Problem 16 - University Physics with Modern Physics

Question

An ideal spring of negligible mass is 12.00 \(\mathrm{cm}\) long when nothing is attached to it. When you hang a \(3.15-\mathrm{kg}\) weight from it, you measure its length to be 13.40 \(\mathrm{cm}\) . If you wanted to store 10.0 \(\mathrm{J}\) of potential energy in this spring, what would be its total length? Assume that it continues to obey Hooke's law.

Step-by-Step Solution

Answer

The spring length would be 21.52 cm to store 10.0 J of energy.

Step 1: Initial Setup and Understanding

First, identify the information given:- Original spring length, \( L_0 = 12.00 \, \text{cm} = 0.1200 \, \text{m}\).- Length with 3.15 kg weight, \( L_1 = 13.40 \, \text{cm} = 0.1340 \, \text{m}\).- Weight, \( m = 3.15 \, \text{kg}\).The spring force constant \( k \) needs to be calculated, and then used to find the length where the spring's potential energy is 10.0 J.

Step 2: Calculate the Spring Constant

The force exerted by the weight is equal to \( F = mg \), where \( g = 9.81 \, \text{m/s}^2 \). This force causes a change in the spring's length of \( \Delta x = L_1 - L_0 = 0.1340 - 0.1200 = 0.0140 \, \text{m} \).Using Hooke's Law \( F = k \Delta x \),\[ k = \frac{mg}{\Delta x} = \frac{3.15 \times 9.81}{0.0140} \approx 2204.25 \, \text{N/m}\].

Step 3: Determine the Required Extension for Desired Potential Energy

The potential energy stored in a spring is given by \( U = \frac{1}{2}k(\Delta x')^2 \).We want this energy to be 10.0 J:\[ 10.0 = \frac{1}{2} \times 2204.25 \times (\Delta x')^2 \].Solve for \( \Delta x' \):\[ 20.0 = 2204.25 \times (\Delta x')^2 \].\[ (\Delta x')^2 = \frac{20.0}{2204.25} \approx 0.009073 \].\[ \Delta x' \approx \sqrt{0.009073} \approx 0.0952 \, \text{m}\].

Step 4: Calculate the Total Length of the Spring

The total length \( L_2 \) of the spring when storing 10.0 J of potential energy can be found using:\[ L_2 = L_0 + \Delta x' = 0.1200 + 0.0952 = 0.2152 \, \text{m} \].Convert to cm:\[ L_2 = 21.52 \, \text{cm} \].

Key Concepts

Spring Potential Energy

Spring potential energy is an essential concept in physics, particularly when studying springs and mechanical systems. This energy is essentially the elastic potential energy stored in a spring when it is either compressed or stretched. According to Hooke's Law, the potential energy in a spring is represented by the formula:\[ U = \frac{1}{2} k \Delta x^2 \]Where:

• \( U \) is the potential energy
• \( k \) is the spring constant
• \( \Delta x \) is the displacement of the spring from its equilibrium position

In practical terms, this means how far the spring is either compressed or stretched from its original length. A higher spring constant signifies a stiffer spring that requires more force to stretch or compress. Conversely, a smaller displacement results in less energy stored. Therefore, understanding the relationship between these variables helps us predict how much energy can be stored in a spring. This study is vital for designing systems where controlling energy storage and release is necessary, such as in automotive suspensions or measuring devices.

Spring Constant Calculation

The spring constant, denoted as \( k \), measures the stiffness of a spring. It defines how much force is required to stretch or compress the spring by a unit distance. One common method to calculate the spring constant is by using Hooke's Law:\[ F = k \Delta x \]Where:

• \( F \) is the force applied to the spring
• \( \Delta x \) is the change in length of the spring
• \( k \) is what we want to find

In the exercise, we calculate \( k \) by measuring the force exerted by a known weight and the corresponding change in length of the spring. Utilizing the formula \( k = \frac{F}{\Delta x} \), we determine the spring constant by dividing the force (due to the weight) by the stretched length. The spring constant helps us understand how resistant a spring is to deformation—higher values mean the spring is more resistant.

Force Exerted by a Weight

The force exerted by a weight is fundamental when dealing with springs because it drives the change in spring length. This force is the gravitational force acting on a mass, calculated by:\[ F = mg \]Where:

• \( m \) is the mass of the object
• \( g \) is the acceleration due to gravity (approximately 9.81 m/s² on Earth)

So, if you have a mass suspended from a spring, the weight's force pulls the spring, causing it to extend. The extent of this extension depends on both the force and the spring constant. By understanding the force exerted due to weight, you can predict how much a spring will stretch, calculate spring constants, and even design systems to handle specific forces. This forms the basis for applications ranging from weighing scales to complex load-bearing systems.