The blocks A and B shown in fig have masses MA =5kg and MB=4kg.The system is released from rest.The speed of B after A has travelled a distance 1m along the incline is
Asked by m.nilu
21st August 2018,
2:45 PM
Answered by Expert
Answer:
Tension in the rope is due to resolved component 5gsin37 of the 5kg block's weight as shown in figure.
Hence Tension in the rope tied to 4 kg block is 2T.
Acceleration of 4 kg block = (2T - 4g) / 4 = (2×5g×sin37 - 4g)/4 = g/2
if 5kg block moves down by 1 m on the inclined surface, then 4 kg block will move 0.5 m upwards.
hence required veleocity v is obtained from the equation, v2 = 2×(g/2)×0.5 or v = 
Answered by Expert
27th August 2018,
6:23 PM
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