how to verify the law of conservation of energy
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Here is ur answer :
LAW OF CONSERVATION OF ENERGY :
According to law of conservation of energy, energy can only be transferred from one form to another, it can neither be created nor be destroyed. the total energy before and after transformation, always remain constant.
CONSERVATION OF ENERGY DURING THE FREE FALL OF A BODY :
Consider a body of mass m, lying at position B. It is made to fall freely from a height (h) above the ground as shown in the figure.
At point B
At the start, the potential energy is mgh and kinetic energy is zero (as its velocity is zero).
i. e. PE = mgh
KE = 0
Therefore,
Total energy, TE = PE + KE = mgh
At point A
As it falls, its potential energy will change into kinetic energy. if v is the velocity of the object at a given instant, its kinetic energy would be
1/2 mv²
PE = mg ( h - x )
From Newton's Third equation of motion,
v² = u² + 2gx
v² = 2gx ( because u = 0 )
KE = 1/2 mv²
= 1/2 * m * 2gx ( because v² = 2gx)
= mgx
Therefore,
Total energy = mg( h - x) + mgx
= mgh - mgx + mgx
= mgh
At point C
As The Fall of the object continuous, the potential energy would decrease while the kinetic energy would increase. When the object is about to reach the ground, h = 0 and v will be the highest point.
PE = 0
KE = 1/2 mv²
= 1/2m ( 2gh) {because v² = 2gh}
= mgh
Therefore,
Total energy, TE = mgh
Thus, the sum of the potential energy and kinetic energy of the object would be same at all points, that is
PE + KE = Constant
Or mgh + 1/2 mv² = constant
This verifies the law of conservation of energy.
HOPE IT HELPS YOU..
:-)
Here is ur answer :
LAW OF CONSERVATION OF ENERGY :
According to law of conservation of energy, energy can only be transferred from one form to another, it can neither be created nor be destroyed. the total energy before and after transformation, always remain constant.
CONSERVATION OF ENERGY DURING THE FREE FALL OF A BODY :
Consider a body of mass m, lying at position B. It is made to fall freely from a height (h) above the ground as shown in the figure.
At point B
At the start, the potential energy is mgh and kinetic energy is zero (as its velocity is zero).
i. e. PE = mgh
KE = 0
Therefore,
Total energy, TE = PE + KE = mgh
At point A
As it falls, its potential energy will change into kinetic energy. if v is the velocity of the object at a given instant, its kinetic energy would be
1/2 mv²
PE = mg ( h - x )
From Newton's Third equation of motion,
v² = u² + 2gx
v² = 2gx ( because u = 0 )
KE = 1/2 mv²
= 1/2 * m * 2gx ( because v² = 2gx)
= mgx
Therefore,
Total energy = mg( h - x) + mgx
= mgh - mgx + mgx
= mgh
At point C
As The Fall of the object continuous, the potential energy would decrease while the kinetic energy would increase. When the object is about to reach the ground, h = 0 and v will be the highest point.
PE = 0
KE = 1/2 mv²
= 1/2m ( 2gh) {because v² = 2gh}
= mgh
Therefore,
Total energy, TE = mgh
Thus, the sum of the potential energy and kinetic energy of the object would be same at all points, that is
PE + KE = Constant
Or mgh + 1/2 mv² = constant
This verifies the law of conservation of energy.
HOPE IT HELPS YOU..
:-)
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