https://file.aiquickdraw.com/mathbot/file/17278804195920zm9xgpz.jpg

To solve this problem, we can use **Hooke's Law**, which is stated as:

\[
F = kx
\]

Where:
- \( F \) is the force applied (in Newtons),
- \( k \) is the spring constant (in Newtons per meter, N/m),
- \( x \) is the extension or compression of the spring (in meters).

### Step 1: Calculate the force applied by the weight
The force exerted by the mass is due to gravity and is given by:

\[
F = mg
\]

Where:
- \( m = 0.5 \, \text{kg} \) (mass),
- \( g = 9.8 \, \text{m/s}^2 \) (acceleration due to gravity).

Substitute the values:

\[
F = 0.5 \times 9.8 = 4.9 \, \text{N}
\]

### Step 2: Use Hooke's Law to find the spring constant \( k \)
We know the extension of the spring \( x = 0.02 \, \text{m} \), so we can solve for \( k \):

\[
k = \frac{F}{x}
\]

Substitute the values:

\[
k = \frac{4.9}{0.02} = 245 \, \text{N/m}
\]

### Final Answer:
The spring constant \( k \) is **245 N/m**.

Would you like more details or have any questions?

Here are 5 follow-up questions for you to explore:
1. How does the spring constant change if the mass is doubled?
2. What happens to the extension if a different spring with a lower spring constant is used?
3. How would you find the spring constant if the spring's mass was not negligible?
4. What is the potential energy stored in the spring at this extension?
5. How would air resistance affect the spring’s extension in real-world conditions?

**Tip:** Always remember that Hooke's Law is valid only within the elastic limit of the spring.

A mass of 0.5 kg hung vertically from an ideal spring stretches the spring 0.02 meters. Find the value of the spring constant (k).

This problem involves calculating the spring constant using Hooke's Law. A mass of 0.5 kg stretches the spring by 0.02 meters, and the spring constant k is found using F = kx and F = mg. Ideal for students in grades 9-12.

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