Answer
The tension in the cord is $~~16~N$
Work Step by Step
We can find the component of the weight of $m_1$ directed down the ramp toward the left:
$F_1 = m_1~g~sin~\theta_1$
$F_1 = (3.0~kg)(9.8~m/s^2)~sin~30^{\circ}$
$F_1 = 14.7~N$
We can find the component of the weight of $m_2$ directed down the ramp toward the right:
$F_2 = m_2~g~sin~\theta_2$
$F_2 = (2.0~kg)(9.8~m/s^2)~sin~60^{\circ}$
$F_2 = 16.97~N$
Since $F_2 \gt F_1$, the system accelerates to the right. We can find the magnitude of acceleration:
$F_2-F_1 = (m_1+m_2)~a$
$a = \frac{F_2-F_1}{m_1+m_2}$
$a = \frac{16.97~N-14.7~N}{3.0~kg+2.0~kg}$
$a = 0.454~m/s^2$
We can consider the forces on $m_2$ to find the tension $F_T$:
$F_2 - F_T = m_2~a$
$F_T = F_2 -m_2~a$
$F_T = (16.97~N) -(2.0~kg)(0.454~m/s^2)$
$F_T = 16~N$
The tension in the cord is $~~16~N$
