# two blocks A and B of mass ma and mb 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

Abhijith A Nair
37 Points
10 years ago
Force acting on B = F
Let Fext be the force exerted by the person
Therefore the acceleration of the system Fext=( Ma+Mb)A     -------------1
Acceleration is common for both the blocks.
Now, the force on B by A = F= Mb * A
A=F/Mb --------- 2
Sub 2 in 1
Fext =(Ma+Mb)F / Mb

ruhi
16 Points
5 years ago
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