Question

derive law of conservation of energy​

Answer 1

Answer:

Law of Conservation of Energy Derivation

v^{2 }= 2g(H – X)\\ \\ \Rightarrow \frac{1}{2}mv^{2}=\frac{1}{2}m.2g(H – X)\\ \\ \Rightarrow K.E=\frac{1}{2}m.2g(H – X) \\ \Rightarrow K.E=mg(H – X) K.E=mg(H-X)——– (3) ...

=> P.E = K.E = \frac {E}{2} ——– (4) E2 is the new energy. ...

mgX=\frac{mgH}{2}\\ \\ \Rightarrow X=\frac{H}{2} H2 is referred to the new height.

Answer 2

Answer:

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In physics and chemistry, the law of conservation of energy states that the total energy of an isolated system remains constant; it is said to be conserved over time.

Derivation:-

Consider a point A, which is at 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.