Answer
F = 115 N
Work Step by Step
We can find the force of kinetic friction $F_f$ as the box moves up the incline.
$F_f = F_N~\mu_k$
$F_f = (mg~cos(\theta)+F~sin(\theta))~\mu_k$
$F_f = ((6.00~kg)(9.80~m/s^2)~cos(37.0^{\circ})+F~sin(37.0^{\circ}))\times (0.30)$
$F_f = 0.18~F + 14.1~N$
We can use a force equation to find the horizontal force $F$.
$\sum F = ma$
$F~cos(\theta) - F_f - mg~sin(\theta) = ma$
$F~cos(\theta)- 0.18~F - 14.1~N = ma + mg~sin(\theta)$
$F = \frac{ma + mg~sin(\theta)+14.1~N}{cos(\theta)-0.18}$
$F = \frac{(6.00~kg)(3.60~m/s^2) + (6.00~kg)(9.80~m/s^2)~sin(37.0^{\circ})+14.1~N}{cos(37.0^{\circ})-0.18}$
$F = 115~N$
