To determine the stretch of the spring when a 200 g mass is hung from it, we can use Hooke's Law:

$F = k \cdot x$

where:

• $F$ is the force exerted by the mass,
• $k$ is the spring constant (185 N/m),
• $x$ is the extension of the spring from its equilibrium length.

1. Convert the mass to kilograms: $200 \, \text{g} = 0.2 \, \text{kg}$.

2. Calculate the gravitational force exerted by the mass: $F = m \cdot g = 0.2 \, \text{kg} \times 9.8 \, \text{m/s}^2 = 1.96 \, \text{N}$

3. Calculate the stretch $x$: $x = \frac{F}{k} = \frac{1.96 \, \text{N}}{185 \, \text{N/m}} \approx 0.0106 \, \text{m} = 1.06 \, \text{cm}$

So, the spring stretches approximately 1.06 cm from its equilibrium length.

Would you like a further breakdown or have any questions?

Here are some related questions for further understanding:

1. How would the stretch change if the spring constant were higher?
2. What happens to the stretch if the mass is doubled?
3. How does the gravitational force affect the extension of the spring?
4. Can Hooke's Law be applied indefinitely, or does it have limitations?
5. What other factors might affect the accuracy of this calculation?

Tip: Remember, Hooke's Law applies only within the elastic limit of the spring; if the force exceeds this limit, the spring may deform permanently.