## University Physics with Modern Physics (14th Edition)

(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) $T = 425~N$
(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. $\sum F = ma$ $490~N - T = (100~kg)(0.653~m/s^2)$ $T = 490~N - 65.3~N = 425~N$