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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
Force acting on B = FLet Fext be the force exerted by the personTherefore the acceleration of the system Fext=( Ma+Mb)A -------------1Acceleration is common for both the blocks.Now, the force on B by A = F= Mb * AA=F/Mb --------- 2 Sub 2 in 1 Fext =(Ma+Mb)F / Mb
let the force exerted by the person is Fp now the acceleration of the system isa=Fp/Ma+Mb -------------1now let the force applied by block A on block B is Ftherefore F=Mb×aa=F/Mb--------------2since, the acceleration of the whole system is same then by eqauting equation 1 and 2 we getF/Mb=Fp/Ma+MbFp=(Ma+Mb)×F/Mbit is also return as Fp=( Ma/Mb+1)×F
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