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.