#### Answer

(a) Since the magnitude of the force directed to the left is greater, the system will move to the left.
(b) $a = 0.653~m/s^2$
(c) $T = 425~N$

#### Work Step by Step

(a) We can find the magnitude of the force directed to the left.
$F = mg~sin(\theta)$
$F = (100~kg)(9.80~m/s^2)~sin(30.0^{\circ})$
$F = 490~N$
We can find the magnitude of the force directed to the right.
$F = mg~sin(\theta)$
$F = (50~kg)(9.80~m/s^2)~sin(53.1^{\circ})$
$F = 392~N$
Since the magnitude of the force directed to the left is greater, the system will move to the left.
(b) $\sum F = Ma$
$490~N - 392~N = (100~kg+50~kg)a$
$a = \frac{98~N}{150~kg} = 0.653~m/s^2$
(c) We can consider the system of the 100-kg block to find the tension $T$ in the cord.
$\sum F = ma$
$490~N - T = (100~kg)(0.653~m/s^2)$
$T = 490~N - 65.3~N = 425~N$