A body is falling freely under the action of gravity alone in vacuum. Which of the following quantities remain constant during the fall?
A
Kinetic energy
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B
Potential energy
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C
Total mechanical energy
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D
Total linear momentum
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Solution
The correct option is C Total mechanical energy
As the body is falling freely under gravity, the potential energy decreases continuously and kinetic energy increases continuously as all the conservative forces are doing work. So, total mechanical energy (PE+KE) of the body will be constant.
At point A total mechanical energy will be EA=K.E+P.E
EA=12mv2+mgH
As velocity will be zero at A, so its kinetic energy will be zero
EA=MgH
Velocity at point B will be
VB=√2gh
So energy at point B will be
EB=KE+PE
EB=12m(2gh)+mg(H−h)
EB=mgh+mgH−mgh
EB=mgH
Now velocity at point C will be
VC=√2gh
So energy at point C will be
EC=KE+PE
EC=12m(2gH)+mg(0)
So, total mechanical energy will remain same (if we neglect the air friction).