A 0.20-kg block and a 0.25-kg block are connected to each other by a string draped over a pulley that is a solid disk of inertia 0.50 kg and radius 0.10 m. When released, the 0.25-kg block is 0.30 m off the ground. What speed does this block have when it hits the ground?

0.65 m/s

Explanation:

Potential energy = Kinetic energy + Potential energy

m₁gh = ½ m₁v² + ½ m₂v² + ½ Iω² + m₂gh

(m₁ - m₂) gh = ½ m₁v² + ½ m₂v² + ½ Iω²

2 (m₁ - m₂) gh = m₁v² + m₂v² + Iω²

For a solid disk, I = ½ Mr². Assuming no slipping, ω = v/r.

2 (m₁ - m₂) gh = m₁v² + m₂v² + (½ Mr²) (v/r) ²

2 (m₁ - m₂) gh = m₁v² + m₂v² + ½ Mv²

4 (m₁ - m₂) gh = 2m₁v² + 2m₂v² + Mv²

4 (m₁ - m₂) gh = (2m₁ + 2m₂ + M) v²

v² = 4 (m₁ - m₂) gh / (2m₁ + 2m₂ + M)

v² = 4 (0.25 kg - 0.20 kg) (9.8 m/s²) (0.30 m) / (2 * 0.25 kg + 2 * 0.20 kg + 0.50 kg)

v² = 0.42 m²/s²

v = 0.65 m/s