A block is fastened to a horizontal spring. The block is pulled to a distance $$x = 10$$ cm from its equilibrium position (at $$x = 0$$) on a frictionless surface from rest. The energy of the block at $$x = 5$$ cm is 0.25 J. The spring constant of the spring is ______ $$\text{N m}^{-1}$$.

A block is fastened to a horizontal spring on a frictionless surface. It is pulled to $$x = 10$$ cm from equilibrium and released from rest. The energy of the block at $$x = 5$$ cm is 0.25 J, and we need to find the spring constant $$k$$.

At the initial position $$x = A = 10$$ cm = 0.1 m, the block is at rest, so all the mechanical energy is stored as spring potential energy. Since $$E_{total} = \frac{1}{2}kA^2 = \frac{1}{2}k(0.1)^2 = 0.005k\,, $$ this expression represents the total energy of the system.

At $$x = 5$$ cm = 0.05 m, the potential energy in the spring is $$PE = \frac{1}{2}k(0.05)^2 = 0.00125k\,. $$ This gives the kinetic energy of the block at that point as $$KE = E_{total} - PE = 0.005k - 0.00125k = 0.00375k\,. $$

Given that the kinetic energy at $$x = 5$$ cm is 0.25 J, substituting into $$0.00375k = 0.25$$ yields $$k = \frac{0.25}{0.00375} = \frac{250}{3.75} = \frac{200}{3} \approx 66.67\text{ N/m}\,. $$

To verify, with $$k = \frac{200}{3}\text{ N/m}$$ the total energy is $$\frac{1}{2}\times\frac{200}{3}\times0.01 = \frac{1}{3}\text{ J}\approx0.333\text{ J}\,, $$ the potential energy at 5 cm is $$\frac{1}{2}\times\frac{200}{3}\times0.0025 = \frac{1}{12}\text{ J}\approx0.083\text{ J}\,, $$ and therefore the kinetic energy at 5 cm is $$\frac{1}{3} - \frac{1}{12} = \frac{3}{12} = 0.25\text{ J}\,, $$ which matches the given value.

The spring constant is $$k = \frac{200}{3} \approx 67\text{ N/m}\,. $$ The answer is $$\boxed{67}$$.