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Ask iit jee aieee pet cbse icse state board experts Expert Question: Kinetic Energy
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sam (2)

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 How to derive the formula K.E= 1/2 mv2

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mukundeng12 (5)

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w=F.S=ma.VxV/2a

hi
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edison (4420)

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Another elegant approach for deriving an expression for Kinetic energy is the
 
Classical Relativistic approach using famous Einstein's mass- Energy
 
relationship i.e.,
 
E = mc2,
 
Where,
 
 E = Total energy of the particle = rest mass energy + Kinetic energy....(1)
 
and 'm' = Relativistic mass of the particle i.e., mass of the particle when its
 
velocity is v
 
'c' = Speed of light in vacuum
 
Rest mass of the particle is taken = m0
 
m = m0/(1 -v2/c2)=m0(1 -v2/c2)-1/2
 
or m = m0(1 + v2/2c2 + neglecting higher terms) =m0(1 + v2/2c2)
 
( as here v<<c)
 
Therefore, E = m0c2(1 + v2/2c2) =m0c2+ m0v2/2 ........(2)
 
Where, m0c2 = Rest mass energy
 
Therefore Kinetic energy of the particle = m0v2/2
 
NOTE: REMEMBER THIS FORMULA FOR KINETIC ENERGY HOLDS GOOD
 
ONLY FOR THE CASES WHERE v<<c, WHAT WE CALL THE NON
 
RELATIVISTIC CASE. WHEREAS, IF WE CONSIDER RELATIVISTIC CASE
 
WHERE, VELOCITY OF PARTICLE IS COMPAREBLE TO THE VELOCITY OF
 
LIGHT, THE KINETIC ENERGY TERM INVOLVES HIGHER POWERS OF
 
v2/c2 THAT WERE OTHERWISE NEGLECTED IN DERIVING ABOVE
 
EXPRESSION.

The most incomprehensible thing about the world is that it is

at all comprehensible.
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udaanparvaz (0)

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K.E is the amount of work to stop a moving body.Therefore W=Fd
But F=-ma (K.E is opposite to workdone)
then,W=-mad
NOw,
-2as=v^2-u^2
-2a[x(t)-x(0)]=v(t)^2-v(0)^2
-2ad=0-v^2
-2ad=-v^2
-ad=-1/2v^2
ad=1/2v^2
but W=-mad
then,putting the value
W=m1/2v^2
therefore k.e=1/2mv^2
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neeraj (149)

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Suppose a constant resultant force F acting on a particle of mass m it will produce a constant acceleration a .
 
Let us choose the x axis to be in the common direction of F and a. Whatever work done by this force will cause a displacement x .
 
For const. Acc.
 
            a = ( v  - v0 ) / t
 
            x = ( v  + v0 ) / 2 . t
 
v0 is the particle speed at t = 0
 
v is the particle speed at t = t
 
work done w = F . x   =  m . a . x
 
                                    = m [ ( v  - v0 ) / t ] . [ ( v  + v0 ) / 2 ].  T
 
                                    = 1 / 2  m v2  -  1/ 2 m v02
 
We call one half the product of the mass of a body and  square of its speed the kinetic energy of the body
 
                        K = 1 / 2 m v2
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