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

## 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) – Yong See Foo · 3 years, 9 months ago

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– Saad Haider · 3 years, 9 months ago

But what if mass is not constant so \(\frac{dm}{dt} \neq 0\)Log in to reply

– Yong See Foo · 3 years, 9 months ago

Then you have to find out at what rate \(m\) is changing wrt to \(t\) and continue with the derivation.Log in to reply

– Saad Haider · 3 years, 9 months ago

Would not that be equivalent to saying \(KE\neq \frac{1}{2}mv^{2}\)Log in to reply

– Yong See Foo · 3 years, 9 months ago

Exactly, that is what relativity is about.Log in to reply

dmv/dt .dr here dr/dt =v and mdv will be left so we can procede in the same way above – Pavansai Dosawada · 3 years, 1 month 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 · 3 years, 9 months ago

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