Force from a Kinetic Energy Function

[Dustinc]

Say you're given a function that represents the kinetic energy of some object, what would you have to do to derive the force from that function? I know that for motion along a straight line a conservative force F(x) is the negative derivative of its associated potential energy function U, but what is there to do if the function is one of kinetic energy?

 

[gneill]

Is your given KE function a function of time or position? That is, do you have KE(t) or KE(x)?

 

[HallsofIvy]

I would phrase it a little differently! Rather than the force being "caused" by the energy, the force is what causes a change in energy. And specifically, the force that causes a change in energy is equal to the rate of change of the energy with respect to distance. In special cases, for example lifting a constant weight a fixed distance, that means dividing the change in energy (change in potential energy) by the distance. In more general cases, the force is the derivative of the energy function with respect to distance.

If, as gneil suggests, the energy is a function of time, t, use $dE/dx= (dE/dt)(dt/dx)= (1/v)(dE/dt)$

 

[Dustinc]

The kinetic energy function is a function of time, my apologies!

 

[Simon Bridge]

The question is repeated here.
https://www.physicsforums.com/showthread.php?p=4548373#post4548373

 

[Simon Bridge]

If, as stated in post #1, the issue is having a kinetic energy function instead of a potential energy function - they you cannot use the force-PE relation directly. Instead you should use the relationship between PE and KE (conservation of energy perhaps?) to get a force-KE relation for your situation.

Having KE(t) instead of KE(x) is an additional complication dealt with above.

 

[LioNiNoiL]

KE(t)

I get the same result as HallsofIvy, by a different route:

KE = ½mv² ,
so d(KE)/dt = mv(dv/dt) = (ma)v = Fv

therefore F = (1/v)d(KE)/dt ... (eq'n A)

and d(KE)/dx = mv(dv/dx) = mv(dv/dt)(dt/dx) = m(dv/dt) = ma = F ,
because dt/dx is the reciprocal of v

so d(KE)/dt = (1/v)d(KE)/dt , from eq'n A

 

[Simon Bridge]

You can also do: $$K=\frac{p^2}{2m}\\ \Rightarrow \frac{d}{dt}K = \frac{p}{m}\frac{dp}{dt} = \frac{p}{m}F = vF$$... because $F=dp/dt$

If you do: $K+U=\text{constant}$ ... then take the gradient - the 1D case comes out:
$$\frac{dK}{dx} = -\frac{dU}{dx} = F$$... which takes you to the same place by using HallsofIvy's post.

Though I notice this is the "homework" section ... technically Dustinc should have been working all this out.
I still think there are details that will depend on the specific situation in the specific problem.

 

[LioNiNoiL]

so d(KE)/dt = (1/v)d(KE)/dt , from eq'n A

That should be (of course)
d(KE)/dx = (1/v)d(KE)/dt

Mistakes happen after a long day.