/s2

B

2 m/s2

C

4m/s2

D

6 m/s2

Answer

4m/s2

Explanation

Solution

For block

T – mg = 2 m/s2

And for the man if a is his acceleration w.r.t. the ground in upward direction

T – mg = ma

a = 2m/s2

(hus w.r.t. the ground, both move up by an acceleration 2 m/s22 \mathrm {~m} / \mathrm { s } ^ { 2 }. Suppose theman moves up by acceleration 'b' w.r.t. rope. As rope is coming down by acceleration 2 m/s22 \mathrm {~m} / \mathrm { s } ^ { 2 }, net acceleration of the man in upward direction =b−2=2= \mathrm { b } - 2 = 2 b=4 m/s2b = 4 \mathrm {~m} / \mathrm { s } ^ { 2 }

Thus w.r.t the ground, both move up by an acceleration 2m/s2. Suppose theman moves up by acceleration ‘b’ w.r.t. rope. As rope is coming down by acceleration 2 m/s2, et acceleration of = b – 2 =3

b = 4m/s2.