Given force as a function of x, how do I find the total energy?

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SUMMARY

The discussion focuses on determining the potential energy function U(x) from the force function F = -kx + kx³/α², where k and α are constants with k > 0. The potential energy is calculated as U(x) = kx²/2 - kx⁴/4α². The user explores the implications of the total energy E = kα²/4, leading to the kinetic energy expression T = kα²/4 - kx²/2 + kx⁴/4α². The user expresses confusion regarding the dependence of kinetic energy on mass and the overall independence of the system from time.

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GriffinC
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Homework Statement


F=-kx+kx32 where k and α are constants and k > 0. Determine U(x) and discuss the motion. What happens when E=kα2/4?

Homework Equations


F=ma=mv2d/dx
U=-∫Fdx

The Attempt at a Solution


The first part is easy.
U(x) = kx2/2-kx4/4α2
Now I'm looking for what happens when E=kα2/4
I know E=T+U, so T=kα2/4-kx2/2+kx4/4α2
To find T, I need to know velocity since T=1/2mv2
Solving for v, F=mv2d/dx, v=(k-3kx22)/2(kx32-kx)3/2

Here's where I'm not sure I'm going in the right direction. If I find T, I introduce an m term. The potential energy is independent of mass, so why would kinetic energy depend on mass? It also seems as though the whole thing is independent of time.
 
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GriffinC said:

Homework Statement


F=-kx+kx32 where k and α are constants and k > 0. Determine U(x) and discuss the motion. What happens when E=kα2/4?

Homework Equations


F=ma=mv2d/dx
U=-∫Fdx

The Attempt at a Solution


The first part is easy.
U(x) = kx2/2-kx4/4α2
Now I'm looking for what happens when E=kα2/4
I know E=T+U, so T=kα2/4-kx2/2+kx4/4α2
To find T, I need to know velocity since T=1/2mv2
Solving for v, F=mv2d/dx, v=(k-3kx22)/2(kx32-kx)3/2

Here's where I'm not sure I'm going in the right direction. If I find T, I introduce an m term. The potential energy is independent of mass, so why would kinetic energy depend on mass? It also seems as though the whole thing is independent of time.

Did you think of drawing a graph of ##U##?

You're right that expressing ##T = \frac12 mv^2## isn't going to help.
 

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