In the arrangement shown in the figure, work done by the string on the block of mass (0.36 ~kg ) during the first second after the blocks are released from state of rest is (Ignore friction and mass of the string.) (Acceleration due to gravity, (g ) (=10 ~ms ⁻² ))

Step-by-step solution

According to the question, Given, mass of block (1, m₁=0.36 ~kg ) mass of block (2, m₂=0.72 ~kg ) and acceleration due to gravity, (g=10 ~m / s ^2 ) Now, Dynamic equation of block (m₁ (m₂ > m₁ ) ). ( T-m₁ g=m₁ a )...(i) Dynamic equation of block (m₂ (m₂ > m₁ ) ). ( m₂ g-T=m₂ a (ii) ) Thus, from adding Eqs. (i) and (ii), we get ( T-m₁ g+m₂ g-T=m₁ a+m₂ a ) (m₂ g-m₁ g=a (m₁+m₂ ) ) or (a= (m₂-m₁ ) g (m₁+m₂ ) ) Hence, acceleration, (a= ( m₂-m₁ m₁+m₂ ) g ) ( ) Acceleration, (a= (0.72-0.36) 1.08 10 ) (a= 3.6 1.08 ~m / s ^