According to this law, energy can only be converted from one form to another; it can neither be created nor destroyed. The total energy before and after the transformation remains the same. The law of conservation of energy is valid in all situations and for all kinds of transformations.
The sum of kinetic energy and potential energy of an object is its total mechanical energy.
During the free fall of the object, the decrease in potential energy, at any point in its path, appears as an equal amount of increase in kinetic energy. (Here the effect of air resistance on the motion of the object has been ignored.) There is thus a continual transformation of gravitational potential energy into kinetic energy.
Activity 11.15 page 155 class 9 Science
An object of mass 20 kg is dropped from a height of 4 m. For simplifying the calculations, take the value of g as 10 m s–2.
Calculate potential energy, kinetic energy and total energy in each case at the following heights 4m, 3m, 2m, 1m, & Just above the ground.
Solution:
At Height 4 meters:
m = 20 Kg, g = 10 ms–2 , h = 4 m, v = 0 ms-1 (as it is just dropped)
So,
Potential Energy = mgh = 20 x 10 x 4 = 800 J
Kinetic Energy = ½ mv2 = 0.5 x 20 x 02 = 0 J
Total Energy = mgh + ½ mv2 = 800 J
At Height 3 meters:
m = 20 Kg, g = 10 ms–2 , h = 4 m, v = has to be calculated
Finding Velocity
Initial velocity, u = 0, Distance travelled, s = 4 – 3 = 1m; a = g = 10 ms–2
Third Equation of Motion is,
v2 – u2 = 2as
v2 – 02 = 2 x 10 x 1 = 20, So, v2 = 20 (Need not to solve it because in formula we need v2 only)
So,
Kinetic Energy = ½ mv2 = 0.5 x 20 x 20 = 200 J
Potential Energy = mgh = 20 x 10 x 3 = 600 J
Total Energy = mgh + ½ mv2 = 800 J
At Height 2 meters:
m = 20 Kg, g = 10 ms–2 , h = 2 m, v = has to be calculated
Finding Velocity
Initial velocity, u = 0 (Velocity at beginning), Distance travelled, s = 4 – 2 = 2m; a = g = 10 ms–2
Third Equation of Motion is,
v2 – u2 = 2as
v2 – 02 = 2 x 10 x 2 = 40, So, v2 = 40 (Need not to solve it because in formula we need v2 only)
So,
Kinetic Energy = ½ mv2 = 0.5 x 20 x 40 = 400 J
Potential Energy = mgh = 20 x 10 x 2 = 400 J
Total Energy = mgh + ½ mv2 = 800 J
At Height 1 meters:
m = 20 Kg, g = 10 ms–2 , h = 1 m, v = has to be calculated
Finding Velocity
Initial velocity, u = 0 (Velocity at beginning), Distance travelled, s = 4 – 1 = 3m; a = g = 10 ms–2
Third Equation of Motion is,
v2 – u2 = 2as
v2 – 02 = 2 x 10 x 3 = 60, So, v2 = 60 (Need not to solve it because in formula we need v2 only)
So,
Kinetic Energy = ½ mv2 = 0.5 x 20 x 60 = 600 J
Potential Energy = mgh = 20 x 10 x 1 = 200 J
Total Energy = mgh + ½ mv2 = 800 J
At Height 0 meters (just above the ground):
m = 20 Kg, g = 10 ms–2 , h = 0 m, v = has to be calculated
Finding Velocity
Initial velocity, u = 0 (Velocity at beginning), Distance travelled, s = 4 – 0 = 4m; a = g = 10 ms–2
Third Equation of Motion is,
v2 – u2 = 2as
v2 – 02 = 2 x 10 x 4 = 40, So, v2 = 80 (Need not to solve it because in formula we need v2 only)
So,
Kinetic Energy = ½ mv2 = 0.5 x 20 x 80 = 800 J
Potential Energy = mgh = 20 x 10 x 0 = 0 J
Total Energy = mgh + ½ mv2 = 800 J
So, using this activity we have proved the Law of Conservation of Energy.
𝘛𝘩𝘢𝘯𝘬𝘴