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Grade: 12
        

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

10 years ago

Answers : (2)

Abhijith A Nair
37 Points
							
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
 
 
 
5 years ago
ruhi
16 Points
							
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
one year ago
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