NEET Class neet Answered
question from Newtons laws

Asked by beuradeshraj | 20 Jan, 2023, 10:40: AM

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

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

Answered by Thiyagarajan K | 21 Jan, 2023, 09:38: AM
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