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Derivation of kinetic energy

by bentley4
Tags: kinetic energy work
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bentley4
#1
Feb23-12, 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?
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Doc Al
#2
Feb23-12, 05:50 PM
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Quote Quote by bentley4 View Post
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?
No. Your step (4) is incorrect.
bentley4
#3
Feb23-12, 05:59 PM
P: 66
Quote Quote by Doc Al View Post
No. Your step (4) is incorrect.
I'm sorry, I don"t understand what I am doing wrong going from (3) to (4). Can you elaborate please?

Doc Al
#4
Feb23-12, 06:02 PM
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Derivation of kinetic energy

Quote Quote by bentley4 View Post
I'm sorry, I don"t understand what I am doing wrong going from (3) to (4). Can you elaborate please?
∫v dv = v2/2 → v12/2 - v02/2
Philip Wood
#5
Feb24-12, 04:17 AM
PF Gold
P: 953
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.
bentley4
#6
Feb24-12, 02:11 PM
P: 66
Thnx for your responses.
juanrga
#7
Feb25-12, 11:49 AM
P: 476
Quote Quote by bentley4 View Post
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?
Question I: Because a derivative is equivalent to a division of differentials. And then by usual algebraic rules you can move the denominator.

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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