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

Sort by:

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 ago

Log in to reply

– Saad Haider · 3 years ago

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

– Yong See Foo · 3 years 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 ago

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

– Yong See Foo · 3 years 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 · 2 years, 4 months ago

Log in to reply

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 ago

Log in to reply