Question

A spring that is stretched \(2.6 \mathrm{~cm}\) stores a potential energy of \(0.053 \mathrm{~J}\). What is the spring constant of this spring?

Short answer

The spring constant is approximately 156.65 N/m.

Step-by-step solution

Understand the formula for spring potential energy

The potential energy stored in a spring is given by the formula: \( PE = \frac{1}{2} k x^2 \), where \( PE \) is the potential energy, \( k \) is the spring constant, and \( x \) is the displacement from the spring's equilibrium position. In this problem, the potential energy is \(0.053 \, \text{J}\) and the displacement \(x\) is \(2.6 \, \text{cm}\).

Convert displacement to meters

Since the formula uses SI units, we need to convert the displacement from centimeters to meters. \(2.6 \, \text{cm} = 0.026 \, \text{m}\).

Rearrange the formula to solve for the spring constant

Rearrange the potential energy formula to solve for the spring constant \(k\):\[ k = \frac{2 \cdot PE}{x^2} \]

Substitute the values into the formula

Substitute the known values into the rearranged formula:\[ k = \frac{2 \times 0.053}{(0.026)^2} \]

Calculate the spring constant

Perform the calculation:\[ k = \frac{2 \times 0.053}{0.000676} = \frac{0.106}{0.000676} \approx 156.65 \] Thus, the spring constant is approximately \(156.65 \, \text{N/m}\).

Key concepts

Potential Energy in Springs

Potential energy is the energy stored in an object due to its position or configuration. In the context of springs, it is the energy stored when the spring is stretched or compressed. The potential energy in a spring follows the formula:

• \( PE = \frac{1}{2} k x^2 \)

where \( PE \) is the potential energy, \( k \) is the spring constant, and \( x \) is the displacement from the spring's equilibrium position.
This relationship arises from Hooke's Law, which we'll discuss further. For the spring in the problem, the potential energy is given as \(0.053 \mathrm{~J}\). Understanding how this energy is stored in the spring will help us determine other attributes like the spring constant.

Hooke's Law and the Spring Constant

Hooke's Law is a fundamental principle that describes the behavior of springs.

• It states that the force \( F \) needed to extend or compress a spring by a distance \( x \) is proportional to that distance.
• The formula is expressed as \( F = kx \).

Here, \( k \) is the spring constant, which measures the stiffness of the spring. A larger \( k \) value indicates a stiffer spring.
In the context of potential energy, Hooke's Law helps us rearrange the potential energy formula to solve for \( k \). With the given potential energy and displacement, we can uncover the spring constant which is crucial to understanding how the spring behaves.

Importance of Unit Conversion

Unit conversion is a vital part of solving physics problems since it ensures that calculations are consistent with SI units. In the given problem, the displacement of the spring is given in centimeters.

• Since the potential energy formula used meters, we convert \(2.6 \text{ cm}\) to \(0.026 \text{ m}\).

Using consistent units prevents errors in calculation and ensures accuracy.
Whether converting length, mass, or time, understanding how to interchange units is essential for solving physics problems uniformly and accurately.

Effective Physics Problem Solving

Solving physics problems efficiently involves a series of strategic steps.

• First, understand the problem and identify known values and desired outcomes.
• Next, choose the appropriate formulas that relate the known and unknown variables.
• Ensure units are consistent to avoid miscalculations.
• Finally, perform algebraic manipulations as needed to solve for unknown quantities.

Following these steps systematically allows for logical reasoning and successful resolution of complex physics scenarios.
Practicing this approach with various problems will develop a keen understanding of physics principles such as those involving potential energy and Hooke's Law.