Describe law of conservation of energy in context to a free falling body(NO COPY PASTE ).
Answers
Law of conservation of energy states that Energy can neither be created nor be destroyed but can be transferred from one form to another.
In the case of free falling body P.E converts into K.E
Case 1:Consider an object of mass(m) which is made to fall freely from a certain height(h).At the starting point A, P.E is maximum & K.E is minimum because the velocity of an object become zero.
Total energy(T.E) = P.E + K.E
T.E = P.E + 0
T.E = mgh
Case 2:At point B(mid-point in air), the velocity of an object will increase & height of the object will decrease hence the P.E will be converted into K.E.
Total energy(T.E) = P.E + K.E
T.E = mgh + 1/2mv^2(square)
Case 3:At point C(slightly above the ground in air), an object is just above the ground & body will have a certain velocity & acquire a K.E.
Total energy(T.E) = P.E + K.E
T.E = 0 + K.E
T.E = 1/2MV^2(square)
Conclusion:The sum of energies is same at all points.
Potential energy(P.E) + K.E = Constant
mgh + 1/2mv^2(square) = Constant = Mechanical energy(M.E)
Important Formulas :
Potential energy U = mgh
Kinetic energy K = 1/2 mv²
Total energy = K + U
At A
U = mgh
K = 0 since v = 0
Total energy = K + U
= mgh + 0
= mgh
At B
U = m g h
h = h - x
U = mg ( h - x )
K = 1/2 m v²
By laws of motion:
v² = 2 gh
= > v²= 2 gx
K = mgx
K + U = mg( h - x ) + mgx
= mgh - mgx + mgx
= mgh
At C
K = 1/2 m v²
= 1/2 m × 2 gh [ v² = 2 gh ]
= mgh
U = 0 since h = 0
Hence K + U = mgh + 0
= mgh
Observations :
In all cases ,
K + U = mgh
There fore the total mechanical energy remains constant.
This verifies the Law of Conservation of Energy and also proves that the energy of a body during free fall is conserved !
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