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Dear Souvik ,
PLease see the above free body diagram.
writing equations--
for mass 4m : Fr = 4ma2 ----------- (1)
for mass m placed above 4m : T - Fr = ma1 ---------------(2)
For hanging mass m : mg -T = ma1 ( acceleration is same as string has to be tight )-------------(3)
Fr = µkmg ---------------------(4)
solving we get , a2 = µkg /4
a1 =g -µkg /2
When only half of the block is still on the 4m mass block , distance covered = 3l/2
time taken = (2s/a)1/2 =(3l/a1)1/2
in that time distance travelled by m mass = 1/2 (a2)t2 = 1/2 *(µkg /4)*(6l/(1-µk) g) = 3µkl/4(1 - µk) m
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Respected Sir,
I have also proceeded like this way and arrived to the same answer provided by you, but the ans. given in the material is something like this: x= 7μl/[8(2-3μ)]
I hope you will clear this confusion.
how didi u come to the conclusion that distance travelled by the m block is 3L/2?
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