A block of mass 8 kg is placed on a rough horizontal table. It is connected by a light inextensible string passing over a smooth pulley to a hanging block of mass 2 kg . When the system is released from rest, it accelerates at 1 m s ⁻² . What is the coefficient of kinetic friction between the 8 kg block and the table? (Take g = 10 m s ⁻² )

Step-by-step solution

Let the mass of the block on the table be m₁ = 8 kg and the hanging mass be m₂ = 2 kg . The driving force for the system is the weight of the hanging mass: W₂ = m₂ g = 2 10 = 20 N . The opposing force is the kinetic friction on the 8 kg block: f_k = _k N = _k (m₁ g) = _k (8 10) = 80 _k . According to Newton's second law for the entire system: Net force = (m₁ + m₂) a m₂ g - f_k = (m₁ + m₂) a 20 - 80 _k = (8 + 2) 1 20 - 80 _k = 10 80 _k = 10 _k = 10 80 = 0.125 The coefficient of kinetic friction is 0.125 . Answer: 0.