state and prove law of conservation of sound
Answers
Proof for Law of Conservation of Energy:
Considering the potential energy at the surface of the earth to be zero. Let us see an example of a fruit falling from a tree.
Consider a point A, which is at some height ‘H’ from the ground on the tree, the velocity of the fruit is zero hence potential energy is maximum there.
E = mgH ———- (1)
When the fruit is falling, its potential energy is decreasing and kinetic energy is increasing.
At point B, which is near the bottom of the tree, the fruit is falling freely under gravity and is at a height X from the ground, and it has a speed as it reaches point B. So, at this point it will have both kinetic and potential energy.
E = K.E + P.E
P.E = mgX ——— (2)
According to third equation of motion,
v2=2g(H–X)⇒12mv2=12m.2g(H–X)⇒K.E=12m.2g(H–X)⇒K.E=mg(H–X)
K.E=mg(H-X)——– (3)
Using (1), (2) and (3)
E = mg(H – X) + mgX
E = mg(H – X + X)
E = mgH
Similarly, if we see the energy at point C which is at the bottom of the tree, it will come out to be mgH. We can see as the fruit is falling to the bottom and here, potential energy is getting converted into kinetic energy. So there must be a point where kinetic energy becomes equal to potential energy. Suppose we need to find that height ‘x’ from the ground. We know that at that point,
K.E = P.E
=> P.E = K.E = E2 ——– (4)
E2 is the new energy
Where, E = mgH2
H2 is the new height.
As the body is at height X from the ground,
P.E = mgX ——— (5)
Using (4) and (5) we get,
mgX=mgH2⇒X=H2
H2 is referred to the new heightProof for Law of Conservation of Energy:
Considering the potential energy at the surface of the earth to be zero. Let us see an example of a fruit falling from a tree.
Consider a point A, which is at some height ‘H’ from the ground on the tree, the velocity of the fruit is zero hence potential energy is maximum there.
E = mgH ———- (1)
When the fruit is falling, its potential energy is decreasing and kinetic energy is increasing.
At point B, which is near the bottom of the tree, the fruit is falling freely under gravity and is at a height X from the ground, and it has a speed as it reaches point B. So, at this point it will have both kinetic and potential energy.
E = K.E + P.E
P.E = mgX ——— (2)
According to third equation of motion,
v2=2g(H–X)⇒12mv2=12m.2g(H–X)⇒K.E=12m.2g(H–X)⇒K.E=mg(H–X)
K.E=mg(H-X)——– (3)
Using (1), (2) and (3)
E = mg(H – X) + mgX
E = mg(H – X + X)
E = mgH
Similarly, if we see the energy at point C which is at the bottom of the tree, it will come out to be mgH. We can see as the fruit is falling to the bottom and here, potential energy is getting converted into kinetic energy. So there must be a point where kinetic energy becomes equal to potential energy. Suppose we need to find that height ‘x’ from the ground. We know that at that point,
K.E = P.E
=> P.E = K.E = E2 ——– (4)
E2 is the new energy
Where, E = mgH2
H2 is the new height.
As the body is at height X from the ground,
P.E = mgX ——— (5)
Using (4) and (5) we get,
mgX=mgH2⇒X=H2
H2 is referred to the new height