Mechanics> laws of motion 1...

two blocks A and B of mass m_aand 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 ?

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chanakya rock ing , 16 Years ago

Grade 12

2 Answers

Abhijith A Nair

Last Activity: 11 Years ago

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

Last Activity: 7 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

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Mechanics> laws of motion 1...

chanakya rock ing , 16 Years ago

Abhijith A Nair

ruhi

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