First, K = 0.5mv^2 is not "average" kinetic energy- it is kinetic energy- in fact it is the definition of kinetic energy- there is no "derivation". At some time, back at the beginning of "physics" someone notice that, in many dynamics problems that quantity was conserved- so they gave it a name!
#4
mani
19
0
First, K = 0.5mv^2 is not "average" kinetic energy- it is kinetic energy- in fact it is the definition of kinetic energy- there is no "derivation". At some time, back at the beginning of "physics" someone notice that, in many dynamics problems that quantity was conserved- so they gave it a name!
Quantities, mv^2, 35mv^2, are also conserved!
#5
physics247
24
0
Originally posted by PrudensOptimus
ΣWork = ΣF * d = mad = 0.5m(v^2 - v_{0}^2) = ke
So ummmm ... This is relative kinetic energy not average kinetic energy?
That is, relative to the kinetic energy of the object when v_{0} = 0
Interesting derivation. Thanks.
#6
lethe
653
0
energy is the ability to do work.
work is force times distance.
kinetic energy is that energy a body has due to its motion.
but by Newton
[tex]
\mathbf{F}=m\mathbf{a}
[/tex]
so
[tex]
W=\int_A^Bm\mathbf{a}\cdot d\mathbf{s} = m\int_{t_{A}}^{t_{B}}\mathbf{a}\cdot\mathbf{v}dt\\
= \frac{m}{2}\int_{t_{A}}^{t_{B}}\frac{d(v^2)}{dt}\! dt = \frac{1}{2}mv_B^2 - \frac{1}{2}mv_A^2
[/tex]
so you see that the work done is equal to the change in kinetic energy between the two points, and this is a derivation of the formula for that kinetic energy.