# prove the formula K.E.=1/2mv2

K.E. = 1/2 mv2

Consider a body of mass "m" starts moving from rest. After a time interval "t" its velocity becomes V.
If initial velocity of the body is Vi = 0 ,final velocity Vf = V and the displacement of body is "d".

 First of all we will find the acceleration of body. Using equation of motion 2aS = Vf2 – Vi2 Putting the above mentioned values 2ad = V2 – 0a = V2/2d Now force is given byF = ma Putting the value of accelerationF = m(V2/2d) As we know that Work done = Fd Putting the value of FWork done = (mv2/2d)(d)Work done = mV2/2OR Work done = ½ mV2 Since the work done is motion is called "Kinetic Energy"i.e. K.E. = Work done OR K.E. =1/2mV2.

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The work done accelerating a particle during the infinitesimal time interval dt is given by the dot product of force and displacement: $mathbf{F} cdot d mathbf{x} = mathbf{F} cdot mathbf{v} d t = frac{d mathbf{p}}{d t} cdot mathbf{v} d t = mathbf{v} cdot d mathbf{p} = mathbf{v} cdot d (m mathbf{v}),,$

where we have assumed the relationship p = m v. (However, also see the special relativistic derivation below.)

Applying the product rule we see that: $d(mathbf{v} cdot mathbf{v}) = (d mathbf{v}) cdot mathbf{v} + mathbf{v} cdot (d mathbf{v}) = 2(mathbf{v} cdot dmathbf{v}).$

Therefore (assuming constant mass), the following can be seen: $mathbf{v} cdot d (m mathbf{v}) = frac{m}{2} d (mathbf{v} cdot mathbf{v}) = frac{m}{2} d v^2 = d left(frac{m v^2}{2}right).$

Since this is a total differential (that is, it only depends on the final state, not how the particle got there), we can integrate it and call the result kinetic energy: $E_text{k} = int mathbf{F} cdot d mathbf{x} = int mathbf{v} cdot d (m mathbf{v}) = int d left(frac{m v^2}{2}right) = frac{m v^2}{2}.$

This equation states that the kinetic energy (Ek) is equal to the integral of the dot product of the velocity (v) of a body and the infinitesimal change of the body's momentum (p). It is assumed that the body starts with no kinetic energy when it is at rest (motionless).

### Rotating bodies

If a rigid body is rotating about any line through the center of mass then it has rotational kinetic energy ( $E_text{r},$) which is simply the sum of the kinetic energies of its moving parts, and is thus given by: $E_text{r} = int frac{v^2 dm}{2} = int frac{(r omega)^2 dm}{2} = frac{omega^2}{2} int{r^2}dm = frac{omega^2}{2} I = begin{matrix} frac{1}{2} end{matrix} I omega^2$

where:

• ω is the body's angular velocity
• r is the distance of any mass dm from that line
• $I,$ is the body's moment of inertia, equal to $int{r^2}dm$.

(In this equation the moment of inertia must be taken about an axis through the center of mass and the rotation measured by ω must be around that axis; more general equations exist for systems where the object is subject to wobble due to its eccentric shape).

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