How energy conservation is not violated here?

If I lift a book of mass $m$ from the ground and put it at height $h$,then initial and final energies with respect to ground are 0 and $mgh$ respectively.

So how energy is conserved here?

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don't forget that you yourself performed work when lifting up the book, your "internal energies" decreased. – Zoltan Zimboras Jan 26 '14 at 7:00
internal energy is stored from the food we eat to be expended as we live and move. – anna v Jan 26 '14 at 7:10

Repeat lifting books to height h several times per minute for an hour or two, and observe who gets tired.

You do work lifting the book, burning up biological energy stored in ATP molecules in your cells, to activate muscles, which apply some force upward on the book (as well as suffer some internal friction). The biological energy used up is converted into: a lifted book, extra body heat, sweat, and groans of displeasure because such smart people as physicists ought to have butlers to take care of book-placing.

All the chemical energy of food you ate and oxygen breathed, and all those energetic outputs mentioned, should add up perfectly. Physiologists do lab experiments to measure these things with athletes, average people, decrepit oldsters, kids, etc.

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you give your muscular energy to book, so when it is in air at $h$ height it is actually having energy $mgh$ given out by your muscles and if you drop that book, the potential energy inside that book will gets converted to kinetic energy slowly and will gain speed. Then the moment before when it reaches ground it has all the energy converted to kinetic energy and after that moment the energy gets transferred to earth. Hence conservation of energy is not violated.

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