Answer
The tension in the cord is $~~13~N$
Work Step by Step
We can consider the forces on block A to find an expression for the acceleration:
$m_A~g~sin~\theta-F_T = m_A~a$
$a = \frac{m_A~g~sin~\theta-F_T}{m_A}$
We can consider the forces on block B to find an expression for the acceleration:
$F_T -m_B~g~\mu_k = m_B~a$
$a = \frac{F_T -m_B~g~\mu_k}{m_B}$
We can equate the two expressions to find the tension $F_T$:
$a = \frac{m_A~g~sin~\theta-F_T}{m_A} = \frac{F_T -m_B~g~\mu_k}{m_B}$
$m_A~m_B~g~sin~\theta-F_T~m_B = F_T~m_A - m_A~m_B~g~\mu_k$
$F_T~(m_A+m_B) = m_A~m_B~g~sin~\theta + m_A~m_B~g~\mu_k$
$F_T = \frac{m_A~m_B~g~(sin~\theta + \mu_k)}{m_A+m_B}$
$F_T = \frac{(4.0~kg)~(2.0~kg)~(9.8~m/s^2)~(sin~30^{\circ} + 0.50)}{4.0~kg+2.0~kg}$
$F_T = 13~N$
The tension in the cord is $~~13~N$
