Kinetic energy is the energy a system has due to its motion. An object in motion is utilizing its kinetic energy: a jogger running past you, a baseball thrown in the air, a coin falling from a table, and a charged particle in an electric field are all examples of kinetic energy in use. In short, anything that moves, vertically or horizontally- has kinetic energy. There are different forms of KE such as rotational, vibrational and translational.

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The kinetic energy KE possessed by such a system, i.e. a massive body in motion, may be estimated as half its mass m times the square of its velocity v. Classical mechanics describes macroscopic systems moving at a small fraction of the speed of light.

Kinetic energy of a point mass m is thus given as

$KE$ = $\frac{1}{2}$ $mv^2$

Kinetic energy is a scalar quantity which is independent of direction and it is evident from the relation above that it is always positive. One can infer the proportionality as follows. When the mass is doubled, the kinetic energy is doubled too; however, when the velocity is doubled, kinetic energy becomes four times larger than its original value.

As its name suggests, kinetic energy, being a form of mechanical energy in a system, is expressed in SI unit of Joules (J) or kiloJoules (kJ).

Rigid bodies are treated in Newtonian mechanics as a collection of point particles. A rigid body in three dimensional space can therefore execute translation or rotation. In contrast to point masses, kinetic energy of rigid bodies due to translation is expressed as

$KE_{tran}$ = $\frac{1}{2}$ $Mv_{c}^{2}$

where M is the mass of the rigid body in question and $v_{c}$ is the velocity of the center of mass of the rigid body with respect to the origin of the frame of reference. This relation is obtained by first dividing the rigid body into a the horde of point particles referred above, and then summing up the kinetic energies due to the linear velocities of the constituent masses.

With the aid of a similar calculation of dividing the rigid body into a the collection of point masses, and then adding the kinetic energies due to the tangential velocities of the constituent masses with respect to the frame of reference, kinetic energy of a rigid body due to its rotation is summed up as

$KE_{rot}$ = $\frac{1}{2}$ $Iw^{2}$

where I is the moment of inertia of the rigid body about the axis of rotation and w is the net angular velocity of the rigid body.

Total kinetic energy of a moving and rotating rigid body must be written as a sum of its translational kinetic energy and rotational kinetic energy.

$KE_{rigid}$ = $KE_{tran}$ + $KE_{rot}$ = $\frac{1}{2}$ $Mv_{c}^{2}$ + $\frac{1}{2}$ $Iw^{2}$

An elastic collision is an encounter between two objects where the sum of kinetic energies of the objects remain same before and after the collision. It can also be said that kinetic energy is conserved in elastic collisions.

For example if there are two objects of mass $m_{1}$ and $m_{2}$ and undergoing head-on collision with velocities $v_{1}$ and $v_{2}$ respectively, they would have velocities $v_{3}$ and $v_{4}$ after the collision such that

$\frac{1}{2}$ $m_{1}v_{1}^{2}$ + $\frac{1}{2}$ $m_{2}v_{2}^{2}$ = $\frac{1}{2}$ $m_{1}v_{3}^{2}$ + $\frac{1}{2}$ $m_{2}$ $v_{4}^{2}$

because kinetic energy of the combined system before and after the collision is conserved. This is a condition often used alongside the momentum conservation relation in elastic collision

$m_{1}v_{1}$ + $m_{2}v_{2}$ = $m_{1}v_{3}$ + $m_{2}v_{4}$

Using two simultaneous equations one can express the final velocities as

$v_{3}$ = $\frac{( m1 - m2 )}{( m1 + m2 )}$ and $v_{4}$ = $\frac{2 m_{1} m_{2}}{( m_{1} + m_{2} )}$

We know that work is done on or by an object when it travels a distance under the influence of an external force. The energy associated with the work done by the net force does not disappear after the net force is removed (or becomes zero), it is transformed into the kinetic energy of the body. This is called the Work-Energy Theorem which is more formally stated as that the work done by the net force acting on a body results in a change of its kinetic energy.

$W$ = $KE_{f}$ - $KE_{i}$

If the body's speed increases, the work done on the body is deemed positive and we say its kinetic energy is raised. However there is a drop in the body's speed, its kinetic energy decreases. In this case the body does positive work on the system slowing it down or alternately the work done on the body is negative. This is an obvious consequence of energy conservation and serves as one of the most powerful relationships in classical physics.

A 1000 kg car is travelling at a steady speed 30 m/s brakes on snow. The coefficient of kinetic friction of ice is 0.05. Use the work energy theorem to find the distance the car travels before coming to a stop.

$F_{N}$ being the normal force which balances the car’s weight vertically,

$F_{N} = mg = (1000 kg)(9.8 m/s^{2}) = 980 N$

Force of kinetic friction acting on the car after braking is $F_{k} = \mu_{k} . F_{N}$ where coefficient of kinetic friction for ice is $\mu_{k} = 0.05$. Therefore,

$F_{k} = \mu_{k} . F_{N} = (0.05)(980N) = 49 N$

If the distance traveled by the car is d, work done by friction on the car is

$W = -F_{k}.d$

Initial KE of the car $KE_{i}$ = $\frac{1}{2}$ $mv^{2}$ = $\frac{1}{2}$ $(1000 kg)(30 m/s)^{2} = 450000 J$

Initial KE of the car $KE_{i}$ = $\frac{1}{2}$ $mv^{2}$ = $\frac{1}{2}$ $(1000 kg)(30 m/s)^{2} = 450000 J$

Final KE of the car $KE_{f} = 0$ , as the car comes to a halt.

$W = KE_{f} - KE_{i}$

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