show that a energy of a freelly falling body is. conserved
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For freely falling body, V^2-U^2= 2gh, where V is final speed and U is initial speed.
At the top all energy is potential energy in form of m*g*h.
So U=0 and at any instant V^2=2gh. Where h is distance from top
hence at any height h=(V^2)/2g.
Thus mgh= mg(V^2)/2g = (1/2)(mV^2) which is the kinetic energy
At the top all energy is potential energy in form of m*g*h.
So U=0 and at any instant V^2=2gh. Where h is distance from top
hence at any height h=(V^2)/2g.
Thus mgh= mg(V^2)/2g = (1/2)(mV^2) which is the kinetic energy
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the energy of freely falling body is conserved only when body is ready to fall then it have highest potential energy but when it fall then it's kinetic energy is increasing. now suppose total p.e is 50 then it's k.e is 0 it's p.e is decreasing 10 and k.e is increasing 10 means p.e is 40 and k.e is 10 and so on... so here energy is conserved in equal manner.
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