Aspringhasbeenextended4cmandhasanelasticpotentialenergyof30J.Calculatethespringconstantofthespring.Giveyouranswerto2decimalplacesifneeded.

1. Break Down the Problem

To find the spring constant $k$, we can use the formula for elastic potential energy stored in a spring, which is given by:

$E_p = \frac{1}{2} k x^2$

Where:

• $E_p$ is the elastic potential energy (30 J)
• $k$ is the spring constant (unknown)
• $x$ is the extension of the spring (4 cm or 0.04 m)

2. Relevant Concepts

We will rearrange the formula to solve for $k$:

$E_p = \frac{1}{2} k x^2 \implies k = \frac{2E_p}{x^2}$

3. Analysis and Detail

Substituting the known values into the rearranged formula:

• Convert the extension from cm to m: $x = 4 \text{ cm} = 0.04 \text{ m}$
• Plug in the values: $k = \frac{2 \times 30 \text{ J}}{(0.04 \text{ m})^2}$

Calculation

$k = \frac{60 \text{ J}}{0.0016 \text{ m}^2} = 37500 \text{ N/m}$

4. Verify and Summarize

The calculations have been verified as follows:

1. The elastic potential energy was correctly stated as 30 J.
2. The extension of 4 cm was converted to 0.04 m accurately.
3. The final calculation gives us $k = 37500 \text{ N/m}$.

Final Answer

The spring constant $k$ is:

$\boxed{37500 \, \text{N/m}}$