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# Types of Kinetic Energy

The word kinetic originated from the Greek word "kinesis" which means motion. Kinetic energy is basically the energy of motion.
Any object, which is in the state of motion, whether in a horizontal or vertical direction will possess some kinetic energy.

The Energy possessed by an object by virtue of its motion.

So, kinetic energy is a form of energy that represents motion. Kinetic energy has magnitude. It has no direction.

 Related Calculators Calculate Kinetic Energy Calculate Kinetic Friction Mass to Energy Activation Energy Calculator

## Kinetic Energy Equation

$V_{f}^{2} - V_{i}^{2}$ = $2ad$

Replace $a$ = $\frac{F}{m}$

$V_{f}^{2} - V_{i}^{2}$ = $\frac{2Fd}{m}$

If $V_{i}$ = 0 and work is force through distance then

W = E = $\frac{1}{2}$$mv^{2} Another way: \Delta K = W =\int F(r) dr= \int madr = m\int \frac{dv}{dt}$$dr$

Evaluating the integral:

$\Delta K$ = $m\int$ $\frac{dv}{dt}$$dr = m\int \frac{dr}{dt}$$dv$ = $m\int vdv$

## Kinetic Energy Units

Kinetic energy is measured in Joules in S.I. system.

Hence average kinetic energy of molecules = $\frac{1}{2}$$mv_{aver}^{2} = \frac{3}{2} K_{b}T ## Translational Kinetic Energy Back to Top Translational motion is the motion along a line or in the space. So, the kinetic energy possessed by the body or the object on account of its translational motion is called translational kinetic energy. It is given by, E = \frac{1}{2}$$mv^{2}$

Here m = mass of the body, v = velocity of motion of the body in m/s.

Derivation:

We know: $work$= $F.dx$ = $F.Vdt$

$dF\times vdt$ = $v.d(mv)$

Now $E_{k}$= $\int F.dx$ = $\int v.d(mv)$ = $\int d$($\frac{mv^{2}}{2}$) = $\frac{1}{2}mv^{2}$

$K$ = $\frac{1}{2}$$(u_{1}^{2} + u_{2}^{2}+u_{3}^{2}) Here u = turbulence normal stress ## Relativistic Kinetic Energy Back to Top If the body’s speed is close to the fraction of the speed of light or 'c' then it is necessary to use the concept of relativistic kinetic energy.To calculate it we use the concept of relative mechanics. We know that, E_{k}= \int vdP = \int v d(myv) [as P = m y c] =myv^{2} - \frac{m}{2} \int yd(v^{2}) Solving and taking Eo as \int E_{k} = myc^{2}–E_{0} Also E_{0} = mc^{2} So E_{k} = myc^{2}-mc^{2} Here y=\sqrt{1 –\frac{v^{2}}{c^{2}}}-1 ## Negative Kinetic Energy Back to Top Negative kinetic energy is not possible if we consider the macroscopic objects because negative kinetic energy would mean that either the mass is negative, which is not possible or the velocity is negative. Both these are not feasible. But the concept of negative kinetic energy comes from the Tunneling Effect of Particles like electron and proton etc. While tunneling if we calculate the momentum and kinetic energy at the tunneling portion, we would find the kinetic energy to be negative. Solving the Schrodinger Equations for tunneling we can mathematically calculate the negative kinetic energy. Kinetic energy can never become negative. But we can see that the change in kinetic energy or delta can be negative. For example, if we throw the ball in the sky then at the peak point the kinetic energy becomes zero and now the change would be negative. Tunneling is a process in which the particle will travel by an imaginary route made by itself and not by the usual route. Tunneling has been proved and observed experimentally. So, this is all about the negative kinetic energy. Understanding will improve once you understand the basics of quantum mechanics. ## Total Kinetic Energy Back to Top The total kinetic energy of a system can be the sum of various kinetic energies. The energy could be the sum of rotational kinetic energy, translational kinetic energy. Translational kinetic energy = \frac{1}{2}$$mv^{2}$
Rotational kinetic energy = $\frac{1}{2}$$Iw^{2} Hence, Total KInetic Energy E_{k} = \frac{1}{2}$$mv^{2}$ + $\frac{1}{2}$$Iw^{2}$