Question

state the principle of conservation of energy.show that the mechanical energy of a body falling freely under the action of gravity is conserved?​

Answer 1

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Principle of conservation of energy says that energy cannot be created not destroyed. It exists in various forms in universe. It can be converted from one energy to another but cannot created.

Also it states that the amount of energy in a body remains same.

Now to verify that mechanical energy of a body under action of gravity is conserved, look at the figure.

Let's say a random point A is at height h from the point C which is just before the ground. There is also a point B between A and C.

Now let's say for instance there is a body on point A, at rest.

Now in this case, initial velocity of body = 0m/s

So now, kinetic energy of body at point A will be :

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

$= \frac{1}{2} \times m \times {0}^{2}$

$= 0$

Thus K.E of body at point A is 0 .

Now potential energy of body at point A is :

$m \times g \times h$

$= mgh$

Where nothing is 0 , thus P.E of body at point A

is mgh.

Now total mechanical energy of body at point A

= K.E + P.E = 0+mgh

= mgh ------------------------(A)

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Now as in the figure the body now travelled from point A to B, let's say with a velocity v.

So according to third equation of motion the velocity of body will be

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

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

Now the kinetic energy of body at point B will be :

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

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

$= mgx$

Now similarly as in previous case let's calculate potential energy of body at point B :

$= m \times g \times h$

Now since the body travelled to Point B thus the height will become (h -x)

$= m \times g \times (h - x)$

Thus P.E of the body at point B is - mg(h-x)

Now the trial mechanical energy of body at point B will be :

K.E + P.E = mgx + mg(h-x)

$= mg(x + 1(h - x)$

$= mgh$

Thus we get the total mechanical energy of body at point B is mgh ---------------- (B)

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Now let's say the body now travelled from point B to Point C. Now as in the figure it travelled a distance (h - x) in this case.

In this case, according to third equation of motion :

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

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

Now Similarly the Kinetic energy of body at point C will be :

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

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

$k.e \: = mgh$

Thus kinetic energy of body at point C is mgh.

Now, Since point C is just before the ground thus here potential energy will decrease to 0.

Thus potential energy of body at point C is 0.

Now total mechanical energy of body at point C :

K.E + P.E = mgh + 0

= mgh ----------------(C).

From this we can observe that results A, B and C all are same, this sates that the mechanical energy of body remains conserved.

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BY √\__________ SCIVIBHANSHU

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