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Mechanics}Chain Rule

Recall that a=\frac{dv}{dt}=\frac{dv}{dx}\frac{dx}{dt}=v\frac{dv}{dx}. \frac{dv}{dx}=-n\beta x^{-n-1}\Rightarrow v\frac{dv}{dx}=-n\beta^2 x^{-2n-1}, as in choice (A).

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Comments
JasonHupp
2018-07-30 11:09:59
Before the energy conversion is a hectic task to me but when i have seen here this simple tutorial on how to solve i was surprised. As i have been looking for a help and thinking is assignmenthelps.com.au legit good to help me out in this problem solving. Mean while got yours.NEC
RayGSD
2006-11-03 15:38:36
You can also do this through energy conservation.

\frac{1}{2} m v(x)^2 + U(x) = E_0

\Rightarrow U(x) = E_0 - \frac{1}{2} m \beta^2 x^{-2n}

and

-\frac{\partial U(x)}{\partial x} = F = m a

So

a = 1/2 \beta^2 \frac{\partial}{\partial x}\left( x^{-2n} \right)

a = -n \beta^2 x^{-2n-1}
FutureDrSteve
2011-11-02 14:36:35
I don't think this is faster or easier than the posted solution, but it's very elegant. Nice job!
NEC
StudyTime
2005-11-11 09:16:50
You can also do it through dimensional analysis.NEC

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Before the energy conversion is a hectic task to me but when i have seen here this simple tutorial on how to solve i was surprised. As i have been looking for a help and thinking is assignmenthelps.com.au legit good to help me out in this problem solving. Mean while got yours.

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