As shown in the figure, a heavy block rests on a frictionless surface and is pulled by two ropes, each having a mass m . The force applied to the left rope is removed at t = 0 , while the force on the right end continues to act. Let F₁ and F₂ represent the magnitudes of the forces exerted on the block by the right rope and the left rope, respectively.

Step-by-step solution

Let M be the mass of the heavy block. For t The system is pulled by equal and opposite forces F . The net external force is zero, so the acceleration of the system is a = 0 . For the left rope of mass m , the equation of motion is F - F₂ = m(0) F₂ = F . For the right rope of mass m , the equation of motion is F - F₁ = m(0) F₁ = F . Thus, for t For t > 0 : The left force is removed. The net force on the entire system is F to the right. The acceleration of the system is a = F M + 2m . For the left rope, the only hori