Two blocks having masses m₁ and m₂ are connected by a spring of spring constant k and placed on a smooth horizontal surface. Each block is pulled by a constant force F in opposite directions. Find the maximum elongation of the spring and the distances moved by the blocks of masses m₁ and m₂ respectively till that instant.

Step-by-step solution

Let the acceleration of the centre of mass be a = F - (-F) m₁ + m₂ = 2F m₁ + m₂ . In the centre of mass frame, each block experiences an effective force stretching the spring. The relative acceleration between the two blocks is: a_ rel = F m₁ + F m₂ = F ( 1 m₁ + 1 m₂ ) = F(m₁ + m₂) m₁ m₂ Using work-energy principle for relative motion: 1 2 v_ rel ^2 = 1 2 k x_ max ^2 , where = m₁ m₂ m₁ + m₂ Also, effective force in relative motion is 2F , so maximum extension occurs when: k x_ max = 2F x_ max = 2F k Distances moved