Law of Conservation of Energy
Hello students, we have already learned about kinetic energy and potential energy. The amount work done required to change the position of object is from one place to another is nothing but the potential energy of object. Kinetic energy is also expressed in term of the work done. This gives the relation between the kinetic energy and potential energy of object. We can say the total energy of system during the change in position of object is sum of its kinetic energy and potential energy. Total energy of body in given system always remains the same or constant, this statement leads to the most unique and important law termed as law of conservation of energy.
Law of conservation of energy states that, “energy can neither be created nor be destroyed. It can be transferred from on form to another.” Hence the energy is always inter convertible. This can be proved by following example.
Imagine boy enjoying the trip of giant wheel. Consider the following position of boy as shown in diagram,
H= higher position at height ‘h’ from ground
M= middle position at height ‘x’ above ground
L= lower position on ground.
u=initial velocity of children at higher position
v= final velocity of children at lower position
g= acceleration due to gravity.
Let ‘m’ is mass of boy which is moving from higher to lower position in time‘t’.
Case:1) When boy is at higher position and reaching towards ground, kinetic energy increase due to gravity at the same instant potential energy goes on decreases. In this case potential energy is given as,
∴ Potential energy = m × g × h …….(1)
Now kinetic energy will be,
∴ Kinetic energy of body = ½ mu2
∴ Kinetic energy of body = 0
Total energy of boy will be sum of kinetic energy and potential energy,
∴ Total energy = Kinetic energy + potential energy
∴ Total energy=0 + mgh
∴ Total energy = mgh ……………………… (A)
Case:2) When boy is at lower position: kinetic energy is maximum and potential energy is minimum and close to zero.
∴ Potential energy = 0 …………….(3)
Now kinetic energy will be,
∴ Kinetic energy of body = ½ mv2
Now from 3rd equation of motion we have,
∴ 2as = v2 – u2
∴ 2gh = v2 – 0
∴ v2 = 2gh
∴ Kinetic energy of body = ½ m × 2gh
∴ Kinetic energy of body = mgh ………….(4)
Total energy of boy will be sum of kinetic energy and potential energy,
∴Total energy = Kinetic energy + potential energy
∴Total energy = mgh + 0
∴ Total energy=mgh ………………………(B)
Case:3) When boy is at midway position: At midway position, kinetic energy and potential energy both will present then,
∴Potential energy = mg (h – x)
∴ Potential energy = mgh – mgx …………….(5)
Now kinetic energy will be,
∴ Kinetic energy of body =1/2 mv2
Now from 3rd equation of motion we have,
∴ 2as = v2 – u2
∴ 2gx = v2 – 0
∴ v2 = 2gx
∴ Kinetic energy of body =1/2 m × 2g (h – x)
∴ Kinetic energy of body = mgx ………….(6)
Total energy of boy will be sum of kinetic energy and potential energy,
∴ Total energy = Kinetic energy + potential energy
∴ Total energy = mgx + mgh – mgx
∴ Total energy = mgh………………………(C)
From equations (A), (B) and (C) we can say that the total energy at all above points is constant. i.e. constant or conserved.