Show that the total mechanical energy of a body falling freely under gravity 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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