I think the negative/positive stuff is referring to the voltage across the membrane. Anyway, there is a high concentration of K inside the cell and a high concentration of Na outside the cell. The pump moves Na and K into areas of higher concentration (across their conc. gradient). If the pump were to stop working, there would be no movement of these ions, therefore there will still be a high concentration of K inside the cell and a high conc. of Na outside the cell. According to the answer, water enters the cell because of an increase of Na ions inside the cell. What increase? Na concentration was lower inside the cell before and after the pump was inhibited.
If water enters the cell via osmosis, then that implies that there is a greater concentration of total solutes inside the cell as it stands "inactive" with respect to the pump. The only explanation I could come up with besides this assumption, that incorporates the excess Na inside is that there are channel proteins or co-transporters that allow the Na to enter the cell along its concentration gradient bringing in whatever else that causes the cell to lyse.
I think I have another idea. When the pump is working, there is a net movement of Ions out of the cell (3 out vs 2 in). This constant movement of ions out the cell keeps the total concentration of ions outside of the cell higher (even though K ions are in higher conc. inside the cell), which in turn, keeps the water from entering the cell. If the pump were to stop working, this net movement of ions out the cell is lost...but the hiccup here for me is that it's still assuming that the total concentration of solutes is higher inside the cell (the cell is hypertonic to the exterior). This is true naturally isn't it? I have a thing for keeping all of my reasoning limited to the info that is provided in the question.
