JEE Class main Answered
Please solve question no.7
Asked by vidyavikram10 | 05 Aug, 2019, 04:43: PM
Expert Answer
Obviously, 10kg block is sliding down and 5 kg block is pulled up.
Forces acting on the blocks are shown in figure. T is tension force pulling the block.
friction force f1 = 0.3×10g cos37 is acting against the motion of 10 kg block as shown in figure.
friction force f2 = 0.1×5g cos53 is acting against the motion of 5 kg block as shown in figure.
( g is acceleration due to gravity )
Let a be the acceleration of both the blocks.
If we apply newtons law to 10 kg block, we get, 10g sin37 -T -0.3×10g cos37 = 10×a ....................(1)
If we apply newtons law to 5 kg block, we get, T -5g sin53 -0.1×5g cos53 = 5×a ....................(2)
By adding eqn.(1) and (2), tension T is eliminated. Then we get,
10g sin37 -0.3×10g cos37 -5g sin53 -0.1×5g cos53 = 15×a ......................(3)
from eqn.(3), We get acceleration a = 0.95 m/s2
Answered by Thiyagarajan K | 05 Aug, 2019, 06:07: PM
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