To find the maximum force exerted by the spring on the block, we can use Hooke's law and the properties of simple harmonic motion.

First, let's find the angular frequency $$\omega$$:

$$\omega = \frac{2\pi}{T}$$

where $$T = 3.14\,\mathrm{s}$$ is the time period.

$$\omega = \frac{2\pi}{3.14} \approx 2\,\mathrm{rad/s}$$

Now, let's find the maximum velocity $$v_{max}$$ of the block:

$$v_{max} = \omega A$$

where $$A = 1\,\mathrm{m}$$ is the amplitude.

$$v_{max} = 2\,\mathrm{rad/s} \times 1\,\mathrm{m} = 2\,\mathrm{m/s}$$

Next, we can find the spring constant $$k$$ using the mass of the block $$m = 5\,\mathrm{kg}$$ and the angular frequency $$\omega$$:

$$\omega^2 = \frac{k}{m} \Rightarrow k = m\omega^2$$

$$k = 5\,\mathrm{kg} \times (2\,\mathrm{rad/s})^2 = 20\,\mathrm{N/m}$$

Finally, we can find the maximum force exerted by the spring on the block. At maximum displacement, the force is given by Hooke's law:

$$F_{max} = kA$$

$$F_{max} = 20\,\mathrm{N/m} \times 1\,\mathrm{m} = 20\,\mathrm{N}$$

The maximum force exerted by the spring on the block is $$20\,\mathrm{N}$$.