GREPhysics.NET: GR9677 #44

GR9677 #44

Problem

GREPhysics.NET Official Solution

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Mechanics $\Rightarrow$ }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.

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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!

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StudyTime
2005-11-11 09:16:50

You can also do it through dimensional analysis.

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Post A Comment!

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

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