Question

Show that mechanical energy of a freely falling
body justifies the Law of Conservation of
Mechanical Energy.​

Answer 1

Answer:

at initial position when body is at height 'h' above the ground

PE= mgh KE=0 (at rest)

it comes down by x (free fall)

PE= mg(h-x) KE= 1/2mv² [v²—u²=2gx. u=0 => u²=0 so v²= 2gx]

KE=1/2mgx so PE+KE = mgh-mgx+mgx=mgh

on the point just before the ground

KE=mgh PE= 0 so KE+PE=mgh

so total mechanical energy remains constant