Question

What Do You Mean By Kinetic Energy?

Answer 1

Hint: Think of kinetic energy as the energy of kinesis i.e. the energy of motion. And now using the basic Newtonian laws of motion and the force formula F=ma derive the formula for kinetic energy.

Complete answer:
The term kinetic energy was coined by Lord Kelvin in 1849.The word kinetic has been derived from the Greek word kinesis meaning motion.Kinetic energy basically means the energy possessed by a moving body. It can also be defined as the work done to accelerate the object to a given specified velocity or speed.In classical mechanics we can say that the kinetic energy of a non-rotating object is, $\dfrac{1}{2}mv^2$
Where $v$=velocity and $m$=mass of the object.

But in relativistic mechanics it is a nice approximation when the velocity of a moving object is way less than the speed of light. The SI unit of Kinetic energy is Joule.So here we go deriving the expression for kinetic energy of non-rotating bodies. Let $F$ be the applied force and x be the distance moved by the body
Work done= \int{\vec{F}\cdot d\vec{x}}
Put $F=\dfrac{dp}{dt}$
$E_k=\int\dfrac{dp}{dt}.dx \\
\Rightarrow E_k=\int\dfrac{dx}{dt}.dp$
Put $v=\dfrac{dx}{dt}$
$E_k=\int v.dp$
Put $p=mv$
$E_k=\int v.d\left(mv\right) \\
\Rightarrow E_k=\int v.m.dv \\
\Rightarrow E_k=m\int v.dv \\
\Rightarrow E_k=\dfrac{1}{2}mv^2 $
Here it is assumed that the object starts with no initial motion i.e. initial velocity is 0.

The relation of kinetic energy with the momentum is,
$\dfrac{p^2}{2m}$
Where $p$=momentum and $m$=mass of the object
In case of purely rotational objects the formula of kinetic energy can be derived as follows
$E_k=\int v.d\left(mv\right)$
As we have obtained in above derivation
$E_k=\int\dfrac{v^2.dm}{2} $
Put $v= \omega r$
$E_k=\int\dfrac{\left(\omega r\right)^2}{2}.dm \\
\Rightarrow E_k=\dfrac{\omega^2}{2}\int r^2.dm$
Put $I=\int r^2.dm$
$E_k=\dfrac{1}{2}I\omega^2$
Where $\omega$= angular velocity, $r$= distance of mass dm from axis of rotation and $I$ = moment of inertia of object.

In the case of rotational objects we also have a rotational kinetic energy along with translational kinetic energy
$E_k=E_t+E_r
E_k=\ \dfrac{1}{2}mv^2+\dfrac{1}{2}I\omega^2$
Where ${E_k}$=Total kinetic energy, ${E_t}$=Translational kinetic energy and ${E_r}$=Rotational kinetic energy.
And the other terms used have been explained in the above points.

Note:So as we can see that if we had known the formulas and laws mentioned above in hint we would have easily derived the formula for kinetic energy under different scenarios whether it is translational or rotational motion and even mix of both. Energy can neither be created nor be destroyed.