The time derivative of kinetic energy

Therefore, ##\frac{dp}{dt} = \vec{F}## and using the chain rule, ##\frac{dT}{dt} = \frac{1}{2m}2p\frac{dp}{dt} = \frac{1}{2m}2p\vec{F} = \vec{v}\cdot \vec{F}##. In summary, we can see that ##\frac{dT}{dt}=\vec{v}\cdot \vec{F}## from considering the equations for kinetic energy and Newton's second law.
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Lets consider [tex]T(\vec{p})=\frac{\vec{p}^2}{2m}=\frac{\vec{p}\cdot \vec{p}}{2m}[/tex]. Then [tex]\frac{dT}{dt}=\vec{v}\cdot \vec{F}[/tex].
And if we consider
[tex]T=\frac{p^2}{2m}[/tex] than [tex]\frac{dT}{dt}=\frac{1}{2m}2p\frac{dp}{dt}[/tex]
Could I see from that somehow that this is [tex]\vec{v}\cdot \vec{F}[/tex]?
 
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  • #2
LagrangeEuler said:
Lets consider [tex]T(\vec{p})=\frac{\vec{p}^2}{2m}=\frac{\vec{p}\cdot \vec{p}}{2m}[/tex]. Then [tex]\frac{dT}{dt}=\vec{v}\cdot \vec{F}[/tex].
And if we consider
[tex]T=\frac{p^2}{2m}[/tex] than [tex]\frac{dT}{dt}=\frac{1}{2m}2p\frac{dp}{dt}[/tex]
Could I see from that somehow that this is [tex]\vec{v}\cdot \vec{F}[/tex]?
Well, ##\vec {p} = m \vec {v}## and ##\vec F = \frac{d\vec p}{dt}## is Newton's second law.
 
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