Answer:
1.59 m/s^2, 65.2°
Explanation:
F1 = 390 N North
F2 = 180 N east
m = 270 kg
Net force is the vector sum of both the forces.
[tex]F = \sqrt{F_{1}^{2}+F_{2}^{2}}[/tex]
[tex]F = \sqrt{390^{2}+180^{2}}[/tex]
F = 429.53 N
Direction of force
tan∅ = F1 / F2 = 390 / 180 = 2.1667
∅ = 65.2°
The direction of acceleration is same as the direction of net force.
The magnitude of acceleration is
a = F / m = 429.53 / 270 = 1.59 m/s^2
