Answer
2 cm. 4 cm each for a total of 8 cm.
Work Step by Step
Hooke’s law says that the stretch is proportional to the applied force, $F \sim \Delta x$. In Case a, each spring supports only 5 N, half the weight. Each spring stretches half as far as before, or 2 cm.
In Case b, each spring still supports 10 N. Each spring stretches 4 cm, so the weight is lowered by a total distance of 8 cm.
This is discussed on page 231.
