Answer:
6.44° north of + x-axis
Explanation:
First find the net force on the block.
Given F1 = 11.8 N, F2 = 22.9 N
Sum of forces in the x direction:
F1x + F2x = (11.8)(cos 53.7) + (22.9)(cos 15.8) = 29.02 N
Sum of forces in the y direction:
F1y + F2y = (11.8)(sin 53.7) - (22.9)(sin 15.8) = 3.275 N
Resultant Force = √29.02² + 3.275² = 29.2 N
F = ma (Newton's 2nd Law)
a = F/m = 29.2 N/15.5 kg = 1.88 m/s² (this step is not necessary to solve the problem, but it's helpful to know)
Now find the direction of the acceleration, which is in the same direction as the resultant Force:
tan⁻¹(Fy/Fx) =
tan⁻¹(3.275/29.02) = 6.44° north of + x-axis