illustrate the law of conservation of energy by discussing the free fall of a body
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definition of free fall is the movement of an object under the force of gravity alone is called as free fall
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LAW OF CONSERVATION OF ENERGY:
Energy can neither be created nor destroyed, but it
is transformed from one form to another. Alternatively,
whenever energy gets transformed, the total energy
remains unchanged.
Proof – In case of a freely falling body
It may be shown that in the absence of external frictional force the total mechanical energy of a body remains constant.
Let a body of mass m falls from a point A, which is at a height h from the ground as shown in fig.
At A,
Kinetic energy kE = 0
Potential energy Ep = mgh
Total energy E = Ep + Ek = mgh + 0= mgh
During the fall, the body is at a position B. The body has moved a distance x from A.At B,velocity v2 = u2 + 2asapplying, v2 = 0 + 2ax = 2axKinetic energy Ek = 1/2 mv2 = 1/2 m x 2gx = mgx
Potential energy Ep = mg (h – x)
Total energy E = Ep + Ek = mg (h-x) + mgx = mgh – mgx + mgx= mgh
If the body reaches the position C.
At C,
Potential energy Ep = 0
Velocity of the body C is
v2 = u2 + 2as
u = 0, a = g, s = h
applying v2 = 0 + 2gh = 2gh
kinetic energy Ek =1/2 mv2=1/2 m x 2gh= mgh
Total energy at C
E = Ep + Ek
E = 0 + mgh
E = mgh
Thus we have seen that sum of potential and kinetic energy of freely falling body at all points remains same. Under the force of gravity, the mechanical energy of a body remains constant.
Energy can neither be created nor destroyed, but it
is transformed from one form to another. Alternatively,
whenever energy gets transformed, the total energy
remains unchanged.
Proof – In case of a freely falling body
It may be shown that in the absence of external frictional force the total mechanical energy of a body remains constant.
Let a body of mass m falls from a point A, which is at a height h from the ground as shown in fig.
At A,
Kinetic energy kE = 0
Potential energy Ep = mgh
Total energy E = Ep + Ek = mgh + 0= mgh
During the fall, the body is at a position B. The body has moved a distance x from A.At B,velocity v2 = u2 + 2asapplying, v2 = 0 + 2ax = 2axKinetic energy Ek = 1/2 mv2 = 1/2 m x 2gx = mgx
Potential energy Ep = mg (h – x)
Total energy E = Ep + Ek = mg (h-x) + mgx = mgh – mgx + mgx= mgh
If the body reaches the position C.
At C,
Potential energy Ep = 0
Velocity of the body C is
v2 = u2 + 2as
u = 0, a = g, s = h
applying v2 = 0 + 2gh = 2gh
kinetic energy Ek =1/2 mv2=1/2 m x 2gh= mgh
Total energy at C
E = Ep + Ek
E = 0 + mgh
E = mgh
Thus we have seen that sum of potential and kinetic energy of freely falling body at all points remains same. Under the force of gravity, the mechanical energy of a body remains constant.
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