Question

A spring has a spring constant of \(80 \mathrm{~N} / \mathrm{m}\). How much potential energy does it store when stretched by \(1.0 \mathrm{~cm} ?\) a) \(4.0 \cdot 10^{-3}\) J b) \(0.40 \mathrm{~J}\) c) 80 d) \(800 \mathrm{~J}\) e) \(0.8 \mathrm{~J}\)

Short answer

Answer: (a) \(4.0 \cdot 10^{-3}\) J

Step-by-step solution

Identify the given information

We are given: - spring constant (k) = \(80 \mathrm{~N/m}\) - stretch length (x) = \(1.0 \mathrm{~cm}\)

Convert units if necessary

Since the spring constant is given in \(\mathrm{N/m}\), we need to convert the stretch length to meters: \(x = 1.0 \mathrm{~cm} \times \dfrac{1 \mathrm{~m}}{100 \mathrm{~cm}} = 0.01 \mathrm{~m}\)

Apply Hooke's Law formula

Using the formula for potential energy (PE) stored in a stretched spring: \(PE = \dfrac{1}{2}kx^2\) Plug in the values of k and x: \(PE = \dfrac{1}{2}(80 \mathrm{~N/m})(0.01 \mathrm{~m})^2\)

Solve for the potential energy

Calculate the potential energy stored: \(PE = \dfrac{1}{2}(80 \mathrm{~N/m})(0.0001 \mathrm{m}^2) = 0.004 \mathrm{J}\)

Compare the result with the given options

The calculated potential energy is \(0.004 \mathrm{J}\) which is the same as \(4.0 \cdot 10^{-3} \mathrm{J}\). Therefore, the correct answer is (a) \(4.0 \cdot 10^{-3}\) J.

Key concepts

Understanding Potential Energy in Springs

When we talk about potential energy, specifically in the context of a spring, we refer to the stored energy due to its deformation. If a spring is compressed or stretched, it possesses potential energy. This is calculated using Hooke's Law for springs, with the formula: \(PE = \dfrac{1}{2}kx^2\), where

• \(PE\) is the potential energy,
• \(k\) is the spring constant, and
• \(x\) is the displacement from its natural length.

This energy is what allows springs to "spring back" to their original position after being deformed, contributing to various applications like storing energy in a mechanical clock or absorbing shock in vehicle suspensions.

The Role of the Spring Constant

The spring constant, symbolized by \(k\), is a fundamental parameter in understanding a spring's behavior. It measures a spring’s stiffness — essentially, how resistant it is to being deformed.
A larger spring constant means the spring is stiffer, requiring more force to achieve the same displacement as a less stiff spring. The unit for the spring constant is Newton per meter (N/m), signifying how many newtons of force are required to stretch or compress the spring by one meter.
When solving for potential energy, knowing the spring constant is crucial, as it directly influences the magnitude of energy stored when the spring is stretched or compressed.

Why Unit Conversion Matters

Unit conversion is pivotal in physics because it ensures that all calculations are consistent and correct. In exercises involving springs, it is common for different units to be used, especially for displacement.
For example, in a spring problem with a spring constant in \(N/m\), displacement should be converted to meters to maintain consistency.

• Converting units ensures that the calculations are valid and the formula can be applied accurately.
• A common conversion factor is converting centimeters to meters, where 1 cm equals 0.01 m.

Therefore, before applying Hooke’s Law, always double-check that your units match those of the spring constant, as seen in this exercise where 1.0 cm was converted to 0.01 m.