prove the formula K.E.=1/2mv2
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
Work done = ½ mV2
Since the work done is motion is called "Kinetic Energy"
K.E. = Work done
The work done accelerating a particle during the infinitesimal time interval dt is given by the dot product of force and displacement:
where we have assumed the relationship p = m v. (However, also see the special relativistic derivation below.)
Applying the product rule we see that:
Therefore (assuming constant mass), the following can be seen:
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:
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).
If a rigid body is rotating about any line through the center of mass then it has rotational kinetic energy () which is simply the sum of the kinetic energies of its moving parts, and is thus given by:
- ω is the body's angular velocity
- r is the distance of any mass dm from that line
- is the body's moment of inertia, equal to .
(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).