Answer
$3.5\frac{m}{s^2}$
Work Step by Step
We know that
$\Sigma F_y=N-mg-Fsin\theta=0$
$\implies N=mg+Fsin\theta $
Now the horizontal component is
$\Sigma F_x=Fcos\theta-\mu_sN=ma_x$
This simplifies to:
$a_x=\frac{1}{m}(Fcos\theta-\mu_sN)$
$\implies a_x=\frac{1}{m}(Fcos\theta-\mu_s(mg+Fsin\theta))$
We plug in the known values to obtain:
$a_x=\frac{330cos21^{\circ}-(0.45)(32)(9.81)+330sin21^{\circ}}{32}$
$a_x=3.5\frac{m}{s^2}$
