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

In summary, based on the homework statement, a particle of mass experiences a force that is inversely proportional to its distance from the center of the system. If k>0, then the real part of the velocity is greater than the cosine of the angle between the velocity and the x-axis.
  • #1
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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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  • #2
Double-check your potential function. Does [tex]\mathbf{F} = -\nabla \Phi[/tex] for the function you got?
 
  • #3
! 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]
 
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FAQ: What are the Values of x, y, z for Mechanical Energy to be Positive?

What is the definition of force?

Force is a physical quantity that can cause an object with mass to accelerate. It is measured in newtons (N) and is represented by the symbol F.

How is force related to mechanical energy?

Force is a component of mechanical energy, along with position and velocity. When a force is applied to an object and causes it to move, it is doing work and therefore adding mechanical energy to the system.

What is the difference between potential and kinetic energy?

Potential energy is the energy an object has due to its position or state, while kinetic energy is the energy an object has due to its motion. Both are forms of mechanical energy and can be converted into each other.

How does mechanical energy conservation apply to real-life situations?

The law of conservation of mechanical energy states that the total mechanical energy in a closed system remains constant. This means that in real-life situations, energy can neither be created nor destroyed, but rather transferred or converted from one form to another.

How does friction affect force and mechanical energy?

Friction is a force that opposes the motion of an object. When applied, it can decrease the amount of force and mechanical energy in a system. This is because some of the energy is converted into heat due to the resistance of the surface the object is moving on.

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