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the blocks a and b are shown in the figure have masses 5 kg and 4kg respectively.the system is released from rest.the speed of b after a has travelled a distance of 1 m along
Asked by preeshitadey07 | 18 Apr, 2019, 08:11: AM
Figure shows the tension forces acting on A and B. Same tension T acting along the rope a-b-c-d-e-f.
Along the rope g-h, tension force 2T is acting because  there are two tension forces T acting on other side of pulley.

Due to movement of block-A , if length increases along a-b by l, then same length l decreases along c-d-e-f.
Since rope c-d-e-f has two folding, then distance moved by block B is l/2 .
Hence if block-A moves downward with acceleration a, then block-B moves upwards with acceleration a/2.

Newtons law for block-B,    2T - 4g = 4(a/2) = 2a ..................(1)

Newtons law for block-A,   5 g sin37 - T = 5a   or  5g (3/5) - T = 5a   or   3g - T = 5a ..............(2)

( please note sin37 = 3/5 )

By solving eqn.(1) and eqn.(2),  we get acceleration a of block-A  as ,   a = g/6  m/s2

acceleration of block-B, a/2 = g/12 m/s2

As stated earlier, if block-A moves a distance 1 m downwards, block-B will move 1/2 m upwards.

Speed of block-B , after bloc-A has moved 1 m = .........................(3)
( In eqn.3, formula "v2 = u2 + 2as " is applied ,  v - final speed, u - initial speed which is zero here, a is acceleration and s is distance moved)
Answered by Thiyagarajan K | 18 Apr, 2019, 08:20: PM

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