Question

Derivation of conservation of energy

Answer 1

Considering the potential energy at the surface of the earth to be zero.

At point A, the velocity of the ball is zero hence only potential energy is there.

 

E = mgH ———- (1)

At point B, the ball is falling freely under gravity so it will gain speed as it reaches point B so at this point it will have both kinetic and potential energy.

 

E = K.E + P.E

P.E = mgX ——— (2)

 

According to third equation of motion,

 

v2=2g(H–X)⇒12mv2=12m.2g(H–X)⇒K.E=12m.2g(H–X)⇒K.E=mg(H–X) ——– (3)

 

Using (1), (2) and (3)

E = mg(H – X) + mgX

E = mg(H – X + X)

E = mgH

 

Similarly if we energy at point C it will come out to be mgH. We can see as the ball is coming down the total energy remains constant, only potential energy is getting converted into kinetic energy. So there must be a point where kinetic energy becomes equal to potential energy. Suppose we need to find that height ‘x’ from the ground. We know that at that point,

 

K.E = P.E

=> P.E = K.E = E2  ——– (4)

Where, E = mgH

 

As the body is at height X from the ground,

P.E = mgX ——— (5)

Using (4) and (5) we get,

mgX=mgH2⇒X=H2