Question

Two masses of 40 kg and 30kg are connected by a weightless string will be $(takeg=9.8m/s^2)$

Answer 1

For $m_1$

$m_{1}g\sin \theta -T=m_{1}a$

And for $m_2$

$T-m_{2}g\sin \theta =m_{2}a$

Adding the above equations-

$(m_1-m_2)g\sin \theta =(m_1+m_2)a$

$⟹a=\frac{(m_1-m_2)g\sin \theta }{m_1+m_2}$

$⟹a=\frac{20\times 9.8\times 0.5}{60}=1.63m/s^2$

$T=m_{2}g\sin \theta +m_{2}a=20\times 9.8\times 0.5+20\times 1.63$

$⟹T=130.66N$