A couple of questions regarding work to velocity exchange I was tasked to express the velocity of the cube of mass $m$ at point B using the constants given in the picture below. Friction could be ignored, and the cube could be presumed to reach B with the given force. To solve it, I need to have a few questions answered.

1) The cube is affected by three forced. Gravitational, tension and a normal-force. Can the normal force be disregarded as an inactive force when calculating the work done?
Even if that is the case, establishing a equation for the resultant force as a function of r to integrate over proved very hard. Since the path taken should be irrelevant, I replaced it with a straight line connecting A to B. This simplified system still gave extremely long and tedious algebraic och trigonometric expressions, which I was then somehow supposed to integrate. This seemed way harder than the other questions, so I assume I overcomplicate the system. 
Which leads me to question 2) Can I instead use the average force during the movement? Such as $F_{avg}=\frac{F+\cos(\theta)F}{2}$. While this seems logical, I understand that it isn't necessarily the case. 
And finally, 3) Is there something majorly flawed with my solution idea? That is:


*

*Place the x-axis along the traveled path

*Express the angle between the traveled path and the line as a function of x

*Express the resultant force in the x-axis using the angle.

*Integrate this expression over the traveled path

*Use $v=\sqrt{\frac{2W}{m}}$


Or is there a much easier way of doing it that I am completely forgetting?
 A: I am giving the solutions of original task (to get the speed at point B). I am not sure if the questions are necessary to perform the task.
If you are sure the path taken does not matter (and I will assume that per your statement). So, let us consider a straight line path. Vertical component of F overcomes gravity and causes vertical move. Only horizontal component of F (let us say F(H)), gives it horizontal motion. You can figure the value of horizontal component. Horizontal distance is l per figure and horizontal acceleration will be F(H)/m. Putting this in Newton's equation of motion and considering u, initial speed = 0, l = u* t + 1/2a* t* t = 
0+.5* t* t* F(H)/m. You get t from this because you have all other values. Horizontal component of speed  = u + a* t = 0 + F(H)* t = F(H)* t (you got t from previous).
In a straight line, case, you will also have a vertical speed. Let us say vertical component of F is F(V). force of gravity is mg. Resultant vertical force is F(V) - mg. Vertical acceleration is (vertical force)/m = (F(V)-mg)/m. So, vertical component of speed is u +  a* t = 0 + t* ((F(V)-mg)/m). As you already have t, you can get the vertical speed. Then you can vector sum horizontal and vertical speeds for the straight line case.
Per your figure, if you consider the curved path, the final speed will be only horizontal which will be = the horizontal component = F(H)* t. The vertical component will be stopped by the support of the bar.
I have assumed you can figure F(H) and F(V) given the angle (or l and h).
