Laws of Motion Part 2

Concept:For constant speed, net force on each block is zero.Friction opposes relative motion and is kinetic friction.Explanation:Block C moves left, so string pulls block B right with same speed.For block A: $N_{AB} = m_A g = 4 \times 10 = 40\ \mathrm{N}$.Friction between A and B: $f_{AB} = \mu N_{AB} = 0.5 \times 40 = 20\ \mathrm{N}$ (opposes relative motion).For block B: vertical forces give $N_{BC} = (m_A + m_B)g = (4+6)\times10 = 100\ \mathrm{N}$.Friction between B and C: $f_{BC} = \mu N_{BC} = 0.5 \times 100 = 50\ \mathrm{N}$ (opposes relative motion).For ground: $N_C = (m_A+m_B+m_C)g = (4+6+8)\times10 = 180\ \mathrm{N}$.Ground friction on C: $f_g = \mu N_C = 0.5 \times 180 = 90\ \mathrm{N}$ (opposes motion of C).Block B moves at constant speed: tension $T = f_{AB} + f_{BC} = 20 + 50 = 70\ \mathrm{N}$.Block C moves at constant speed: applied force $F = T + f_{BC} + f_g = 70 + 50 + 90 = 210\ \mathrm{N}$.Answer:$F = 210\ \mathrm{N}$ (Option A).