Why Is the Derivation of Kinetic Energy Confusing?

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Discussion Overview

The discussion revolves around the derivation of kinetic energy from work, focusing on specific steps in the mathematical process and the reasoning behind them. Participants raise questions about the validity of certain transformations and the representation of changes in kinetic energy.

Discussion Character

  • Technical explanation
  • Debate/contested
  • Mathematical reasoning

Main Points Raised

  • One participant questions the validity of changing the division in the transition from step (2) to step (3) in the derivation.
  • Another participant expresses confusion about the representation of the change in kinetic energy and suggests an alternative expression after step (4).
  • A participant asserts that the step (4) presented by the original poster is incorrect.
  • Another response suggests that viewing the derivative as a division of differentials may clarify the transformation from (2) to (3).
  • One participant insists that the representation of the change in kinetic energy is always correct as stated, countering the original poster's claim of an alternative form.

Areas of Agreement / Disagreement

Participants express differing views on the correctness of the steps in the derivation, with some asserting that certain transformations are valid while others contest those claims. The discussion remains unresolved regarding the accuracy of the original poster's steps and the interpretations of kinetic energy changes.

Contextual Notes

There are unresolved assumptions regarding the mathematical steps taken in the derivation, particularly concerning the treatment of derivatives and the integration process. The discussion reflects varying interpretations of these concepts.

bentley4
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Hi everyone,

There are 2 things I do not understand in the derivation of kinetic energy from work:
https://www.physicsforums.com/showthread.php?t=111162

(1) W = \int\vec{F}(t).d\vec{r}(t)=

(2) m.\int\frac{d\vec{v}(t)}{dt}.d\vec{r}(t)=

(3) m.\intd\vec{v}(t).\frac{d\vec{r}(t)}{dt}=

(4) \frac{m}{2}.(v(t1) - v(t0))^{2}

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 \frac{1}{2}.m.v(t1)^{2} - \frac{1}{2}.m.v(t0)^{2}
E.g. like in https://www.physicsforums.com/showthread.php?t=194461.

After (4), it should be \frac{m}{2}.v(t1)^{2} - m.v(t1).v(t0) + \frac{m}{2}.v(t0)^{2} , right?
 
Last edited:
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bentley4 said:
Question II: I don't understand why the change in kinetic energy between t1 and t0 is sometimes written as \frac{1}{2}.m.v(t1)^{2} - \frac{1}{2}.m.v(t0)^{2}
E.g. like in https://www.physicsforums.com/showthread.php?t=194461.

After (4), it should be \frac{m}{2}.v(t1)^{2} - m.v(t1).v(t0) + \frac{m}{2}.v(t0)^{2} , right?
No. Your step (4) is incorrect.
 
Doc Al said:
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?
 
bentley4 said:
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
 
Re your QI...
I respect your unwillingness to regard \frac{d\vec{v}}{dt} as one thing divided by another, but you might be happier to do this with \frac{Δ\vec{v}}{Δt}, and then associate Δt with Δ\vec{r} to make \vec{v}. 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.
 
Thnx for your responses.
 
bentley4 said:
Hi everyone,

There are 2 things I do not understand in the derivation of kinetic energy from work:
https://www.physicsforums.com/showthread.php?t=111162

(1) W = \int\vec{F}(t).d\vec{r}(t)=

(2) m.\int\frac{d\vec{v}(t)}{dt}.d\vec{r}(t)=

(3) m.\intd\vec{v}(t).\frac{d\vec{r}(t)}{dt}=

(4) \frac{m}{2}.(v(t1) - v(t0))^{2}

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 \frac{1}{2}.m.v(t1)^{2} - \frac{1}{2}.m.v(t0)^{2}
E.g. like in https://www.physicsforums.com/showthread.php?t=194461.

After (4), it should be \frac{m}{2}.v(t1)^{2} - m.v(t1).v(t0) + \frac{m}{2}.v(t0)^{2} , 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.
 
Last edited:

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