Question

Show that for freely falling body the sum of its kinetic energy and potential energy remains constant at all points during its fall ?

Answer 1

At A,velocity $=0=u$

$KE=\frac{1}{2}m{v}^2=0$

$PE=mgh,TE=PE+KE=mgh$

At B,velocity $v:$

$v^2=u^2+2as$

$v^2=2gx$

$KE=\frac{1}{2}m{v}^2=mgx$

$PE=mg(h-x)$

Total energy $TE=PE+KE$

$=mg(h-x)+mgx=mgh$

At C,velocity $v:$

$v^2=u^2+2as$

$v^2=2gh$

$KE=\frac{1}{2}m{v}^2=mgh$

$PE=0$

$TE=PE+KE=mgh$