Force, Work, and Energy

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)

To reinforce this concept, if I'm given that a particle is sliding down the path y=x^2 from a height of 4 to a height of 0 in a uniform gravitational field F=<0,-mg>, then it doesn't matter if I parametralize this as r(t)=<t,t^2>, even though the particle wouldn't really fall at that rate?

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.

I spent an hour thinking about this. If the dot product between the force and the unit tangent vector is the magnitude of the projection of the force onto the tangent vector, then the dot product between the force and the unit normal vector is the magnitude of the projection of the force onto the normal vector.

Let's say that a particle is sliding down an irregular ramp. Since friction is F= -μFn, then would the energy lost due to friction be this?

E= -∫ μF • N ds

E= -∫ μF(t) • T'(t)/||T'(t)|| ||r'(t)|| dt

I'm not sure about this, because I've never taken vector calculus before, but I do know that doing this process over a over a surface gives the flux，or the amount of liquid that passes through the surface. My equation seems to be correct for the frictional force, but because of flux, the equation also seems to be counter-intuitive. Also, is there any way to simplify N ds? It seems really messy.

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

Could you please elaborate on this? How would you use

W= ∫ F • dl = ∫ F cos(θ dl

on a general curve, say... r(t)=<e^t, t^2>? I thought this equation only works when you're dragging a box on a flat surface and the force or angle the force applied isn't changing.


Thanks a bunch for helping out.