In the figure shown, the left block of mass 1 kg is struck sharply with an impulse J, horizontal to the right. The spring of stiffness 200 N/m was initially in its natural state. The floor has friction coefficient µ = 1/4 and the mass of the right block is
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When impulse J is applied to the 1 kg block, it gains initial velocity v = J/1. The block moves right, compressing the spring. Friction opposes motion with force f = μN = (1/4)(1)(10) = 2.5 N. The spring force F = kx = 200x. The block stops when work done against friction equals initial kinetic energy: (1/2)(1)v² = f × distance. Maximum compression occurs when spring force overcomes friction. The right block (mass m) remains stationary if friction force μmg ≥ maximum spring force 200x_max. Solve for m using energy and force balance.
Final answer: The mass m of the right block is 4 kg.