A rectangular block of mass 5 kg attached to a horizontal spiral spring executes simple harmonic motion of amplitude 1 m and time period 3.14 s. The maximum force exerted by spring on block is _______ N.

We need to find the maximum force exerted by the spring on a block executing simple harmonic motion (SHM). Next, we note that the mass is $$m = 5$$ kg, the amplitude is $$A = 1$$ m, and the time period is $$T = 3.14$$ s.

We begin by finding the angular frequency $$\omega$$. The angular frequency is related to the time period by:
$$\omega = \frac{2\pi}{T} = \frac{2\pi}{3.14} = \frac{2 \times 3.14}{3.14} = 2 \text{ rad/s}$$

Next, we find the spring constant $$k$$. For a mass-spring system executing SHM, the angular frequency is related to the spring constant and mass by:
$$\omega = \sqrt{\frac{k}{m}}$$
Squaring both sides: $$\omega^2 = \frac{k}{m}$$, so:
$$k = m\omega^2 = 5 \times (2)^2 = 5 \times 4 = 20 \text{ N/m}$$

Then, by Hooke's Law, the force exerted by the spring is $$F = kx$$, where $$x$$ is the displacement from the equilibrium position. The maximum force occurs at the maximum displacement, which is the amplitude $$A$$:
$$F_{max} = kA = 20 \times 1 = 20 \text{ N}$$

Therefore, the maximum force exerted by the spring on the block is 20 N.