Problem with solving for force constant k of spring?

[David Day]

The question is stated as the following:

When a 3.60 kg object is placed on top of a vertical spring, the spring compresses a distance of 2.83 cm. What is the force constant of the spring?

The correct answer was acquired by using the equation F = mg = -kx, where k is the spring constant and x is the displacement from equillibrium. Using this equation, k is solved as 1,246.6 N/m.

However, in my first attempt at solving, I tried a different method by calculating the amount of work done on the spring and setting that equal to the equation for the elastic potential energy of a spring, U = .5kx². Calculating the work F*(y_f - y_i), W = mg*(.0283) = .998 J.
Therefore, .998 = .5kx². But solving with this method gives k = 2,493.2 N/m, which is not only different from the first result but is exactly double the amount.

What is the reason for this? Why would using this second method not give the correct result for k? Any help/explanation would be greatly appreciated.

 

[Orodruin]

Energy is not conserved. In order to come to rest, the energy must dissipate out of the system. Otherwise the mass will oscillate around the equilibrium. What you have computed is the second turning point of such oscillations, which naturally has the spring twice as compressed as the equilibrium.

 

[Chestermiller]

The elastic force F is not constant while you are stretching the spring. It is only equal to mg at the very end. Part of $mg\Delta y$ goes into kinetic energy,

 

[haruspex]

When a 3.60 kg object is placed on top of a vertical spring, the spring compresses a distance of 2.83 cm

That is somewhat open to interpretation. If it is placed on top of the relaxed spring and let go, and the compression distance quoted refers to the maximum distance, then your energy-based method and answer are correct.
But more likely it is intended that the object is only let go gradually, allowing the spring to compress to the equilibrium position without leaving the object any KE.