Answer:
The mass of the second block is 0.94 kg
Step-by-step 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
![k = (mg)/(x) \\\\(m_1g)/(x_1) = (m_2g)/(x_2) \\\\But, x_2 \ is \ triple \ of \ x_1,\\x_2 = 3x_1\\\\(m_1g)/(x_1) = (m_2g)/(3x_1)\\\\(m_1)/(x_1) = (m_2)/(3x_1)\\\\Substitute \ in \ the \ value \ of \ m_1\\\\(0.47)/(x_1) = (m_2)/(3x_1)\\\\(0.47)/(1) = (m_2)/(3)\\\\m_2 = 3 *0.47\\\\m_2 = 1.41 \ kg](https://img.qammunity.org/2021/formulas/physics/college/dpfpsob4mmjzlegmop9zzn5oapasuvpa7k.png)
mass of the second block alone = 1.41 - 0.47 = 0.94 kg
Therefore, the mass of the second block is 0.94 kg