Answer
(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
Work Step by Step
(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
