Question

show that total mechanical energy remains constant when a body is dropped from some height?​

Answer 1

Answer:

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Answer 2

Explanation:

At all the points , throughout the journey

Initial K.E + initial P.E. = final K.E+ final P.E

K.E =

$m \times v ^{2} \div 2$

P.E = m x g x h

Mechanical Energy = K.E + P.E

For initial Position when ball is at height h:

At the highest point , velocity is 0, so K.E = 0

THEREFORE Mechanical energy =0 + m x g x h

For final position when ball is Just before hitting the ground:

height from the ground of the body = 0 , so P.E = 0

THEREFORE Mechanical Energy = mv²/2 + 0

As we can see that K.E and P.E is changing at different positions throughout the journey but the sum of K.E and P.E Is remaining same(conserved).

In other words,Mechanical Energy is being conserved .