Two blocks A and B of equal mass are connected by a spring of force constant 10 Nm–1 and separated by a distance 1 m. A constant force 5 N is acting on block A. Find the maximum separartion during the motion.
(Neglect the frictional force between the block and surface).
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Two blocks connected by a spring experience a constant force on block A. The center of mass accelerates, but the relative motion between the blocks is simple harmonic about the center of mass. The maximum separation occurs when the spring is at its maximum extension from the equilibrium position under the force.
The force stretches the spring to a new equilibrium length. The extension from the natural length is given by Hooke's Law: , so . The maximum extension in SHM is equal to the amplitude. The initial condition (spring unstretched at 1m separation) gives the amplitude A = x₀ = 0.5m. Therefore, the maximum separation is the natural length plus the total stretch: 1m + x₀ + A = 1m + 0.5m + 0.5m = 2m.
Final Answer: 2 m