A pressing problem with Kinetic Energy. ALL PHYSICISTS HELP!!

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

It'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)

Yong See Foo - 7 years, 8 months ago

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

Saad Haider - 7 years, 8 months ago

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

@Yong See Foo – Would not that be equivalent to saying $KE\neq \frac{1}{2}mv^{2}$

@Saad Haider – Exactly, that is what relativity is about.

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

Fahad Shihab - 7 years, 8 months ago

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

Pavansai Dosawada - 7 years ago

Math	Appears as
Remember to wrap math in `\(` ... `\)` or `\[` ... `\]` to ensure proper formatting.
`2 \times 3`	$2 \times 3$
`2^{34}`	$2^{34}$
`a_{i-1}`	$a_{i-1}$
`\frac{2}{3}`	$\frac{2}{3}$
`\sqrt{2}`	$\sqrt{2}$
`\sum_{i=1}^3`	$\sum_{i=1}^3$
`\sin \theta`	$\sin \theta$
`\boxed{123}`	$\boxed{123}$