1) A stone of mass m is dropped from a Height H. Prove that the
mechanical energy remains conserved under gravitational
force.
2) A stone of mass 20 kg is dropped from a top of a tower of height
150 m. Calculate the Mechanical Energy, Potential Energy and
Kinetic Energy attained by this stone after covering a distance
of 100 m from the top of the tower.(Take g=10 m/s2)
Answers
1) Energy can neither be created nor destroyed, it can only be transformed from one form to another. The total energy before and after transformation remains the same.
A body of mass'm is raised to height b. At A its potential energy is maximum and kinetic energy is 0 as it is stationary.
When body is allowed to free fall and it reaches at B, h is decreasing hence potential energy decreases and V is increasing hence kinetic energy is increasing.
When the body is about to reach the ground level, h = 0, v will be maximum hence kinetic energy > potential energy
∴ Decrease in potential energy = Increase in kinetic energy
This shows the continual transformation of gravitational potential energy into kinetic energy. Potential energy + Kinetic energy = Constant (Mechanical energy)
__________________________
2) when the body is released from a height of 2 meters then all the potential energy will be converted to kinetic energy.
assuming base as a reference
mgh+0=K.E.+0
so K.E.=400J
(a). Mass of the Body = 10 kg.
Height = 10 m.
Acceleration due to gravity = 9.8 m/s².
Using the Formula,Potential Energy = mgh
= 10 × 9.8 × 10 = 980 J.
(b). Now, By the law of the conservation of the Energy, Total amount of the energy of the system remains constant.
∴ Kinetic Energy before the body reaches the ground is equal to the Potential Energy at the height of 10 m.
∴ Kinetic Energy = 980 J.
(c). Kinetic Energy = 980 J.
Mass of the ball = 10 kg.
∵ K.E. = 1/2 × mv²
∴ 980 = 1/2 × 10 × v²
∴ v² = 980/5
⇒ v² = 196
∴ v = 14 m/s.