Two blocks having masses of 400 g and 200 g are connected by a light string passing over a pulley that is free to rotate about its axis. The pulley possesses a moment of inertia of 1.6 10⁻⁴ kg-m ^2 and a radius of 2.0 cm . Determine the kinetic energy of the system when the 400 g block falls through a distance of 50 cm .

Step-by-step solution

By the principle of conservation of mechanical energy, the gain in the total kinetic energy of the system is equal to the net loss in its potential energy. Let m₁ = 400 g = 0.4 kg and m₂ = 200 g = 0.2 kg . When the heavier block m₁ falls through a distance h = 50 cm = 0.5 m , the lighter block m₂ rises by the same distance h . The net loss in potential energy of the system is given by: U = m₁ g h - m₂ g h = (m₁ - m₂) g h Substituting the given values and taking g = 9.8 m/s ^2 : U = (0.4 - 0.2) 9.8 0.5 U = 0.2 9.8 0