Answer
$F = 43.7~N$
Work Step by Step
We can find the required acceleration up the ramp.
$x = \frac{1}{2}at^2$
$a = \frac{2x}{t^2} = \frac{(2)(8.00~m)}{(6.00~s)^2}$
$a = 0.444~m/s^2$
We can use a force equation to find the required force $F$.
$\sum F = ma$
$F - mg~sin(\theta) - mg~cos(\theta)~\mu_k = ma$
$F = ma + mg~sin(\theta) + mg~cos(\theta)~\mu_k$
$F = (5.00~kg)(0.444~m/s^2) + (5.00~kg)(9.80~m/s^2)~sin(30.0^{\circ}) + (5.00~kg)(9.80~m/s^2)~cos(30.0^{\circ})(0.40)$
$F = 43.7~N$
