Questions: A 0.68 kg block is hung from and stretches a spring that is attached to the ceiling. A second block is attached to the first one, and the amount that the spring stretches from its unstretched length triples. What is the mass of the second block?

Solution Steps

In the first scenario, a block of mass \(m_1 = 0.68 \, \text{kg}\) is hung from a spring, causing it to stretch a distance \(x_1\). The force acting on the spring is the weight of the block, given by \(F_1 = m_1 g\), where \(g\) is the acceleration due to gravity. Hooke's Law states that the force exerted by a spring is proportional to its displacement from its equilibrium position: \(F_1 = kx_1\), where \(k\) is the spring constant. Thus, we have \(m_1 g = kx_1\).

In the second scenario, a second block of mass \(m_2\) is attached to the first block. The total mass hanging from the spring is now \(m_1 + m_2\). The spring stretches a distance \(x_2 = 3x_1\). The force acting on the spring is now \(F_2 = (m_1 + m_2)g\). Applying Hooke's Law again, we have \(F_2 = kx_2\), so \((m_1 + m_2)g = kx_2\).

We have two equations:

1. \(m_1 g = kx_1\)
2. \((m_1 + m_2)g = kx_2\)

Since \(x_2 = 3x_1\), we can substitute this into the second equation:

\((m_1 + m_2)g = k(3x_1)\) \((m_1 + m_2)g = 3kx_1\)

From the first equation, we know that \(kx_1 = m_1g\), so we can substitute this into the above equation:

\((m_1 + m_2)g = 3m_1g\)

Divide both sides by \(g\):

\(m_1 + m_2 = 3m_1\)

Subtract \(m_1\) from both sides:

\(m_2 = 2m_1\)

Substitute the value of \(m_1\):

\(m_2 = 2(0.68 \, \text{kg})\) \(m_2 = 1.36 \, \text{kg}\)

Final Answer