We are tasked with finding the spring constant $k$ using the equation:

$k = \frac{W}{\Delta L}$

Where:

• $W$ is the weight of the attached mass (force due to gravity), given by $W = mg$ where $m$ is the mass and $g$ is the acceleration due to gravity.
• $\Delta L$ is the extension of the spring (the difference between the stretched and the unstretched length).

Step 1: Calculate the weight $W$

The mass of the object is $m = 0.80 \, \text{kg}$, and the acceleration due to gravity is $g = 9.8 \, \text{m/s}^2$.

$W = mg = (0.80 \, \text{kg}) (9.8 \, \text{m/s}^2) = 7.84 \, \text{N}$

Step 2: Calculate the extension $\Delta L$

The unstretched length of the spring is $0.60 \, \text{m}$, and the stretched length is $0.70 \, \text{m}$.

$\Delta L = 0.70 \, \text{m} - 0.60 \, \text{m} = 0.10 \, \text{m}$

Step 3: Solve for the spring constant $k$

Now we can plug in the values for $W$ and $\Delta L$ into the equation for $k$:

$k = \frac{W}{\Delta L} = \frac{7.84 \, \text{N}}{0.10 \, \text{m}} = 78.4 \, \text{N/m}$

Thus, the spring constant is:

$k = 78.4 \, \text{N/m}$

The correct answer is: $\boxed{B. 78.4 \, \text{N/m}}$

Would you like further explanation or clarification on any step?

Here are 5 related questions:

1. How would the spring constant change if the mass were doubled?
2. What would happen to the spring's extension if the spring constant were halved?
3. How does the spring constant relate to the stiffness of the spring?
4. What is Hooke's Law and how does it apply to this scenario?
5. Can you find the potential energy stored in the spring when it is stretched?

Tip: Always check that the units of force (N) and extension (m) are consistent before calculating the spring constant!