NEET Class neet Answered

question from Newtons laws

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

Expert Answer

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

begin mathsize 14px style a space equals space fraction numerator open parentheses m subscript 2 minus mu space m subscript 1 close parentheses over denominator open parentheses m subscript 1 plus m subscript 2 close parentheses end fraction space g end style

................................... (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

begin mathsize 14px style T space equals space open parentheses fraction numerator m subscript 1 space m subscript 2 over denominator m subscript 1 plus m subscript 2 end fraction close parentheses left parenthesis 1 plus mu right parenthesis g end style

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

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