Answer:
The mass of the second block is 0.94 kg
Explanation:
Given;
mass of the first block, m₁ = 0.47 kg
let the mass of the first block and second block = m₂
According to Hook's law;
F = kx
where;
F is the applied load (force)
k is the elastic constant
x is the extension of the elastic material
According to Newton's law of motion;
F = ma = mg
Thus,
mg = kx
[tex]k = \frac{mg}{x} \\\\\frac{m_1g}{x_1} = \frac{m_2g}{x_2} \\\\But, x_2 \ is \ triple \ of \ x_1,\\x_2 = 3x_1\\\\\frac{m_1g}{x_1} = \frac{m_2g}{3x_1}\\\\\frac{m_1}{x_1} = \frac{m_2}{3x_1}\\\\Substitute \ in \ the \ value \ of \ m_1\\\\\frac{0.47}{x_1} = \frac{m_2}{3x_1}\\\\\frac{0.47}{1} = \frac{m_2}{3}\\\\m_2 = 3 *0.47\\\\m_2 = 1.41 \ kg[/tex]
mass of the second block alone = 1.41 - 0.47 = 0.94 kg
Therefore, the mass of the second block is 0.94 kg
