Question

4.48 q: two masses are hung from a frictionless pulley by a massless spring. if m, is 5 kg, and m, is 7 kg, determine the acceleration of the system. 4.49 q: two masses, m, and m,, are connected by a light string over a massless pulley as shown. assuming a frictionless surface, find the acceleration of m,. assume m, is 4 kg and m, is 3 kg.

Explanation:

Step1: Analyze forces for each mass

For \(m_1\) (horizontal motion), the only horizontal force is tension \(T\). Using \(F = ma\), we have \(T=m_1a\).
For \(m_2\) (vertical motion), the net force is \(m_2g - T\). Using \(F = ma\), we have \(m_2g - T=m_2a\).

Step2: Substitute \(T\) from the first equation into the second

Substitute \(T = m_1a\) into \(m_2g - T=m_2a\). We get \(m_2g - m_1a=m_2a\).

Step3: Solve for \(a\)

Rearrange \(m_2g - m_1a=m_2a\) to \(m_2g=(m_1 + m_2)a\). Then \(a=\frac{m_2g}{m_1 + m_2}\).
Given \(m_1 = 3\space kg\), \(m_2 = 4\space kg\), and \(g = 9.8\space m/s^2\).
Substitute values: \(a=\frac{4\times9.8}{3 + 4}\).
Calculate \(a=\frac{39.2}{7}=5.6\space m/s^2\).

Answer:

The acceleration of \(m_2\) is \(5.6\space m/s^2\).