Why is the net work of a hiker carrying a 15 kg backpack upwards 10 meters = 0 J (Giancoli)? According to the/a physics textbook by Douglas C. Giancoli, the net work done on a $15$ $kg$ backpack carried upwards $10$ m by a hiker is $0$ J since the work done by the hiker is $147$ J and the work done by gravity is $-147$ J.
But the potential energy of the backpack is increased by $147$ J , so what's going on? (to make it short)
Edit: The work is $+/-$ $1470$ J, not $+/-$ $147$ J , I'm sorry for this error. (It does not change the issue though).
 A: The keyword here is net work.
Earlier in the exercise the author computes the work (no "net") the hiker has done on the backpack, which is the expected amount depending on altitude difference and mass. That's what the hiker did.
Now for the net. The hiker can attest that gravity is a bitch: It counteracted the hiker's force on the backpack the entire time! Witness to that is that the backpack didn't accelerate (if for simplicity we consider a stretch in the middle of the hike). No net forces, no change in kinetic energy.
The work the hiker did didn't end up in the backpack, it ended up in the gravitational field, in what Newtonian physics calls "potential energy". A similar situation would arise if a worker pushed a box across a cement floor, with constant velocity. They certainly perform work, also in the physics sense; but none goes into the box: The forces on the box cancel each other out! All the mechanical work is transformed into heat.
An opposite example would be the same scenario, but without gravity: The hiker displaces a weight with a constant force F over some distance s. Because there is no gravity, we have a net force: It is exactly F, and it all goes into kinetic energy, which is simply F*s if the force pointed in the direction of travel. (Because we have a net force, the backpack also accelerates all the time, as opposed to the gravity scenario.)
A: As the backpack is carried up, the net force, work, and change in kinetic energy are all zero (as it would be if it were carried back down). The potential energy is the work that can be done by the field (alone) in moving the object from its current position to a chosen reference position. (If the pack were tossed back down, there would be a change in the kinetic energy equivalent to the work done by the hiker in taking it up.)
A: The work-energy theorem says that the total work done on a body is equal to  the change in its kinetic energy. Since the pack starts at rest and ends at rest no net work has been done on it.  The total force on the pack (The lifting force  minus the weight) was slightly upwards at the beginning to get it started moving, and slightly downwards at the top to stop it moving.
A: Gravitational potential energy is basically another name for the work done by gravity.
So you can describe the scenario as

*

*you providing energy via work that is stored as potential energy, or as

*you doing work that is balanced out by work done by gravity.

The net work, meaning the work done by all involved parties is indeed zero in your case when potential energy is considered as work. If you aren't losing any energy to heat and aren't speeding something up (the backpack is assumed moving at constant speed, I reckon) then this will have to be the case due to the energy conservation law. All energy supplied has been taken from somewhere else.
A: 
But the potential energy of the backpack is increased by 147 J, so
what's going on? (to make it short)

The positive 147 J of work done by the hiker transfers energy to the backpack. The negative -147 J of work done by gravity takes away the energy the hiker transferred to the backpack and stores it as gravitational potential energy (GPE) of the Earth-backpack system. The fact that the net work done is zero simply means that the backpack does not have kinetic energy after lifting it 10 m.
The underlying principle is the work energy theorem which states that the net work done on an object equals its change in kinetic energy. Since the backpack presumably starts at rest and ends at rest after moving up 10 m its change in KE is zero. All of the work done by the hiker ends up as GPE.
Hope this helps.
A: Would like to add that you can understand this sitaution mathematically by modelling the system of the bag and the Earth as a non-isolated one (for energy).
The reduced equation for the conservation of energy then gives: $$\Delta K + \Delta U + \Delta E_{\text{int}} = \sum T \rightarrow \Delta K + \Delta U_g = W_{\text{ext}}$$
Now assuming that the bagpack comes to rest after the hike, we must have $\Delta K = 0$, and hence, the increased potential energy (gravitational) of the bag-Earth system ($\Delta U_g = mg(\Delta y )$ > $0$) comes from the work done on the bag by the hiker which is external to the system.
If you model only the bag as the system, then there is no $U_g$ term (as there are no interacting objects) and the conservation of energy gets reduced to the familiar work - kinetic energy theorem.
Again, we find that $$W_{\text{ext, net}} = \Delta K = 0$$ which proves the assumption in the first assertion, and also gives the meaning of the bag having no net work done on it ; the bag has gained no additional kinetic energy.
Hope this helps.
A: I think you probably got your answer from the previous ones. I just want to add something.
Potential energy of a system changes only when when the internal conservative forces in a system do work. In this case the force applied by the man on the bag is not conservative but gravitational force between bag and the earth is conservative, thus this causes change in the potential energy of the bag. 
$$\Delta U = -W_c = -\int\vec F_c.d\vec s$$
