Question

Prove the law of conservation of energy for a stone moving vertically down​

Answer 1

$\boxed{ \bold{LAW \: OF \: CONSERVATION \: OF \: ENERGY}}$

◼It states that energy can neither be created nor can be destroyed it can be converted to its one form to another.

$\boxed{ \bold{PROOF}}$

Let a freely falling body a stone of mass m is thrown from a height h It initial velocity be u and final velocity v and acceleration due to gravity be g .Lets take its kinetic energy be K and potential U.

It will be noted that at each point the sum of potential and kinetic enwrgy will be same.

◼When the body is at h height above the ground

Kinetic energy K = 0

Potential energy U = mgh

Total energy will be K + U = 0 + mgh

= mgh

◼When the stone had reached to a certain height x between from the point it had falled and from ground.

We know

${v}^{2} = {u}^{2} +2 as$

where s is displacement x.

${v}^{2} = 0 +2 gx$

Then kinetic energy will be

K = $\frac{1}{2} \times m \times {v}^{2}$

K = $\frac{1}{2} \times m \times {2gx}^{2}$

K= mgx

Where potential energy U is,

U = mg(h-x)

So, total Energy,

K + U = mgx + mg(h- x)

= mgh.

◼When the stone is at ground

Here displacement will be h ,

${v}^{2} = {u}^{2} +2 as$

${v}^{2} = 0+2 gh$

Then kinetic energy will be

K = $\frac{1}{2} \times m \times {v}^{2}$

Then kinetic energy will be

K = $\frac{1}{2} \times m \times 2 gh$

K = mgh

Where potential energy is U bevause the height is zero ,

so, total energy = K + U

mgh + 0

=mgh.

Hence it can be noticed the energy at all the instances is mgh .

Therefore its proved that total energy always remains constant.

Answer 2

Heya...

Here's your answer.........

Let the stone of mass m be at a height h and be dropped from rest at t = 0 sec.

Initial potential energy = PE1 = m g h

initial kinetic energy = 1/2 m 0^2 = 0

Let the stone drop by a distance s in time  t.  As per the laws of kinetics we have

      2 g s = (v^2 - u^2 ) = v^2

Potential energy at time t and height of (h - s) is   PE2 = m g (h - s)

Kinetic energy = KE2 = 1/2 m v^2 = 1/2 m (2 g s) = m g s

Total energy = m g (h - s) + m g s = m g  h

All along the path ,  from t = o to  t = root(2h/g) ,  and height h to 0, the total energy remains as  m g h.

The other forms of energy in the object are assumed to be constant and not varying.

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✨✨✨✨BY BeWaFa ✨✨✨

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Thanks...!!!

XD

Sorry baby 'wink'