# Derivation of kinetic energy

by bentley4
Tags: kinetic energy work
 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 = $\int$$\vec{F}$(t).d$\vec{r}$(t)= (2) m.$\int$$\frac{d\vec{v}(t)}{dt}$.d$\vec{r}$(t)= (3) m.$\int$d$\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 http://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?
Mentor
P: 41,489
 Quote by bentley4 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 http://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.
P: 66
 Quote by Doc Al 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?

Mentor
P: 41,489
Derivation of kinetic energy

 Quote by bentley4 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
 PF Gold P: 956 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.
 P: 66 Thnx for your responses.
P: 476
 Quote by bentley4 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 = $\int$$\vec{F}$(t).d$\vec{r}$(t)= (2) m.$\int$$\frac{d\vec{v}(t)}{dt}$.d$\vec{r}$(t)= (3) m.$\int$d$\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 http://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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