Question

4.48 q: two masses are hung from a frictionless pulley by a massless spring. if m₁ is 5 kg, and m₂ is 7 kg, determine the acceleration of the system.

Explanation:

Step1: Apply Newton's second law

Let the acceleration of the system be \(a\). For mass \(m_1 = 5\space kg\), the force equation is \(T=m_1a\) (where \(T\) is the tension in the string). For mass \(m_2=7\space kg\), the force equation is \(m_2g - T=m_2a\).

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

Substitute \(T = m_1a\) into \(m_2g - T=m_2a\). We get \(m_2g=m_1a + m_2a\). Factor out \(a\): \(a=\frac{m_2g}{m_1 + m_2}\).

Step3: Plug in the values

Given \(m_1 = 5\space kg\), \(m_2=7\space kg\), and \(g = 9.8\space m/s^{2}\). Then \(a=\frac{7\times9.8}{5 + 7}=\frac{68.6}{12}\approx5.72\space m/s^{2}\)

Answer:

The acceleration of the system is approximately \(5.72\space m/s^{2}\)