Answer
The mark on the ruler at the lowest position will be 74 cm.
Work Step by Step
The gravitational potential energy of the mass at the 15-cm mark will equal the potential energy in the spring at the lowest position.
Let y be the distance the spring stretches at the lowest position. Then,
$\frac{1}{2}ky^2 = mgy$
$y = \frac{2mg}{k}$
$y = \frac{(2)(2.5 ~kg)(9.8 ~m/s^2)}{83 ~N/m}$
$y = 0.59 ~m$
The mark on the ruler at the lowest position will be 15 cm + 59 cm which is 74 cm.
