Answer
$F_2 = -(32.0~N)\hat{i}-(20.8~N)\hat{j}$
Work Step by Step
We can find the horizontal component of the second force:
$F_x + F_1 = -ma~sin~\theta$
$F_x = -F_1 - ma~sin~\theta$
$F_x = -(20.0~N) - (2.00~kg)(12.0~m/s^2)~sin~30.0^{\circ}$
$F_x = -32.0~N$
We can find the vertical component of the second force:
$F_y = -ma~cos~\theta$
$F_y = -(2.00~kg)(12.0~m/s^2)~cos~30.0^{\circ}$
$F_y = -20.8~N$
We can express the second force in unit-vector notation:
$F_2 = -(32.0~N)\hat{i}-(20.8~N)\hat{j}$
