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7 years ago

68 Points
```										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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```
7 years ago
souvik das
33 Points
```										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.
```
7 years ago
Rishabh Dabral
32 Points
```										how didi u come to the conclusion that distance travelled by the m block is 3L/2?
```
7 years ago
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