Question

If an object of mass m is released from a height h,then show that the total mechanical energy is conserved during it’s motion?

Answer 1

Proof :

As per the given data,

An object of mass m is released from a height h

The total energy will be conserved throughout its motion as no external force is acting on the object.

In order to prove this let us find out the total energy on any 2 points during its motion. Let, these points be A and B.

Please refer to the attachment first !!

Point A :

⇒ PE (max) = mgh

⇒ KE(min) = ½ mu²  

⇒ KE(min) = ½ x m x 0

⇒ KE(min) = 0

⇒ Total Energy = PE + KE

⇒ Total Energy = mgh + 0

⇒ Total energy = mgh

________________________

Point B :

⇒ PE (min) = mgh

⇒ PE (min) = m x g x 0

⇒ PE (min) = 0

⇒ KE (max) = ½ mv²

By using third equation of motion,

⇒ v² =u² + 2gh

⇒ v² = 0 + 2gh

⇒ v² = 2gh

Now,

⇒ KE(max) = ½ m x 2gh

⇒ KE(max) = mgh

⇒ Total Energy = PE + KE

⇒ Total Energy = 0 + mgh

⇒ Total Energy = mgh

Total Energy at point A = Total Energy at point B

$\boxed{\sf{\huge{Hence \: proved}}}$