Work done on object being carried upwards If you carry a book in your hands, and you walk up stairs with a change in height of $h$, the net work on both you and the book would be $-M_{\mathrm{total}}gh$ since $W = - \Delta U$. This would be due to gravity.
However, when considering the book alone, the work done by the normal force, i.e. your hands, would be $M_{\mathrm{book}}gh$. Furthermore, the work done by gravity solely on the book would be $-M_{\mathrm{book}}gh$. This means that the net work done on the book through the process of walking up the stairs is $0$. Since work is equal to negative change in gravitational potential energy this means that the change in GPE of the book is $0$? But then doesn't the book have a change in gravitational potential energy of  $M_{\mathrm{book}}gh$?
Am I missing something regarding the kinetic energy of the book?
 A: 
This means that the net work done on the book through the process of
walking up the stairs is 0.

That is correct, because gravity does negative work of $-m_{book}gh$ equal to the positive work done by you of $+m_{book}gh$ for a net work of zero. All that means, per the work energy theorem, is the change in kinetic energy of the book is zero (it starts at rest and ends at rest). It doesn't mean there is no change in potential energy of the book.

Since work is equal to negative change in gravitational potential
energy this means that the change in GPE of the book is 0??

That is not correct. The energy you transferred to the book in doing positive work is not lost. Gravity, in doing negative work, takes the energy you supplied to the book and stores it as gravitational potential energy of the book-earth system.
Hope this helps.
A: Yes , you are . Net work done on an object is equal to change in kinetic energy and not in potential energy.
$\therefore$$W_g+W_n=0$ since the book is rest from your frame .I am assuming you hold the book still during your climb.
$W_g=-W_n$
