
#1
Feb2312, 05:43 PM

P: 66

Hi everyone,
There are 2 things I do not understand in the derivation of kinetic energy from work: http://www.physicsforums.com/showthread.php?t=111162 (1) W = [itex]\int[/itex][itex]\vec{F}[/itex](t).d[itex]\vec{r}[/itex](t)= (2) m.[itex]\int[/itex][itex]\frac{d\vec{v}(t)}{dt}[/itex].d[itex]\vec{r}[/itex](t)= (3) m.[itex]\int[/itex]d[itex]\vec{v}[/itex](t).[itex]\frac{d\vec{r}(t)}{dt}[/itex]= (4) [itex]\frac{m}{2}[/itex].(v(t1)  v(t0))[itex]^{2}[/itex] Question I: I don't understand why you can just change the division like that between (2) and (3). I know multiplication and division have the same order but they are calculated from left to right. So why can you do that from (2) to (3)? http://en.wikipedia.org/wiki/Order_of_operations Question II: I don't understand why the change in kinetic energy between t1 and t0 is sometimes written as [itex]\frac{1}{2}[/itex].m.v(t1)[itex]^{2}[/itex]  [itex]\frac{1}{2}[/itex].m.v(t0)[itex]^{2}[/itex] E.g. like in http://www.physicsforums.com/showthread.php?t=194461. After (4), it should be [itex]\frac{m}{2}[/itex].v(t1)[itex]^{2}[/itex]  m.v(t1).v(t0) + [itex]\frac{m}{2}[/itex].v(t0)[itex]^{2}[/itex] , right? 



#3
Feb2312, 05:59 PM

P: 66





#4
Feb2312, 06:02 PM

Mentor
P: 40,905

Derivation of kinetic energy 



#5
Feb2412, 04:17 AM

P: 863

Re your QI...
I respect your unwillingness to regard [itex]\frac{d\vec{v}}{dt}[/itex] as one thing divided by another, but you might be happier to do this with [itex]\frac{Δ\vec{v}}{Δt}[/itex], and then associate Δt with Δ[itex]\vec{r}[/itex] to make [itex]\vec{v}[/itex]. Only then take it to the limit. I don't suppose this is rigorously valid, but I suspect this is how a lot of physicists justify this sort of transposition. 



#6
Feb2412, 02:11 PM

P: 66

Thnx for your responses.




#7
Feb2512, 11:49 AM

P: 476

Question II: It is not sometimes, but always, because that is the result of the integration of (3). You own result (4) is incorrect. 


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