15-28.

In general we have the energy relationship

$$E=K+U$$

where $K$ is the kinetic, $U$ the potential, and $E$ the total energy. Kinetic energy is always non-negative ($K \propto v^2$), and for a spring so too is the potential ($U \propto x^2$).

Thus, if we set $K=0$ we get

$$E=U_{max}=\frac{1}{2}kx_{max}^2=\frac{1}{2}kA^2,$$

while if we set $U=0$ we get

$$E=K_{max}=\frac{1}{2}mv_{max}^2.$$

Solving the first eqation for $k$ yeilds

$$k=\frac{2E}{A^2}=200\;\textrm{N/m},$$

and solving the second for $m$ yields

$$m=\frac{2E}{v_{max}^2}=1.39\;\textrm{kg}.$$