Question

4.48 q: two masses are hung from a frictionless pulley by a massless spring. if ( m_1 ) is 5 kg, and ( m_2 ) is 7 kg, determine the acceleration of the system.
4.49 q: two masses, ( m_1 ) and ( m_2 ), are connected by a light string over a massless pulley as shown. assuming a frictionless surface, find the acceleration of ( m_2 ).
assume ( m_2 ) is 4 kg and ( m_1 ) is 3 kg.

Explanation:

Step1: Analyze forces using Newton's second law

For the first problem, let \(m_1 = 5\space kg\), \(m_2=7\space kg\). The net force \(F_{net}=(m_2 - m_1)g\), and the total mass \(M = m_1 + m_2\). By \(F = Ma\), we have \(a=\frac{(m_2 - m_1)g}{m_1 + m_2}\).
Substitute \(m_1 = 5\space kg\), \(m_2 = 7\space kg\), \(g = 9.8\space m/s^2\) into the formula:
\(a=\frac{(7 - 5)\times9.8}{5 + 7}=\frac{2\times9.8}{12}\approx1.63\space m/s^2\)

Step2: Analyze the second - problem

For the second problem, let \(m_1 = 3\space kg\), \(m_2 = 4\space kg\). The net force \(F_{net}=m_2g\), and the total mass \(M=m_1 + m_2\). By \(F = Ma\), we have \(a=\frac{m_2g}{m_1 + m_2}\)
Substitute \(m_1 = 3\space kg\), \(m_2 = 4\space kg\), \(g = 9.8\space m/s^2\) into the formula:
\(a=\frac{4\times9.8}{3 + 4}=\frac{39.2}{7}=5.6\space m/s^2\)

Answer:

The acceleration of the first system is approximately \(1.63\space m/s^2\), and the acceleration of \(m_2\) in the second problem is \(5.6\space m/s^2\)