Question

Show that the total mechanical energy of a freely falling body under gravity is conserved.​

Answer 1

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Let us understand this principle more clearly with the following example. Let us say, a ball of mass m is dropped from a cliff of height H, as shown above.

At height H:

Potential energy (PE) = m×g×H

Kinetic energy (K.E.) = 0

Total mechanical energy = mgH

At height h:

Potential energy(PE) = m×g×h

Kinetic energy (K.E.) =1/2(mv^2)

Using the equations of motion, the velocity v1 at a height h for an object of mass m falling from a height H can be written as Conservation Of Mechanical Energy

Hence, the kinetic energy can be given as,

Conservation Of Mechanical Energy

Total mechanical energy = (mgH – mgh) – mgh = mgH

At height zero:

Potential energy: 0

Kinetic energy: 1/2(mv^2)

Using the equations of motion we can see that velocity v at the bottom of the cliff, just before touching the ground is Conservation Of Mechanical Energy

Hence, the kinetic energy can be given as,

Conservation Of Mechanical Energy

Total mechanical energy: mgH

We saw the total mechanical energy of the system is constant throughout.

Answer 2

Answer:

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