two blocks A and B of mass m a and m b respectivly are kept in contact on a frictionless table .the experimenter pushes the block A from behind so that the blocks accelerate if the block A exerts a force F on the block B .what is the force exerted by the experimenter on A ? KINDLY SEND ME ANS FR THIS QUESTION WITH F.B.D

Question

Abhijith A Nair · Answer

Force acting on B = F Let F ext be the force exerted by the person Therefore the acceleration of the system Fext=( M a +M b )A -------------1 Acceleration is common for both the blocks. Now, the force on B by A = F= M b * A A=F/M b --------- 2 Sub 2 in 1 F ext =(M a +M b )F / M b

ruhi · Answer

let the force exerted by the person is Fp now the acceleration of the system is a=Fp/Ma+Mb -------------1 now let the force applied by block A on block B is F therefore F=Mb×a a=F/Mb--------------2 since, the acceleration of the whole system is same then by eqauting equation 1 and 2 we get F/Mb=Fp/Ma+Mb Fp=(Ma+Mb)×F/Mb it is also return as Fp=( Ma/Mb+1)×F

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