IIT JEE , AIEEE Forum - Physics , Mechanics - CHALLENGE-3

frenied

hey thats not my solution....ur unnecessarliy complimenting me..the very nice steps were told by my sir....my original method is a bit different....and i really dunno if its correct..

rhd92781

so what was that u're calling "unworldly"

Betu

why two displacements x1 & x2. It has to be same x.

now consider mass m1 upon whom F1 ia acting.

let's say the accelaratio of the system is a (= [F1-F2]/[m1+m2])

F1-kx=m1.a

Kx=F1-m1a=F1-m1(F1-F2)/(m1+m2)

x= 1/k[(m2.F1+m1.F2)/ (m1+m2)]

you can check the correctness of the answer by putting m1=m2 and/or F1=F2

when F1=F2,there will be no acceleration and extention x= F/k , simple. If m1=m2 , x= (F1+F2)/2k, again simple.

rohith291991

ok ive solved it like this.... consider the frame of the center of mass....now in this frame no net force acts on center of mass and hence it is stationary..
now acc of center of mass in ground frame is a=(F2-F1)/(m1+m2).....now net force on m1 apart from spring force= F1+m1a....let the displacement of this block from
the center of mass be x1... then,,,work done=(F1+m1a)x1...similarly for m2....(F2-m2a)x2...now total work done = elastic energy of the spring...hence....(F1+m1a)x1+(F2-m2a)x2=k(x1+x2)²/2....now in order to solve we need one last vital relation between x1 and x2...see...since center of mass is always at rest in this frame...we get that x1=(m2/m1)x2...now on substituting we get the answer as 2/k[(F1m2+F2m1)/(m1+m2)].....i believe this method is flawless...

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