Question Detail
Question
A 1.5 kg cylinder descends vertically from a pulley, causing a 5 kg block to accelerate on a horizontal table. A string rotates with the pulley of mass 0.4 kg without sliding on the surface. The coefficient of kinetic friction between the block and the table is 0.2. Determine the value of the acceleration of the block and cylinder?
Correct Answer
Option A is the correct answer.
Detailed Explanation
This is a complex system with pulley inertia and friction. Let a be acceleration. For hanging mass m₁=1.5 kg: m₁g - T₁ = m₁a. For block m₂=5 kg on table: T₂ - μ m₂g = m₂a. For pulley I = ½MR², torque = (T₁ - T₂)R = Iα = (½MR²)(a/R) = ½MRa. So T₁ - T₂ = ½Ma. Substitute T₁ = m₁g - m₁a, T₂ = m₂a + μ m₂g. Then m₁g - m₁a - m₂a - μ m₂g = ½Ma → m₁g - μ m₂g = a(m₁ + m₂ + ½M) → a = (m₁g - μ m₂g)/(m₁ + m₂ + ½M) = (1.5×9.8 - 0.2×5×9.8)/(1.5 + 5 + 0.2) = (14.7 - 9.8)/(6.7) = 4.9/6.7 = 0.7313 m/s². Option C is 0.7313 m/s².
Hint
Use Newton's second law for each mass and torque equation for pulley.