A light inextensible string that goes over a smooth fixed pulley connect two blocks of masses 0.36kg and 0.72kg taking g=10m/second square find the work done in joules by the string on the block of 0.36kg during the first second after the system is released from rest
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the answer is in the attachment
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Answer:
2mg − T = 2ma
T − mg = ma
⇒ a=g/3
T = 4mg/3 m
W=T.s=T.\[12at2\] =8Joules
Explanation:
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