Dynamics Of Material Point

May 2019
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i attempted to solve it with energy formulas . but couldn't find a proper solution.
 

topsquark

Forum Staff
Apr 2008
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On the dance floor, baby!


i attempted to solve it with energy formulas . but couldn't find a proper solution.
Hint: Since a constant force is applied I'd use the constant acceleration formulas...
\(\displaystyle s = s_0 + v_0t + (1/2)at^2\)
\(\displaystyle s = s_0 + (1/2)(v_0 + v)t\)
\(\displaystyle v = v_0 + at\)
\(\displaystyle v^2 = v_0^2 + 2a(s - s_0)\)

How do you find the acceleration?

-Dan
 
Aug 2010
434
174
I am not sure why you would attempt "to solve it using energy methods" when you are given forces. If there is no driving force them "force= mass times acceleration" gives $m\frac{dv}{dt}= -m(\alpha v+ \beta v^2)$. The "m"s cancel and we can "separate" the variables v and t as $\frac{dv}{\alpha v+ \beta v^2}= dt$. Integrate both sides.

$\int \frac{dv}{\alpha v+ \beta v^2}= \int \frac{dv}{v(\beta v+ \alpha)}$.
Use "partial fractions".

("decreased twice" is a very strange phrase. Going from $v_0$ to $\frac{v_0}{2}$ would be "decreased by half".)
 
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