Forest UCM Energy TimeDepPE

Time dependent force.

What happens if you have a time dependent force that still manages to satisfy

[math]\vec \nabla \times \vec {F}(\vec r, t) = 0[/math]?

Because of the above, and Stoke's Theorem , you would be able to find a close loop where zero work is done at some given time.

If we consider the work energy theorem

[math]\Delta T = W = \int \vec F \cdot d \vec r[/math]

or

[math]d T = \frac{dT}{dt} dt = (m \vec \dot v \cdot v) dt = \vec F \cdot d \vec r[/math]

If a potential U for the force exists such that

[math]\vec F = - \vec \nabla U(r,t) \cdot d \vec r[/math]

or

[math]dU(r,t) = \frac{\partial U}{\partial x} dx +\frac{\partial U}{\partial y} dy +\frac{\partial U}{\partial z} dz +\frac{\partial U}{\partial t} dt [/math]

[math]= - \vec F \cdot d \vec r + \frac{\partial U}{\partial t} dt [/math]

[math]= - dT + \frac{\partial U}{\partial t} dt [/math]

or

[math]dT + dU = \frac{\partial U}{\partial t} dt [/math]

or

[math]d(T + U) = \frac{\partial U}{\partial t} dt \ne [/math]constant

Mechanical Energy is only conserved if the potential is not time dependent.

Example

Leaky charge

Forest_UCM_Energy#Time_Dependent_PE