Undergrad The time derivative of kinetic energy

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The discussion focuses on the time derivative of kinetic energy, expressed as T(𝑝) = 𝑝²/(2m), and its relationship to force. It establishes that the derivative dT/dt equals 𝑣·𝑭, linking kinetic energy to Newton's second law where 𝑝 = m𝑣 and 𝑭 = d𝑝/dt. The calculations show that the change in kinetic energy can be derived from the momentum and force relationship. This reinforces the connection between energy, momentum, and force in classical mechanics. The thread emphasizes the mathematical derivation and conceptual understanding of these physical principles.
LagrangeEuler
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Lets consider T(\vec{p})=\frac{\vec{p}^2}{2m}=\frac{\vec{p}\cdot \vec{p}}{2m}. Then \frac{dT}{dt}=\vec{v}\cdot \vec{F}.
And if we consider
T=\frac{p^2}{2m} than \frac{dT}{dt}=\frac{1}{2m}2p\frac{dp}{dt}
Could I see from that somehow that this is \vec{v}\cdot \vec{F}?
 
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LagrangeEuler said:
Lets consider T(\vec{p})=\frac{\vec{p}^2}{2m}=\frac{\vec{p}\cdot \vec{p}}{2m}. Then \frac{dT}{dt}=\vec{v}\cdot \vec{F}.
And if we consider
T=\frac{p^2}{2m} than \frac{dT}{dt}=\frac{1}{2m}2p\frac{dp}{dt}
Could I see from that somehow that this is \vec{v}\cdot \vec{F}?
Well, ##\vec {p} = m \vec {v}## and ##\vec F = \frac{d\vec p}{dt}## is Newton's second law.
 
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For simple comparison, I think the same thought process can be followed as a block slides down a hill, - for block down hill, simple starting PE of mgh to final max KE 0.5mv^2 - comparing PE1 to max KE2 would result in finding the work friction did through the process. efficiency is just 100*KE2/PE1. If a mousetrap car travels along a flat surface, a starting PE of 0.5 k th^2 can be measured and maximum velocity of the car can also be measured. If energy efficiency is defined by...

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