the equation for kinetic energy is actually used as $\frac{1}{2}mv^{2}$ as it is a very useful quantity and actually is derived from the work done by a force, without any changes to internal energy or potential energy of the system.

$W = \int \Sigma\vec{F}\cdot d\vec{r}$

Subbing in $\Sigma\vec{F} = m\vec{a}$, you will get

$W = \int m\vec{a}\cdot d\vec{r}$

Subbing in $\vec{a} = \frac{d\vec{v}}{dt}$, you will get

$W = \int m\frac{d\vec{v}}{dt}\cdot d\vec{r}$

Since $\vec{v}=\frac{d\vec{r}}{dt}$, it implies that $d\vec{r} = \vec{v}dt$

$W = \int m\frac{d\vec{v}}{dt}\cdot\vec{v}dt$

$W = \int_{\vec{v}_i}^{\vec{v}_f} m\vec{v}\cdot d\vec{v}$

$W = \frac{1}{2}m(\vec{v}_f \cdot \vec{v}_f) - \frac{1}{2}m(\vec{v}_i\cdot\vec{v}_i) = \frac{1}{2}mv_f {}^{2} - \frac{1}{2}mv_i{}^{2}$

This showed that the quantity $\frac{1}{2}mv^{2}$ can be denoted as the energy of a moving object, and any change in kinetic energy must be from an external work done

However we know that the general definition of force is $\Sigma\vec{F}=\frac{d\vec{p}}{dt}$

Hence work would be $W = \int \frac{d\vec{p}}{dt}\cdot d\vec{r}$.

Then from that kinetic energy should be derived, as it more accurately depicts motion of an object of changing mass. Then how do we derive $\int \frac{d\vec{p}}{dt}\cdot d\vec{r}$

PLEASE HELP THIS QN IS KILLING ME

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## Comments

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TopNewestIt's similar. Hint: Product law for derivatives. $\frac{dp}{dt}=\frac{d(mv)}{dt}=m\cdot\frac{dv}{dt}+v\cdot\frac{dm}{dt}$. Note that $\frac{dm}{dt}=0$ for constant mass, that is the non-relativistic case. Note that $F=ma$ is not true for relativistic cases. (Sorry lazy to put vector arrows and stuff =P)

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But what if mass is not constant so $\frac{dm}{dt} \neq 0$

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Then you have to find out at what rate $m$ is changing wrt to $t$ and continue with the derivation.

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$KE\neq \frac{1}{2}mv^{2}$

Would not that be equivalent to sayingLog in to reply

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Actually according to relativity, our mass is increasing with velocity but not that much. So, the kinetic energy formula holds. Also, the formula is m_{real}=\frac{m_{rest}}{\sqrt{1-\frac{v^{2}}{c^{2}}}

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dmv/dt .dr here dr/dt =v and mdv will be left so we can procede in the same way above

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