What are the Values of x, y, z for Mechanical Energy to be Positive?

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SUMMARY

The discussion focuses on determining the values of x, y, and z for which the mechanical energy E of a particle under the influence of a conservative force \overrightarrow{F} = k (-x\hat{i} -y\hat{j} -z\hat{k}) remains positive. The conservation of mechanical energy equation, \frac{1}{2}mv^2 + V = E, is utilized, leading to the potential energy function V = \frac{k}{2}(x^2+y^2+z^2) + C. The final inequality derived indicates that |\overrightarrow{v}|^2 must satisfy |\overrightarrow{v}|^2 > -\frac{k}{m}(x^2+y^2+z^2), ensuring that the mechanical energy remains greater than zero.

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  • Understanding of conservative forces and potential energy
  • Familiarity with the conservation of mechanical energy principle
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  • Basic differential equations and inequalities
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Homework Statement


A particle of mass m experiences a force
[tex]\overrightarrow{F} = k (-x\hat{i} -y\hat{j} -z\hat{k})[/tex]

Find values of x, y, z for which the mechanical energy E > 0.

Homework Equations


Conservation of mechanical energy
[tex]\frac{1}{2}mv^2 + V = E = constant[/tex]
where [tex]\frac{1}{2}mv^2[/tex] is the kinetic energy and V the potential energy of the system. (This is only valid in an isolated system, ie. no external forces)

The Attempt at a Solution


First checked that the force was conservative (curl is zero), then used a line integral to find a potential function.
I get [tex]V = \frac{1}{2}kz^2[/tex].
Then using the conservation law I get [tex]\frac{1}{2}mv^2 + \frac{1}{2}kz^2 = E > 0[/tex]

From there I'm not sure what to do. I tried decomposing the velocity,
[tex]\overrightarrow{v} = \dot{x}\hat{i} + \dot{y}\hat{j}+ \dot{z}\hat{k}[/tex]
Then get 3 differential 'inequalities'. I'm not sure, but is it valid to say [tex]\dot{x}^2 = 2\ddot{x}[/tex]?
Just confirming this approach before going on
 
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Double-check your potential function. Does [tex]\mathbf{F} = -\nabla \Phi[/tex] for the function you got?
 
! Yes it was wrong. Here's my second try:
[tex]V = -\int_{0,0,0}^{x,0,0}-kx dx + -\int_{x,0,0}^{x,y,0}-ky dy + -\int_{x,y,0}^{x,y,z}-kz dz[/tex]
[tex]V = \frac{k}{2}(x^2+y^2+z^2) +C[/tex]
verifying
[tex]-\nabla V = -\frac{\partial }{\partial x} \frac{kx^2}{2} -\frac{\partial }{\partial y} \frac{ky^2}{2} -\frac{\partial }{\partial z} \frac{kz^2}{2}[/tex]
[tex]= -k(x\hat{i}+y\hat{j}+z\hat{k}) = \overrightarrow{F}[/tex]

Then the conservation law is
[tex]\frac{1}{2}m\left |\overrightarrow{v} \right |^2 + \frac{k}{2}(x^2+y^2+z^2) > 0[/tex]

[tex]|\overrightarrow{v}| ^2 > -\frac{k}{m}(x^2+y^2+z^2)[/tex]

Then square root and let [tex]\overrightarrow{v} = \dot{r} \;\;\; \text{and} \;\; r = \sqrt{x^2+y^2+z^2}[/tex]

[tex]\dot{r} + \sqrt{-\frac{k}{m}}r > 0[/tex]

Simplest solution I get is [tex]\displaystyle{r > Ce^{-\gamma t} \; \; \text{where} \; \; \gamma = \sqrt{-\frac{k}{m}}}[/tex]

If k>0 then the real part, by Euler's formula is [tex]r > Acos(\sqrt{\frac{k}{m}} t)[/tex]
 
Last edited:

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