Force, Work, and Energy

The problem I am having is conceptual. Say that a particle of mass m starts from rest, and then moves along the path r(t) = <x(t), y(t)>. Does the forcing function always have to equal F(t) = m <x''(t), y''(t)>? (Just asking for confirmation on this part.)

Yes.

To calculate the work done by a force field and a path, the line integral is given by :

W= ∫ (F•T) ds

T is the unit tangent vector, r'(t)/||r'(t)||, and ds is the length, ||r'(t)|| dt. Therefore,


Because of the way I defined F(t), I can plug it in directly:

W= ∫ m <x''(t), y''(t)> • <x'(t), y'(t)> dt

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Additionally, accounting for the force supplied against gravity (in this part, I am not sure whether the sign is correct. Someone PLEASE PLEASE check this):

You need to define which direction you're taking gravity in with regards to the rest of the system. Then the correct sign for g should just drop out. Obviously its recommended to take g parallel to one of the axis. Normally its defined as in the direction of -y (or - z in 3d), but you need to specify.

My second question is, what if I'm given that the initial position of a particle is (0,0), and that the forcing function is F = m <x''(t), y''(t)>? Then do I assume that the particle always travels the path r(t) = <x(t), y(t)>? What if my path isn't time dependent?

Yes you assume it travels the path r(t) if the path is time dependent. If the path isn't time dependent?
Then you have to consider a path that IS time Dependant to perform the integral, but because of the independence, no matter what path you choose you'll get the same answer. (F would be a conservative force)

Third, what if there is friction along this path? How do I account for this?

If there is friction on the path, generally it will be *hidden* in the F= (term) since if force isn't constant, one of the reasons for that could be the friction.

Fourth, my book (Giancoli, 4th ed.) states that the "work done by a varying force is":

W= ∫ F • dl = ∫ F cos(θ dl

They just expanded the dot product of work , a•b = ab cos(theta), except, here a= F and b=dl

Fifth, what if the path isn't time dependent?

Again, if a path isn't time dependent, you have to create a path that is, but you'll get the same answer regardless of the path you choose.