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# 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

Note by Saad Haider
3 years, 5 months ago

## Comments

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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) · 3 years, 5 months ago

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But what if mass is not constant so $$\frac{dm}{dt} \neq 0$$ · 3 years, 5 months ago

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Then you have to find out at what rate $$m$$ is changing wrt to $$t$$ and continue with the derivation. · 3 years, 5 months ago

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Would not that be equivalent to saying $$KE\neq \frac{1}{2}mv^{2}$$ · 3 years, 5 months ago

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Exactly, that is what relativity is about. · 3 years, 5 months ago

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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 · 2 years, 9 months ago

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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 $$m_{real}=\frac{m_{rest}}{\sqrt{1-\frac{v^{2}}{c^{2}}}$$ · 3 years, 5 months ago

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