Question

how is energy conserved during a free fall of an object from certain height ?​

Answer 1

Energy at A

At Point A body is at rest

Kinetic Energy = 1/2mv²

➺ 1/2×m×0

${\boxed{\red{\tt{ Kinetic Energy = 0}}}}$

At point 'A' body is height of 'h'

So,

${\boxed{\red{\tt{Potential Energy = mgh}}}}$

Then

Total energy = K.E + P.E

➺ 0+ mgh

➺ mgh

Energy at A = mgh

Energy at B

Let, Distance of point B from A = x

So, Distance of point B from C = h-x

Potential energy = mg(h-x)

${\boxed{\red{\tt{P.E = mgh-mgx }}}}$

Let, velocity of body at B = v

So, Acceleration to third equation of motion

v² = u²+2gh

v² = 0+2gx

v² = 2gx

Kinetic energy = 1/2mv²

➺ 1/2×m× 2gx

${\boxed{\red{\tt{K.E = mgx }}}}$

Total energy at B = K.E + P.E

➺ mgx +mgh -mgx

➺ mgh

Energy at C

height = 0

Potential energy = mgh

${\boxed{\red{\tt{P.E = 0 }}}}$

And

Final velocity of body = v

So , Acceleration to third equation of motion

v²= u²+2gh

v²= 2gh

Kinetic energy = 1/2mv²

➺ 1/2×m×2gh

${\boxed{\red{\tt{K.E = mgh}}}}$

Total energy = K.E + P.E

➺ mgh+0

➺ mgh

$\bf\underline{\underline{\blue{Thus:-}}}$

The total energy of the body at each of the three different point along the direction of motion is the same

So,

the energy of free falling body is conserved during its motion

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Answer 2

Answer:

This is because when the body falls from a height, then its potential energy changes into kinetic energy progressively. A decrease in the potential energy is equal to an increase in the kinetic energy of the body. During the process, total mechanical energy of the body remains conserved.

Step-by-step explanation:

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