Question Detail
Question
A 1.5 kg cylinder descends vertically from a pulley, causing a 5kg block to accelerate on a horizontal table. A string rotates with the pulley of mass 0.4kg 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 C is the correct answer.
Detailed Explanation
This is a complex system with two masses and a massive pulley. Let a be acceleration. For 5kg block on table: T₁ - f_k = 5a, where f_k = μ_k × 5g = 0.2×5×9.8=9.8N. So T₁ - 9.8 = 5a. For 1.5kg hanging cylinder: 1.5g - T₂ = 1.5a => 14.7 - T₂ = 1.5a. For pulley (disk I=½MR²): (T₂ - T₁)R = Iα = (½×0.4×R²)(a/R) = 0.2Ra => T₂ - T₁ = 0.2a. Solve: From first, T₁ = 5a + 9.8. From second, T₂ = 14.7 - 1.5a. Substitute into third: (14.7 - 1.5a) - (5a + 9.8) = 0.2a => 14.7 - 1.5a - 5a - 9.8 = 0.2a => 4.9 - 6.5a = 0.2a => 4.9 = 6.7a => a = 4.9/6.7 ≈ 0.7313 m/s².
Hint
Set up equations for each mass and the pulley considering torque and force. The pulley's rotation links the tensions.