## College Physics (4th Edition)

(a) Block $m_1$ accelerates upward at a rate of $2.45~m/s^2$ Block $m_2$ accelerates downward at a rate of $2.45~m/s^2$ (b) The tension in the cord is 36.75 N
(a) Consider the system of $m_1$ and $m_2$. $\sum F = (m_1+m_2)~a$ $m_2~g-m_1~g = (m_1+m_2)~a$ $a = \frac{(m_2-m_1)~g}{m_1+m_2}$ $a = \frac{(5.0~kg-3.0~kg)(9.80~m/s^2)}{3.0~kg+5.0~kg}$ $a = 2.45~m/s^2$ Block $m_1$ accelerates upward at a rate of $2.45~m/s^2$ Block $m_2$ accelerates downward at a rate of $2.45~m/s^2$ (b) Let $F_T$ be the tension in the cord. We can consider the forces on block $m_1$: $\sum F = m_1~a$ $F_T-m_1~g = m_1~a$ $F_T = (m_1)~(a+g)$ $F_T = (3.0~kg)(2.45~m/s^2+9.80~m/s^2)$ $F_T = 36.75~N$ The tension in the cord is 36.75 N