Question

15. A 6.0 kg body rests on a smooth horizontal tabletop. A horizontal cord attached to the body passes over a light frictionless pulley at the edge of the table to 12.0 kg body hanging freely. Find the acceleration and the tension in the cord when the system is released.

Answer 1

Given:

$m_1=6.0\:\rm kg$

$m_2=12.0\:\rm kg$

The Newton's second law says

$$m_1a=T,\\ m_2a=m_2g-T.$$

Hence

$$a=\frac{m_2g}{m_1+m_2}=\frac{12.0*9.81}{12.0+6.0}=6.54\:\rm m/s^2$$

$$T=m_1a=6.0*6.54=39.2\:\rm N$$