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

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saljudieh07Best ResponseYou've already chosen the best response.1
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 years ago

saljudieh07Best ResponseYou've already chosen the best response.1
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?
 2 years ago

saljudieh07Best ResponseYou've already chosen the best response.1
Junkiejim to the rescue!!!!!!!!!!!!
 2 years ago

JunkieJimBest ResponseYou've already chosen the best response.1
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.
 2 years ago

saljudieh07Best ResponseYou've already chosen the best response.1
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)
 2 years ago

JunkieJimBest ResponseYou've already chosen the best response.1
oh, i see your issue, you want to not use force then, I think you should be using conservation of energy.
 2 years ago

saljudieh07Best ResponseYou've already chosen the best response.1
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)?
 2 years ago

shubhamBest ResponseYou've already chosen the best response.1
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)
 2 years ago

shubhamBest ResponseYou've already chosen the best response.1
V(1,1,1)=\[\sqrt{2(m*Vo^2 /2  \alpha\beta\gamma)/m}\]
 2 years ago

saljudieh07Best ResponseYou've already chosen the best response.1
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....
 2 years ago

saljudieh07Best ResponseYou've already chosen the best response.1
does what i am saying makes sense? coz kenetic = 0.5m(velocity)^2
 2 years ago

JunkieJimBest ResponseYou've already chosen the best response.1
you can split up your dv/dt derivative to look like this: \[\frac{dv}{dx}\frac{dx}{dt}\hat i+\frac{dv}{dy}\frac{dy}{dt}\hat j+\frac{dv}{dz}\frac{dz}{dt}\hat k\]
 2 years ago

saljudieh07Best ResponseYou've already chosen the best response.1
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..
 2 years ago

saljudieh07Best ResponseYou've already chosen the best response.1
however, is the idea right?
 2 years ago

JunkieJimBest ResponseYou've already chosen the best response.1
I think so, when you integrate you should have a line integral, looks something like this \[\int\textbf {F} ~\textbf{dl} = \int mv dv\] \[\text{where}~~ \textbf{dl}= dx\hat i +dy \hat j +dz\hat k\]
 2 years ago

saljudieh07Best ResponseYou've already chosen the best response.1
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!
 2 years ago
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