Two blocks connected by a rope of negligible mass are being dragged by a horizontal force (see figure below). Suppose F = 65.0 N, m1 = 14.0 kg, m2 = 26.0 kg, and the coefficient of kinetic friction between each block and the surface is 0.098
For mass m₁: The normal reaction is m₁g. The resisting force is R₁ = μm₁g.
For mass m₂: The normal reaction is m₂g. The resisting force is R₂ = μm₂g.
Let a = the acceleration of the system. Then (m₁ + m₂)a = F - (R₁ + R₂) (14+26 kg)*(a m/s²) = (65 N) - 0.098*(9.8 m/s²)*(14+26 kg) 40a = 65 - 38.416 = 26.584 a = 0.6646 m/s²
