Shopping Carts, Finding Horizontal Force

[physics114]

Correct? Please let me know if my logic is off.

Given Problem
A shopper pushes a 6.1 kg shopping cart up a 13° incline, as shown in the figure. Find the magnitude of the horizontal force, F, needed to give the cart an acceleration of 1.51 m/s².

Relevant equations
F = ma
F_y = F₁cos13°
F_x = F₁sin13°

Attempted Solution
Given acceleration as 1.51m/s², I found the net force (?)

F = ma
F = (6.1kg) * (1.51m/s²)
F = 9.211 N

Since the problem is asking for the horizontal force, I would only be focusing on the F_1,x direction. With this i would use the F_`,x = F₁sin13° formula (?)

Use this force found earlier with Newton's Second Law to find the force in the horizontal direction.

F_1,x = F₁sin13°
F₁ = F_1,x / sin13°
F₁ = 9.211N / sin13°
F₁ = 40.9467 N

ADJUSTMENT 1:
Calculated with F_x = Fcos13° instead of Fsin13°
F₁ = (9.121 N) / (cos13°)
F₁ = 9.35989N
Answer is still incorrect. The website tells me:
"Your response differs from the correct answer by more than 10%. Double check your calculations"

 

[The legend]

Fx = Fcos(13) not Fsin(13)

 

[godtripp]

This problem is easier if you rotate your axis 13 degrees. So draw the x-axis to be parallel with the acceleration. With your current axis ... Fx = F
and don't forget your normal force and force due to gravity.Now, if you redraw your axis along the direction of acceleration
Fx = Fcos(13) -mg sin(13) = ma

So now the answer is trivial algebra
F = [m(a+g sin(13))] / cos(13)