A force of 60N is required to keep a spring stretched 5m from the equilibrium position. How much work in Joules is done to stretch the spring 9m from equilibrium? Round your answer to the nearest hundredth if necessary.

To find the work done in stretching the spring, we can use Hooke's law and the work formula for springs.

First, let's identify the spring constant \( k \). According to Hooke's law, the force \( F \) needed to stretch or compress a spring by a distance \( x \) from its equilibrium position is given by:
\[ F = kx \]

Given:
\[ F = 60 \text{ N} \]
\[ x = 5 \text{ m} \]

We can solve for \( k \):
\[ k = \frac{F}{x} = \frac{60 \text{ N}}{5 \text{ m}} = 12 \text{ N/m} \]

Next, we need to find the work \( W \) done to stretch the spring by 9 meters. The work done on a spring is given by the formula:
\[ W = \frac{1}{2} k x^2 \]

Here, \( x = 9 \text{ m} \) and \( k = 12 \text{ N/m} \):
\[ W = \frac{1}{2} \times 12 \text{ N/m} \times (9 \text{ m})^2 \]
\[ W = \frac{1}{2} \times 12 \times 81 \]
\[ W = 6 \times 81 \]
\[ W = 486 \text{ J} \]

Therefore, the work done to stretch the spring 9 meters from the equilibrium position is \( 486 \text{ Joules} \).

Would you like more details or have any questions about this solution?

Here are 5 numbered questions you might ask:
1. How do you calculate the spring constant \( k \) from given force and displacement?
2. What is Hooke's law and how is it applied?
3. How is the work formula for a spring derived?
4. What are the units of the spring constant \( k \)?
5. How would the work done change if the spring constant were different?

**Tip:** When dealing with springs, always remember that the work done is proportional to the square of the displacement, not just the displacement.

Learn how to calculate the work done by a spring when a 60N force stretches it 5m from equilibrium. This problem involves applying Hooke's Law and the work formula for springs. Suitable for advanced high school students.

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