I The time derivative of kinetic energy

Click For Summary
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
Messages
711
Reaction score
22
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}?
 
Physics news on Phys.org
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.
 
Last edited:
  • Like
Likes vanhees71, gmax137, Ibix and 1 other person
Thread 'Energy requirements of a 4 wheel drive boat'
4 wheel drive amphibious boat (swamp buggy?) My problem is trying to calculate the amount of energy, or a good estimate, that would be required to keep the boat levitated above the water at the approximate water line shown in the second drawing. I don’t believe it will require as much energy as a helicopter as the blades are pushing against a semi-solid surface but I really don’t know? A planing hull has a lift / drag ratio of about 5:1 with skin friction becoming increasingly more...