Work energy thm and conservation

[rattan5]

Since the change of kinetic energy, K2-K1, ALWAYS equals the integral of F.dr along any path, how can that integral depend upon the path? I realize that the integral is ONLY equal to the change in potential energy (F is the derivative of the potential) at the end points when F is a conservative force. But the integral always equals the same value K2-K1, regardless of the path from 1 to 2. Isn't that what is meant by independence of path?

 

[Doc Al]

Just because ∫F.dr equals K2-K1 (when F is the net force) doesn't mean that it's path independent. For non-conservative forces, both ∫F.dr and K2-K1 depend on the path.

 

[rattan5]

Since KE depends only on the magnitude of the vector, any path I take having the same starting and ending velocities will of course have the same work.

 

[Doc Al]

Since KE depends only on the magnitude of the vector, any path I take having the same starting and ending velocities will of course have the same work.

Well, that's certainly true. But not all paths will have the same starting and ending velocities.

 

[rattan5]

Thanks for your replies. My confusion I think lies with the derivation I watched from a Yale video which went like (please notice dot product and vectors)

K = mv.v/2
dK/dt = m dv/dt.v
dK = ma.v = ma.dr/dt so
∫dK = ∫F.dr and then the lhs was set to K2-K1 to get
K2-K1 = ∫F.dr

From what you're saying, which must be true, the last step is only true for conservative forces.

 

[Doc Al]

From what you're saying, which must be true, the last step is only true for conservative forces.

No. As long as F is the only force acting, ∫F.dr will always equal ΔKE.

(If F is a conservative force, then ∫F.dr will be independent of the particular path chosen between the given endpoints.)