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state and prove law of conservation of energy in case of a freely falling bady
Asked by dhanapolla | 28 Jan, 2024, 10:40: AM

Let us consider an object initially at rest at a height H above ground.

Total mechanical energy EH at the height H is only potential energy that is given as

EH = ( m g H ) ...................... (1)

where m is mass and g is acceleration due to gravity.

Let the object is released from the height H . When it is reaches a point that is at a

height h above ground level, the object has kinetic energy and potential energy.

Total mechanical energy Eh at the poitn that is at a height h above ground levele is

Eh = (1/2) m v2 + ( m g h ) .....................(2)

First term in above expression is kin etic energy and second term is potential energy.

Velocity of particle is considered as v .

From equation of motion , velocit v is determined as " v2 = u2 + 2 g s " , where u = 0 is

initial velocity i.e. zero because the object starts from rest and s = (H-h) is the distance travelled.

Hence kinetic energy = (1/2) m v2 = (1/2) m × [ 2 g (H-h) ] = [ m g (H-h) ]

Hence eqn.(2) is rewritten as

Eh = [ m g (H-h) ] + ( m g h ) = m g H  ..............................(3)

Let V be the velocity of object just before reaching the ground.

Potential energy of object is zero just before reaching ground.

Hence total mechanical energy of object just before reaching the ground is

Eo = (1/2) m V2 ...........................(4)

Velocity V determined from equaion of motion is ,  V2 = ( 2 g H )

Hence eqn.(4) is rewritten as

Eo = (1/2) m ( 2 g H ) = m g H  ............................(5)

From eqn.(1) , (3) and (5) , we see that total mechanical energy remainis constant.

Hence , it can be concluded that energy is conserved .

Answered by Thiyagarajan K | 28 Jan, 2024, 05:06: PM

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