To find the work done by the spring, we can use the formula for the elastic potential energy stored in a spring, which is also the work done by the spring as the mass moves back to its initial position:

$W = \frac{1}{2} k x^2$

Where:

• $W$ is the work done by the spring (in joules, J),
• $k$ is the spring constant (in newtons per meter, N/m),
• $x$ is the displacement from the equilibrium position (in meters, m).

Given:

• $k = 10 \, \text{N/m}$,
• $x = 20 \, \text{cm} = 0.20 \, \text{m}$.

Now, plugging these values into the formula:

$W = \frac{1}{2} (10) (0.20)^2$

Let me calculate this.The work done by the spring is approximately $0.20 \, \text{J}$.

Would you like further clarification or details on this? Here are five related questions to expand on this concept:

1. How does the spring constant affect the work done by the spring?
2. What would happen if the spring were compressed instead of stretched?
3. How would the work change if the spring were stretched twice as far?
4. What is the relationship between potential energy and kinetic energy in this system?
5. How can we calculate the velocity of the mass as it returns to the equilibrium point?

Tip: Remember that work done by a spring is proportional to the square of the displacement, so even small changes in displacement can significantly affect the work.