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question from Newtons laws
Figure shows the forces acting on both the blocks A and B.

Let m1 be the mass of block-A and g is acceleration due to gravity.

( m1 g ) is the weight of block-A that is placed on table . Hence block-A gets equal reaction force R = m1 g .

Let μ be the friction coefficient between the bottom surface of block-A and top surface of table.

( μ R ) = ( μ m1 g ) is the friction force acting in the direction opposite to motion of block-A .

Let T be the tension force acting along the string that pulls the block-A .

Let the system is in accelerated motion with acceleration a

By Newton's second law , we have

T - ( μ m1 g ) = m1 a  ............................................ (1)

Similarly for block-B, we have

( m2 g ) - T = m2 a  ......................................(2)

By adding eqn.(1) and eqn.(2) , we get

(m2 g ) - ( μ m1 g ) = ( m1 + m2) a

Hence acceleration a is obtained from above expression as

................................... (3)
We get tension force T from eqn.(2) as

T = m2 ( g - a ) ............................(4)

By ubstituting acceleration a from eqn.(3) , we get tension force from eqn.(4) after simplification as

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