22. Prove that the total energy of a falling object is the same at every point of its font
Answers
Explanation:
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law of conservation of energy states that the energy can neither be created nor be destroyed but can be transformed from one form to another....
let us now prove that the above law holds good in the case of a freely falling body....
let a body of mass m placed at a height h above the ground, start falling down from rest...
in this case we have to show that the total energy ( potential energy + kinetic energy) of the body at A, B and C remains constant i.e, potential energy is completely transformed into kinetic energy...
AT A--
potential energy=mgh
kinetic energy= 1/2mv square = 1/2*m* 0 kinetic energy= 0 ( the velocity is zero as the object is initially at rest)./// total energy at a = potential energy + kinetic energy = mgh + 0
Total energy at a= mgh ............1eq.....
AT B --
potential energy= mgh
=mgh( h- x ) height from the ground is ( h-x)
potential energy= mgx
kinetic energy= 1/2mv square
the body covers the distance x with a velocity vv. we make use of the third equation of motion to obtain velocity of the body.....
v2- u2u2= 2a here, u=0, a=g, s=x
v2- 0= 2 ax
here, u= 0 ,a= g and s= x
v2 - 0= 2gx..
v2 = 2gx
kinetic energy= 1/2mv square
= 1/2m2gx..
kinetic energy= mgx
Total energy at b= potential energy+ kinetic energy= mgh- mgx+ mgx
Total energy at B = mgh.....2eq....
AT C :-
potential energy = m x g x 0 ( h = 0 )
potential energy = 0 kinetic energy =1/2mv2
the distance covered by the body is hv2 - u2 = 2as
here, u = 0 ,a= g, and s= h
v2 - 0= 2gh
v2 = 2gh
kinetic energy = 1/2mv2
= 1/2* 2gh
total energy at C = potential energy + kinetic energy = 0 + mgh
Total energy at C = mgh....3eq....
its clear from equation 1,2 and 3 that the total energy of the body remains constant at every point..thus, we conclude that law of conservation of energy holds good in the case of a freely falling body....
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