Question

Prove law of conservation of energy for a stone moving vertically down ( explain energy at B & C )

Answer 1

As per given condition of point B we can see that height at point B is "h/2" from the ground

So we know that potential energy is given as

U = mgh

so here we have to put height h = h/2

so potential energy is U = mgh/2

now for kinetic energy we need to find the speed of it after falling the distance h/2

now by kinematics we will have

$v_f^2 - 0^2 = 2(g)(h/2)$

now for kinetic energy

$KE = \frac{1}{2}mv^2 = \frac{1}{2}m(gh)$

$KE = 1/2mgh$

now total energy will be given as

$E = mgh/2 + mgh/2 = mgh$

now for point C we can say that it is the point near to ground

So here height is ZERO

now potential energy will also be zero

U = 0

now for kinetic energy we need to find speed

$v^2 - 0^2 = 2(g)(h)$

now kinetic energy

$KE = \frac{1}{2}mv^2 = \frac{1}{2}m(2gh)$

$KE = mgh$

now again we have total energy

$E = 0 + mgh = mgh$