CBSE Class 11-science Answered

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

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

E_H = ( 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 E_h at the poitn that is at a height h above ground levele is

E_h = (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

E_h = [ 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

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

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

Hence eqn.(4) is rewritten as

E_o = (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 .