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?
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Ответ:
The speed of this block when it hits the ground is equal to 0.648 m/s.
Given the following data:
Mass of block 1 = 0.20 kgMass of block 2 = 0.25 kgMass of solid disk (pulley) = 0.50 kgRadius = 0.10 m.Height = 0.30 mScientific data:
Acceleration due to gravity = 9.8To 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:
For a solid disk:
For angular speed:
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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Ответ:
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
Ответ:
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