A 2 kg block is positioned over a 4 kg block, and both are placed on a smooth horizontal surface. The coefficient of friction between the blocks is 0.20 . Determine the acceleration of the two blocks if a horizontal force of 12 N is applied to the upper block. Take g = 10 m/s ^2 .

Step-by-step solution

The maximum friction force between the two blocks is given by f_ max = m₁ g f_ max = 0.20 2 10 = 4 N Assuming both blocks move together, their common acceleration would be a = F m₁ + m₂ a = 12 2 + 4 = 2 m/s ^2 The force required to move the lower block with this acceleration is f = m₂ a = 4 2 = 8 N Since f > f_ max , the blocks will slip over each other. The friction force acting between them will be f_k = 4 N . For the upper block, the equation of motion is F - f_k = m₁ a₁ 12 - 4 = 2 a₁ a₁ = 4 m/s ^2 For the lower