Question

prove that the energy remains constant in case of freely falling body.​

Answer 1

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The body covers the distance x with a velocity v. We make use of the third equation of motion to obtain velocity of the body. Total energy at C = mgh … ... Thus, we conclude that law of conservation of energy holds good in the case of a freely falling body.

Answer 2

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prove that the energy remains constant in case of freely falling body.

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The total energy of an object in "free fall" or "free rise" is;

E = (1/2)mv^2 + mgy

The time derivative is;

dE/dt = mv(dv/dt) + mg(dy/dt)

Now dv/dt is the acceleration, which is always -g (if up is positive).

And dy/dt is the velocity which may be positive (rising object) or negative(falling object). But if its negative it has to be negative in both terms. So we can write;

dE/dt = mv(-g) + mgv

Thus dE/dt =0 , and E=constant

#hope it helps..