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saljudieh07

  • 2 years ago

A particle of mass m moving in 3 dimensions under the potential energy function....

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  1. saljudieh07
    • 2 years ago
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    Rest of Question: A particle of mass m moving in 3 dimensions under the potential energy function \[V(x,y,z)=\alpha x+\beta y^2+ \gamma z^3\] has speed Vnaught when it passes through the origin. a) what will its speed be if and when it passes through the point (1,1,1)?

  2. saljudieh07
    • 2 years ago
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    This is what I have done so far: F=-gradient (dot) potential energy function \[= - (\alpha i +2y \beta j + 3z^2\gamma k )\] and I make the expression above equal to= mdv/dt but now i am study, when i break down the dv/dt into 3 dimensions how do I integrate both sides with respect to x, y, and z?

  3. saljudieh07
    • 2 years ago
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    Junkiejim to the rescue!!!!!!!!!!!!

  4. JunkieJim
    • 2 years ago
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    okay, having F you should be able to do this: \[\textbf{F}=m \frac{\delta \textbf{v}}{\delta t} \]\[\textbf{F} \delta t = m \delta \textbf{v}\]\[\int\textbf{F}\delta t = \int m \delta \textbf{v}\] and you integrate each component of F with respect to t, the right side just becomes mv.

  5. saljudieh07
    • 2 years ago
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    yeh but that is force with respect to time, i need force with respect to position that is the problem! coz if you read a, it gives me a value of (1,1,1)

  6. JunkieJim
    • 2 years ago
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    oh, i see your issue, you want to not use force then, I think you should be using conservation of energy.

  7. saljudieh07
    • 2 years ago
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    yeh, i have the force equation, i can make it equal to m (dv/dx dx/dt) if i break dv/dt, then i get mv dv/dx, and i can find the equation of velocity wrt to position, but the problem is, this question is three dimensional not only wrt x... so would it be F= mv (dv/dx +dv/dy + dv/dz)?

  8. shubham
    • 2 years ago
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    OR Considering the system to be closed and forces to be conservative, you can deduce that sum of KE and PE would be constant. We know, KE+PE at origin = KE + PE at (1,1,1) 1/2*m*Vo^2 + 0 = 1/2*m*V(1,1,1) ^2 + alpha+beta+gamma Hence, you can calculate V(1,1,1)

  9. shubham
    • 2 years ago
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    V(1,1,1)=\[\sqrt{2(m*Vo^2 /2 - \alpha-\beta-\gamma)/m}\]

  10. saljudieh07
    • 2 years ago
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    when you say 1/2*m*Vo^2 + 0 = 1/2*m*V(1,1,1) ^2 + alpha+beta+gamma 1/2*m*V(1,1,1) ^2 << that V is the velocity with respect to position, and position is in terms of x, y, and z, so how do i find that equation, this is what i am struggling with....

  11. saljudieh07
    • 2 years ago
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    does what i am saying makes sense? coz kenetic = 0.5m(velocity)^2

  12. JunkieJim
    • 2 years ago
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    you can split up your dv/dt derivative to look like this: \[\frac{dv}{dx}\frac{dx}{dt}\hat i+\frac{dv}{dy}\frac{dy}{dt}\hat j+\frac{dv}{dz}\frac{dz}{dt}\hat k\]

  13. saljudieh07
    • 2 years ago
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    so then i will have: let us say this is the expression you wrote (dvdxdxdtiˆ+dvdydydtjˆ+dvdzdzdtkˆ ) would be equal to = -1/m (Force) where the force is: =−(αi+2yβj+3z2γk) but i still don't know how to solve that integral lol..

  14. saljudieh07
    • 2 years ago
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    however, is the idea right?

  15. JunkieJim
    • 2 years ago
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    I think so, when you integrate you should have a line integral, looks something like this \[\int\textbf {F} ~\textbf{dl} = \int mv dv\] \[\text{where}~~ \textbf{dl}= dx\hat i +dy \hat j +dz\hat k\]

  16. saljudieh07
    • 2 years ago
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    hmmm, i see... iight thanks a lot man, if i only could give u more than one star i would of haha.. but yeh.. anyways, amma go sleep on this, when i wake up 2morrow am sure something good will come up with me, if not i will come back 2morrow. thanks guys!

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