Constant force applied parallel to the surface

Constant force applied parallel to the surface

Question: A 4.04kg box sits at rest at the bottom of a ramp that is 8.62m
long and that is inclined at 40.0∘ above the horizontal. The
coefficient of kinetic friction is 0.40, and the coefficient of static
friction is 0.50.
What constant force F , applied parallel to the surface of the ramp, is
required to push the box to the top of the ramp in a time of 3.67s ?
I have $a_x=\frac{2(x-x_0)}{t^2}$ and
$F=m(a_x+gcos(\theta)+\mu_kmgsin(\theta))$.
So I can combine both equations to have $F=m(\frac{2(x-x_0)}{t^2} +
gcos(\theta) + \mu_kmgsin(\theta))$
When I plug in all my values into the the equation above I get 76.7N,
however mastering physics is telling me that my answer is wrong. Am I or
not?