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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?

Respuesta :

MathPhys

Answer:

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

Lanuel

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 kg
  • Mass of block 2 = 0.25 kg
  • Mass of solid disk (pulley) = 0.50 kg
  • Radius = 0.10 m.
  • Height = 0.30 m

Scientific data:

  • Acceleration due to gravity = 9.8 [tex]m/s^2[/tex]

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:

[tex]M_2gh = \frac{1}{2} M_1V^2 + \frac{1}{2} M_2V^2 + \frac{1}{2} I\omega^2 + M_1gh\\\\M_2gh -M_1gh= \frac{1}{2} M_1V^2 + \frac{1}{2} M_2V^2 + \frac{1}{2} I\omega^2\\\\(M_2-M_1)gh= \frac{1}{2} M_1V^2 + \frac{1}{2} M_2V^2 + \frac{1}{2} I\omega^2\\\\(M_2-M_1)2gh =M_1V^2+M_2V^2 + I\omega^2[/tex]  .....equation 1.

For a solid disk:

[tex]I =\frac{1}{2} Mr^2[/tex]    .....equation 2.

For angular speed:

[tex]\omega = \frac{V}{r}[/tex]   .....equation 3.

Substituting the value I and [tex]\omega[/tex] into eqn. 1, we have:

[tex](M_2-M_1)2gh =M_1V^2+M_2V^2 + (\frac{1}{2} Mr^2)(\frac{V}{r}) ^2\\\\(M_2-M_1)2gh =M_1V^2+M_2V^2 + \frac{1}{2} MV^2\\\\(M_2-M_1)4gh =2M_1V^2+2M_2V^2 + MV^2\\\\(M_2-M_1)4gh =(2M_1+2M_2 + M)V^2\\\\V^2=\frac{(M_2-M_1)4gh}{2M_1\;+\;2M_2 \;+\; M} \\\\V=\sqrt{\frac{(M_2-M_1)4gh}{2M_1\;+\;2M_2 \;+\; M}}[/tex]

Substituting the given parameters into the formula, we have;

[tex]V=\sqrt{\frac{(0.25-0.20)\times 4\times 9.8 \times 0.3}{2(0.20)\;+\;2(0.25) \;+\; 0.50}}\\\\V=\sqrt{\frac{0.05\times 4\times 9.8 \times 0.3}{0.40\;+\;0.50 \;+\; 0.50}}\\\\V=\sqrt{\frac{0.588}{1.40}}\\\\V=\sqrt{0.42}[/tex]

Speed, V = 0.648 m/s.

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