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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
Expert Answer
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)
.........................(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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