A mass is attached to a spring with spring constant *k* = 10 N/m. The spring is stretched to 30 cm past its resting position. How much work (in J) does the spring do when the object is released and the mass travels back to its initial position?

The work done by a spring is related to the change in potential energy stored in the spring. The formula for the potential energy of a spring is:

\[
PE_{\text{spring}} = \frac{1}{2} k x^2
\]

Where:
- \(k\) is the spring constant in N/m,
- \(x\) is the displacement from the equilibrium position in meters.

### Given:
- \(k = 10 \, \text{N/m}\),
- The spring is stretched 30 cm = 0.30 m past its resting position.

When the object is released from this stretched position, the spring will do work equal to the initial potential energy stored in the spring. This is because the potential energy converts to work as the spring returns to its natural length (equilibrium position).

### Work done by the spring:
\[
W = PE_{\text{spring}} = \frac{1}{2} k x^2
\]

Substituting the values:
\[
W = \frac{1}{2} (10 \, \text{N/m}) (0.30 \, \text{m})^2
\]

\[
W = \frac{1}{2} (10) (0.09) = 0.45 \, \text{J}
\]

So, the work done by the spring is **0.45 joules**.

Would you like further details or have any questions?

Here are 5 related questions:
1. What happens to the kinetic energy of the mass as it moves back?
2. How is work affected if the displacement is doubled?
3. What is the spring's potential energy when compressed?
4. How does the spring constant affect the energy stored in the spring?
5. What are the factors that could lead to energy losses in real-life spring systems?

**Tip:** The energy stored in a spring depends on the square of the displacement, meaning small changes in stretching or compression can significantly affect the work done.

A mass is attached to a spring with spring constant k = 10 N/m. The spring is stretched to 30 cm past its resting position. How much work (in J) does the spring do when the object is released and the mass travels back to its initial position?

Learn how to calculate the work done by a spring using the spring constant and displacement. In this problem, a spring with k = 10 N/m is stretched by 30 cm. The solution demonstrates how to apply Hooke's law and the potential energy formula for springs. Suitable for students in Grades 9-12.

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