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A spring of equilibrium length L1 and spring constant k1 hangs from the ceiling. Mass m1 is suspended from its lower end. Then a second spring, with equilibrium length L2 and spring constant k2, is hung from the bottom of m1. Mass m2 is suspended from this second spring. How far is m2 below the ceiling?

Respuesta :

Answer:

[tex]d = L_1 + L_2 + \frac{(m_1 + m_2)g}{K_1} + \frac{m_2g}{K_2}[/tex]

Explanation:

For first spring the total extension is given as

[tex]F_{net} = K_1 x_1[/tex]

here net force is due to weight of both masses

[tex](m_1 + m_2)g = K_1 x_1[/tex]

so extension of first spring is

[tex]x_1 = \frac{(m_1 + m_2)g}{K_1}[/tex]

now similarly the extension in second spring is only due to the weight of mass m2

so here we will have

[tex]m_2g = K_2 x_2[/tex]

[tex]x_2 =\frac{m_2g}{K_2}[/tex]

so the total distance from the ceiling for mass m2 is given as

[tex]d = L_1 + L_2 + \frac{(m_1 + m_2)g}{K_1} + \frac{m_2g}{K_2}[/tex]

onyebuchinnaji

The total distance of m₂ from the ceiling is [tex]d = L_1 + L_2 + \frac{g(m_1 + m_2)}{k_1} + \frac{m_2g}{k_2}[/tex]

The given parameters;

  • equilibrium of length of the spring, = L₁
  • spring constant, = K₁
  • mass on the spring = m₁
  • length of the first spring, = L₂
  • spring constant = K₂
  • mass on the second spring, = m₂

The extension of the first spring due to mass m₁ and m₂ is calculated as;

[tex]F = kx\\\\ g(m_1+m_2) = k_1x_1\\\\x_1 = \frac{g(m_1+m_2)}{k_1}[/tex]

The extension of the second spring due to mass (m₂) is calculated as;

[tex]m_2g = k_2x_2\\\\x_2 = \frac{m_2g}{k_2}[/tex]

The total distance of m₂ from the ceiling is calculated as follows;

[tex]d = L_1 + L_2 + x_1 + x_2\\\\d = L_1 + L_2 + \frac{g(m_1 + m_2)}{k_1} + \frac{m_2g}{k_2}[/tex]

Thus, the total distance of m₂ from the ceiling is [tex]d = L_1 + L_2 + \frac{g(m_1 + m_2)}{k_1} + \frac{m_2g}{k_2}[/tex]

Learn more here:https://brainly.com/question/22499689

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