Derivation of Kinetic Energy

francescopadormo

Hi,

I've always tken for granted that KE=(mv^2)/2, but never seen the derivation! I think it probably comes from U= - inegral of the force, but I can't see which force to use. Any help anyone? Maybe dimensional analysis was used?

Thanks!

Doc Al

I suspect you are thinking of the "work-energy" theorem, in which one integrates the net force on an object over the distance traveled to obtain the change in KE.

Start with Newton's 2nd law:
[tex]F = m \frac{dv}{dt}[/tex]

[tex]\int F dx = \int m \frac{dv}{dt} dx = m \int \frac{dx}{dt} dv = m \int v dv[/tex]

I'll leave the last step to you.

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francescopadormo	Derivation of Kinetic Energy Tweet Hi, I've always tken for granted that KE=(mv^2)/2, but never seen the derivation! I think it probably comes from U= - inegral of the force, but I can't see which force to use. Any help anyone? Maybe dimensional analysis was used? Thanks!

Doc Al Mentor Blog Entries: 1	I suspect you are thinking of the "work-energy" theorem, in which one integrates the net force on an object over the distance traveled to obtain the change in KE. Start with Newton's 2nd law: [tex]F = m \frac{dv}{dt}[/tex] [tex]\int F dx = \int m \frac{dv}{dt} dx = m \int \frac{dx}{dt} dv = m \int v dv[/tex] I'll leave the last step to you.