Calculate the spring constant using Hooke's Law and explain impulse using Newton's Second Law.

The extension is the difference between the extended length and the natural length.

$$e = L - L_0$$ $$e = 3.5 \text{ cm} - 2 \text{ cm}$$ $$e = 1.5 \text{ cm}$$

Convert the extension to meters:

$$e = 1.5 \text{ cm} \times \frac{1 \text{ m}}{100 \text{ cm}} = 0.015 \text{ m}$$

Hooke's Law states $F = ke$. Rearranging for $k$:

$$k = \frac{F}{e}$$ $$k = \frac{5 \text{ N}}{0.015 \text{ m}}$$ $$k = 333.33 \text{ N/m}$$

The spring constant is $\boxed{\text{333.33 N/m}}$.

Newton's Second Law states that the net force ($F$) acting on an object is equal to the rate of change of its momentum.

$$F = \frac{\Delta p}{\Delta t}$$

where $\Delta t$ is the time interval over which the force acts.

From Newton's Second Law, we can rearrange the equation to solve for the change in momentum:

$$\Delta p = F \Delta t$$

Since impulse $I = \Delta p$, we can substitute to show:

$$I = F \Delta t$$

This demonstrates that the impulse on an object can be expressed as the product of the force applied and the time interval over which it acts.