An ideal spring stretched 0.7 meters with a spring constant of  k= 50 N/m. What is the applied force?

The applied force on the spring can be calculated using Hooke's Law, which states:

\[
F = k \cdot x
\]

Where:
- \( F \) is the applied force,
- \( k \) is the spring constant, and
- \( x \) is the displacement (stretch) of the spring.

Given:
- \( k = 50 \, \text{N/m} \)
- \( x = 0.7 \, \text{m} \)

Substituting the values into the formula:

\[
F = 50 \, \text{N/m} \times 0.7 \, \text{m} = 35 \, \text{N}
\]

So, the applied force is **35 Newtons**.

Would you like more details or have any questions? 

Here are some related questions you might find useful:
1. What is Hooke's Law, and how is it derived?
2. How does the spring constant affect the force applied to the spring?
3. What happens if the spring is stretched beyond its elastic limit?
4. How would the applied force change if the spring were compressed instead of stretched?
5. Can this spring constant be used to determine the potential energy stored in the spring?

**Tip:** When dealing with springs, always ensure the displacement is within the elastic limit to accurately apply Hooke's Law.

Learn how to calculate the applied force on a spring stretched 0.7 meters with a spring constant of 50 N/m using Hooke's Law. This problem covers the fundamental concept of Hooke's Law and its application in physics. Suitable for advanced high school students.

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