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

the equation for kinetic energy is actually used as 12mv2\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=ΣFdrW = \int \Sigma\vec{F}\cdot d\vec{r}

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

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

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

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

Since v=drdt\vec{v}=\frac{d\vec{r}}{dt}, it implies that dr=vdtd\vec{r} = \vec{v}dt

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

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

W=12m(vfvf)12m(vivi)=12mvf212mvi2W = \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 12mv2\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 ΣF=dpdt\Sigma\vec{F}=\frac{d\vec{p}}{dt}

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


Note by Saad Haider
6 years, 6 months ago

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

Yong See Foo - 6 years, 6 months ago

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But what if mass is not constant so dmdt0\frac{dm}{dt} \neq 0

Saad Haider - 6 years, 6 months ago

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Then you have to find out at what rate mm is changing wrt to tt and continue with the derivation.

Yong See Foo - 6 years, 5 months ago

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@Yong See Foo Would not that be equivalent to saying KE12mv2KE\neq \frac{1}{2}mv^{2}

Saad Haider - 6 years, 5 months ago

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

Yong See Foo - 6 years, 5 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}}}

Fahad Shihab - 6 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

Pavansai Dosawada - 5 years, 10 months ago

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