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souvik das Grade: 12
        

226_21405_IMG.jpg

6 years ago

Answers : (3)

AJIT AskiitiansExpert-IITD
68 Points
										

Dear Souvik ,


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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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6 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.

6 years ago
Rishabh Dabral
32 Points
										

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

6 years ago
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