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68 Points
10 years ago

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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souvik das
33 Points
10 years ago

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.

Rishabh Dabral
32 Points
10 years ago

how didi u come to the conclusion that distance travelled by the m block is 3L/2?