Question

I'm having difficulty deriving the formula for kinetic energy:

Answer 1

\[\LARGE \int\limits_0^x \bar F \cdot d \bar x=\int\limits_0^x m \frac{d \bar v}{dt} \cdot d \bar x\] Now how do we justify using the chain rule to move here: \[\LARGE m \int\limits_0^x \frac{d \bar v}{dt} \cdot d \bar x=m \int\limits_0^x \frac{d \bar x}{dt} \cdot d \bar v\] This seems really interesting to me, and I'd like to understand physically why the dot product of acceleration and a little movement in position should be the same as the dot product of velocity and a little change in velocity.

Answer 2

http://books.google.com.qa/books?id=h4bAAgAAQBAJ&pg=PA335&dq=derivation+formula+of+kinetic+energy&hl=ar&sa=X&ei=WyIEVIvcDoOn0QXfg4GACA&ved=0CBwQ6AEwAA#v=onepage&q=derivation%20formula%20of%20kinetic%20energy&f=false

Answer 3

No, I'm looking for calculus.

Answer 4

\[dx = vdt\]\[\int\limits \frac{dv}{dt}.dx = \int\limits \frac{dv}{dt}. vdt = \int\limits v.dv = \int\limits v dv\]

thats how we usually do it mathematically :P

Answer 5

so you are looking at the first and last terms.. and thinking of \[\int\limits a.dx = \int\limits v.dv\] and asking if there is some physical meaning to it? (some sort of intuition rather than just using math?)
hmmm .. never thought about it that way.

Answer 6

Yes exactly @Mashy.

Answer 7

I'm not really sure what to do to reason out the justification of this dot product. The only thing that comes to mind is that when I think of position and acceleration of a circular path, the vectors are antiparallel.

Answer 8

How that becomes v . dv is beyond me.

Answer 9

I guess I had to take away your medal @Mashy until someone can answer the question unless you have any more thoughts to help figure this out lol

Answer 10

\[dt = \frac{ dx }{ v }\]
\[a = \frac{ dv }{ dt } = \frac{ dv }{ \frac{ dx }{ v } } = \frac{ vdv }{ dx }.\]

Answer 11

acc of the car times change of its position.------------>1
velocity of the car times change of its velocity.----------->2
divide by dt for both
1 = acc time velocity.
2 = velocity times acc.
Because its for the same car
thus 1 = 2 when ever you divided both multiplied both by the same amount :D

Answer 12

Why they are equal ?
Because the definition of each one alone will reach the same result.

Answer 13

I would like to tel you a esy wy , use , dimensional analysis , and figure out constant from work energy therem

Answer 14

in the vector diagram of acceleration and the distance over which the acceleration is ocurring we see that the dot product of the two is equal to the component of the change in velocity per unit time in the direction of the distance dx times that distance. But the time interval over which the the velocity changed is the same over which the change in distance ocurred. Thus we can determine the velocity in the direction of dx due to the change in distance over this time intervaL which is the same as dVx.\[\frac{ dV _{x} }{dt }dx= \frac{ dx }{ dt }dV _{x}=V _{x}dV _{x}\]

Answer 15

"The only thing that comes to mind is that when I think of position and acceleration of a circular path, the vectors are antiparallel. "

the vectors are not anti parallel, They are perpendicular..

a is perpendicular to dx hence the dot product yields zero
also v is perpendicular to dv which also gives the same result of zero..

"I guess I had to take away your medal @Mashy until someone can answer the question unless you have any more thoughts to help figure this out lol"

I don't care about medals :P.. concepts is what matters :D

@Catch.me yes yes.. but he wants to get a physical intuition behind it.. not just using math..

why acceleration (dot) infinitesimal displacement = velocity (dot) the infinitesimal change in velocity

Answer 16

Sorry to be a nuisance, but for circular motion, it is correct to say that the position vector is antiparallel to the acceleration vector. (Ask yourself what is the centripetal acceleration in vector notation.)
The position vector is perpendicular to the instantaneous change in position, but that is a different vector.

Answer 17

ohh.. sorry.. @ProfBrainstorm

position vector.. i thought of it as displacement vector.. lol :D
besides.. that is only true.. provided we take the centre of the circle to be the origin.. (more nuisance i guess xD)