Asked by Shreyas Pande | 14th Feb, 2014, 04:22: PM

Expert Answer:

Consider a body of mass m freely falling under gravity from a height h.
 
 

At position A:

Initial velocity of body = 0.

Thus, kinetic energy = K =0

Potential energy = U = mgh

Hence, total energy = K + U = mgh

 

At position B:

Let v1 be the velocity acquired by the body after falling through distance x.

Then , u = 0, S = x, a = g

From equation v2 = u2 + 2aS, we have

 V1= 0 + 2gx = 2gx

Thus, kinetic energy is

K = 1/2 m V12

   = 1/2 x m x 2gx 

   = mgx

Now at B, height of body above the ground = h – x

Thus, potential energy is = mg (h - x)

Hence, total energy = K + U = mgx + mg(h -x) = mgh

 

At position C:

Let v be the velocity acquired by the body on reaching the ground.

Then , u = 0, S = h, a = g

From equation v2 = u2 + 2aS, we have

 V1= 0 + 2gh = 2gh

Thus, kinetic energy is,

K = 1/2 m V12

   = 1/2 x m x 2gh

   = mgh

Now at C, height of body above the ground = 0

Thus, potential energy is 0.

Hence, total energy = K + U = mgh + 0 = mgh

 

Thus, the mechanical energy always remains constant. Hence, mechanical energy is conserved.

Answered by  | 17th Feb, 2014, 07:26: PM

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