Why Is the Derivation of Kinetic Energy Confusing?

AI Thread Summary
The discussion centers around confusion regarding the derivation of kinetic energy from work. Participants question the validity of transitioning from one mathematical expression to another in the derivation process, particularly between steps (2) and (3). Clarifications indicate that derivatives can be treated as divisions of differentials, allowing for algebraic manipulation. Additionally, it is emphasized that the change in kinetic energy is consistently expressed as the difference of squared velocities, countering an incorrect interpretation of the final expression. The conversation highlights the importance of understanding calculus principles in physics derivations.
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?
 
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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.
 
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