Mass on a ramp with friction discrepancy

[Questlove]

Homework Statement

A mass of 50kg is on a ramp 40° above the ground. The coefficient of kinetic friction for the surface of the ramp is .1. If the mass is pulled by a force 15° above the ramp and accelerates at 1 m/s^2 up the ramp, what is the magnitude of the force?

Homework Equations

Ff=μk*N
F=ma

The Attempt at a Solution

First I separated each net force into x (horizontal to ramp) and y (perpendicular to ramp) components. Also, the net force in the x direction must be 50N up the ramp, since 50kg*1 m/s^2=50N. Also, the net force in the y direction must be 0 since the mass is staying on the ramp.
Fnet x: 50 = F*cos(15°)-(Ff+sin(40°)*50*9.8) = F*cos(15°)-(Ff+314.97)
Fnet y: N + F*sin(15°) = cos(40°)*50*9.8 = 375.36

Since μk=.1 is given, Ff=.1*N

I replaced Ff with .1*N in the x equation, getting:
50=F*cos(15°)-.1*N-314.97

Then I replaced N with 375.36-F*sin(15°), getting:
50=F*cos(15°)-.1*(375.36-F*sin(15°))-314.97

From here, I solved for F to get an answer of about 405.8N, which was matched by 2 other students who both did the same problem individually. The discrepancy is that 6 other students, as well as my teacher, got answers of about 350N. Who is right, and why is the wrong group wrong?

 

[Funkmaster W]

Your math seems to check out. i feel as though the group of six and the teacher may be wrong due to a sign error.