A0.20-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?

The speed of this block when it hits the ground is equal to 0.648 m/s.

Given the following data:

Scientific data:

To determine the speed of block 2 when it hits the ground, we would apply the law of conservation of energy:

Potential energy of block 2 = Total kinetic energy of blocks and solid disk + potential energy of block 1.

Mathematically, this is given by the formula:

 .....equation 1.

For a solid disk:

   .....equation 2.

For angular speed:

  .....equation 3.

Substituting the value I and into eqn. 1, we have:

Substituting the given parameters into the formula, we have;

Speed, V = 0.648 m/s.

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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

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