Answer
v = 12.3 m/s
Work Step by Step
We can find the force of kinetic friction $F_f$ as the box moves down the ramp.
$F_f = F_N~\mu_k$
$F_f = (mg~cos(\theta)-F~sin(\theta))~\mu_k$
$F_f = ((8.00~kg)(9.80~m/s^2)~cos(33.0^{\circ})-(26.0~N)~sin(33.0^{\circ}))\times (0.300)$
$F_f = 15.5~N$
We can find the acceleration as the box moves down the ramp.
$\sum F = ma$
$F~cos(\theta)+mg~sin(\theta)-F_f = ma$
$a = \frac{F~cos(\theta)+mg~sin(\theta)-F_f}{m}$
$a = \frac{(26.0~N)~cos(33.0^{\circ})+(8.00~kg)(9.80~m/s^2)~sin(33.0^{\circ})-15.5~N}{8.00~kg}$
$a = 6.13~m/s^2$
We can use the acceleration to find the speed $v$ after 2.00 seconds.
$v = v_0+at = 0+at = (6.13~m/s^2)(2.00~s)$
$v = 12.3~m/s$
