Answer:
Spring constant: [tex]\rm 447\; N \cdot m^{-1}[/tex].
Work done: [tex]\rm 0.0438\; J[/tex].
Explanation:
Convert all values to SI units.
Assume that the spring-mass system is vertical and is placed on the surface of the earth. The gravitational acceleration constant will be equal to [tex]\rm 9.81\; N\cdot kg^{-1}[/tex].
Gravitational pull on the weight:
[tex]W = m\cdot g = \rm 0.638\; kg \times 9.81\; N\cdot kg^{-1} = 6.25878\; N[/tex].
That's also the size of the force on the spring. [tex]F = \rm 6.25878\; N[/tex].
The spring constant is the size of the force required to deshape the spring (by stretching, in this case) by unit length.
[tex]\displaystyle k_{\text{spring}} = \frac{F}{\Delta x} = \rm \frac{6.25878\; N}{0.014\; m} = 447.056\; N\cdot m^{-1}[/tex].
Assume that there's no energy loss in this process. The work done on the spring is the same as the elastic potential energy that it gains:
[tex]\begin{aligned} &\text{EPE} \\=& \frac{1}{2}k\cdot x^{2} \\=&\rm \frac{1}{2} \times 447.056\; N\cdot m^{-1} \times (0.014\; m)^{2}\\ =& \rm 0.0438\; J\end{aligned}[/tex].
