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Particle moving in conservative force field

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Screenshot (257).png

Screenshot (258).png

I don't get why ##F \cdot dr = \frac{mv^2}{2}##

I know this has to be really easy but don't see it.

Thanks.
 

Orodruin

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What is the derivative of ##\vec v^2## with respect to time?
 
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I see what you mean but that is the vector and not the magnitude
 

Orodruin

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I see what you mean but that is the vector and not the magnitude
No it is not. ##\vec v^2 = \vec v \cdot \vec v = v^2##.
 

ehild

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I think the author intended to use
## F= \frac{m}{2}\frac{d(v^2)}{dr}##.
It comes from the definition of force
##F=m\frac{d^2r}{dt^2}=m\frac{dv}{dt}##
as ##v=\frac{dr}{dt}##.
Applying chain rule when using derivative with respect to r instead of t:
##F=m\frac{d v}{d r} \frac{d r}{d t}=m v \frac{d v}{d r}=\frac{m}{2}\frac{d(v^2)}{dr}## .
 

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