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 – 0
a = V2/2d
Now force is given by
F = ma
Putting the value of acceleration
F = m(V2/2d)
As we know that
Work done = Fd
Putting the value of F
Work done = (mv2/2d)(d)
Work done = mV2/2
OR
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.

  • 17

thx .. !! 

  • 2

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).

[edit]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:

(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).

  • 9
What are you looking for?