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 lower block. Take g = 10 m/s ^2 .

Step-by-step solution

Let m₁ = 2 kg be the mass of the upper block and m₂ = 4 kg be the mass of the lower block. The maximum static friction between the two blocks is given by f_ max = m₁ g = 0.20 2 10 = 4 N . Assuming both blocks move together as a single system, their common acceleration a is: a = F m₁ + m₂ = 12 2 + 4 = 2 m/s ^2 The frictional force required to provide this acceleration to the upper block is: f_ req = m₁ a = 2 2 = 4 N Since the required friction f_ req is equal to the maximum available static friction f_ max , the blo