illustrate law of conservation of energy in case of a ball dropped from a cliff of height 'h'
Answers
Case 1- At the top of the cliff (height - h)
Initially when the ball is at the top of the cliff , its initial velocity is equal to zero. So, kinetic energy of the ball will be zero.
All the energy will be in the form of potential energy , Potential Energy = mgh.
Total Energy = K.E + P.E
=> 0 + mgh = mgh
Case 2- At a distance x from the cliff
At this position the ball will attain certain velocity and will have both kinetic energy and potential energy.
v = √(2gx)
Kinetic Energy = 1/2 mv² = mgx
Potential Energy = mg(h-x)
T.E = P.E + K.E
=> mg(h-x) + mgx
=> mgh
Case 3 - Ball has reached the ground (h=0)
When the ball reaches the ground it's potential energy will become zero and the ball has only kinetic energy .
v = √2gh
Kinetic Energy = 1/2 mv² = mgh
Potential Energy = 0
T.E = K.E + P.E
=> mgh
The total energy of the ball in all the three cases is equal to mgh.
Therefore , the energy of the ball is conserved throughout it's motion.