Please give answer of c
Dear student,
If work done by external forces acting on a body is zero and all the internal forces are consider,then total mechanical energy of body remains conserved.
ui + Ki = uf +Kt
At point A,KEi = 0
ui=mgH
At point B,ut = 0
Ktf = 1/2 mv2
putting the value of V
K= 1/2m(2gh)
kEt = mgH ( TE = Kf + ut = mgH)
V2 = u2 + 2gH
V2 = 0 + 2gH
V = root over of 2gH
So,the initial potential energy was mgH and final kinetic energy also comes out to be mgH.
Hence,it is proved that the kinetic energy of freely falling body on reaching the ground is nothing but transformation of its initial potential energy.
Regards,
If work done by external forces acting on a body is zero and all the internal forces are consider,then total mechanical energy of body remains conserved.
ui + Ki = uf +Kt
At point A,KEi = 0
ui=mgH
At point B,ut = 0
Ktf = 1/2 mv2
putting the value of V
K= 1/2m(2gh)
kEt = mgH ( TE = Kf + ut = mgH)
V2 = u2 + 2gH
V2 = 0 + 2gH
V = root over of 2gH
So,the initial potential energy was mgH and final kinetic energy also comes out to be mgH.
Hence,it is proved that the kinetic energy of freely falling body on reaching the ground is nothing but transformation of its initial potential energy.
Regards,