CBSE Class 11-science Answered
9th sum
Asked by lovemaan5500 | 23 Jan, 2019, 08:19: PM
Expert Answer
top portion of figure gives the free body diagram of both the blocks when force is applied on 10-gram block to hold it.
Tension in the string = 0.03×9.8 = 0.294 N ;
required force to hold the 10-gram block, F = 0.03×9.8 - 0.01×9.8 = 0.02×9.8 = 0.196 N
Free body diagrams of both the blocks are given in the bottom portion of figure.
10-gram block is moving upwards with acceleration a m/s2 .
30-gram block is moving downwards with acceleration a m/s2
at 10-gm block, T+0.01a = 0.01g ...........(1)
at 30-gm block, T-0.03a = 0.03g .............(2)
By subtracting (2) from (1), we get a = - g/2 = -4.9 m/s2 ( -ve sign indicates acceleration is downwards)
by substituting a in eqn.(1), we get T = 0.147 N
Answered by Thiyagarajan K | 23 Jan, 2019, 11:38: PM
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