Placed on a smooth horizontal surface, two blocks of masses m₁ and m₂ are joined by a spring having a spring constant k . A constant force F₁ pulls the block of mass m₁ , while the other block is pulled by a constant force F₂ . Determine the maximum elongation the spring will undergo.

Step-by-step solution

Let the displacements of the blocks m₁ and m₂ be x₁ (to the left) and x₂ (to the right) respectively. The total elongation of the spring is x = x₁ + x₂ . The equations of motion for the two blocks are: m₁ x ₁ = F₁ - kx m₂ x ₂ = F₂ - kx Dividing by their respective masses, we get the accelerations: x ₁ = F₁ m₁ - kx m₁ x ₂ = F₂ m₂ - kx m₂ The relative acceleration of the blocks (rate of change of elongation) is: x = x ₁ + x ₂ = ( F₁ m₁ + F₂ m₂ ) - kx ( 1 m₁ + 1 m₂ ) x = m₂ F₁ + m₁ F₂ m₁ m₂ - k(m₁ + m₂) m₁ m₂ x Let th