An object of mass 'm' is raised to height 'H' from the ground and released from there, prove that it's total mechanical energy remains same -
(i) At the lowest point
(ii) At height H/4 from the ground

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Solution

Dear Student ,
$W h e n t h e b o d y o f m a s s m i s r a i s e d t o h e i g h t H t h e n t h e t o t a l m e c h a n i c a l e n e r g y i s t h e s u m o f t h e k i n e t i c a n d p o t e n t i a l e n e r g y . S o a t t h i s p o i n t p o t e n t i a l e n e r g y = m g H k i n e t i c e n e r g y = 0 S o t o t a l m e c h a n i c a l e n e r g y = m g H N o w a t t h e l o w e s t p o i n t, k i n e t i c e n e r g y = m g H (a s t h e t o t a l p o t e n t i a l e n e r g y i s c o n v e r t e d i n t o k i n e t i c e n e r g y) . S o t h e t o t a l m e c h a n i c a l e n e r g y a t t h e l o w e s t p o i n t = m g H N o w a t t h e h e i g h t \frac{H}{2}, p o t e n t i a l e n e r g y = m g \frac{H}{2} a n d k i n e t i c e n e r g y = \frac{1}{2} m v^{2} = \frac{1}{2} m \times 2 g \frac{H}{2} = m g \frac{H}{2} S o t h e s u m o f t h e k i n e t i c e n e r g y a n d p o t e n t i a l e n e r g y i s t h e t o t a l m e c h a n i c a l e n e r g y = m g \frac{H}{2} + m g \frac{H}{2} = m g H S o a t h e i g h t \frac{H}{2} a n d a t t h e h e i g h e s t p o i n t a n d a t t h e l o w e s t p i n t t h e t o t a l m e c h a n i c a l e n e r g y r e m a i n s s a m e .$

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