Problem 14 - University Physics with Modern Physics

Question

An ideal spring of negligible mass is 12.00 cm long when nothing is attached to it. When you hang a 3.15-kg weight from it, you measure its length to be 13.40 cm. If you wanted to store 10.0 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 total length of the spring is 21.52 cm.

Step 1: Determine Spring Constant

The spring obeys Hooke's Law, which states \( F = kx \), where \( F \) is the force applied and \( x \) is the extension from its original length.First, calculate the extension \( x \) when a 3.15 kg weight is attached: \[ x = 13.40 \, \text{cm} - 12.00 \, \text{cm} = 1.40 \, \text{cm} = 0.0140 \, \text{m} \]The force is given by \( F = mg \), where \( g = 9.8 \, \text{m/s}^2 \) is the acceleration due to gravity:\[ F = 3.15 \, \text{kg} \times 9.8 \, \text{m/s}^2 = 30.87 \, \text{N} \]Thus, the spring constant \( k \) is:\[ k = \frac{F}{x} = \frac{30.87 \, \text{N}}{0.0140 \, \text{m}} = 2205 \, \text{N/m} \]

Step 2: Use the Potential Energy Formula

The potential energy stored in a spring is given by the formula \( PE = \frac{1}{2}kx^2 \), where \( PE \) is the potential energy.Given that we want to store \( 10.0 \, \text{J} \) of potential energy, set \( PE = 10.0 \, \text{J} \) and solve for the displacement \( x \):\[ 10.0 = \frac{1}{2} \times 2205 \, \text{N/m} \times x^2 \]Solve for \( x^2 \):\[ x^2 = \frac{10.0 \times 2}{2205} = \frac{20.0}{2205} \approx 0.00907 \]Now, calculate \( x \):\[ x = \sqrt{0.00907} \approx 0.0952 \, \text{m} \]

Step 3: Calculate the Total Length of the Spring

The total length of the spring when storing 10.0 J of potential energy is the original length plus the calculated extension \( x \):Original length: \( 12.00 \, \text{cm} = 0.120 \, \text{m} \)Thus, the total length is:\[ \text{Total length} = 0.120 \, \text{m} + 0.0952 \, \text{m} = 0.2152 \, \text{m} \]Convert the length back to centimeters:\[ 0.2152 \, \text{m} = 21.52 \, \text{cm} \]

Key Concepts

Spring Constant

The spring constant, denoted as \( k \), is a measure of the stiffness of a spring. In essence, it tells us how much force is necessary to stretch or compress the spring by a certain amount. According to Hooke's Law, the force applied to a spring is directly proportional to the displacement of the spring from its natural, unstressed length. The formula is \( F = kx \), where \( F \) represents the force applied, \( x \) represents the displacement from equilibrium, and \( k \) is the spring constant.

To find the spring constant, you need to know both the force applied to the spring and the displacement caused by this force. For example, if you hang a weight from a spring, the force is equal to the weight due to gravity \( F = mg \), where \( m \) is the mass, and \( g \) is the acceleration due to gravity. The extension or compression of the spring is measured from the spring's original length. Thus, the spring constant \( k \) can be calculated using the formula \( k = \frac{F}{x} \). This constant is crucial in calculations involving the stretching or compressing of springs.

Potential Energy

Potential energy in the context of a spring is the energy stored when it is compressed or extended. This energy is based on the idea that the force applied to the spring does work, which is then converted into energy stored in the spring. The formula for the potential energy stored in a spring is \( PE = \frac{1}{2}kx^2 \). Here, \( PE \) represents the potential energy, \( k \) is the spring constant, and \( x \) is the displacement from the unstressed position.

When solving problems involving potential energy in springs, we often need to find the total length of the spring when a certain amount of potential energy is stored. In such cases, after determining the potential energy, we set \( PE \) equal to the desired energy and solve for \( x \). The displacement \( x \) calculated can then be added to the spring's natural length to find the total length of the spring at that potential energy level.

Mechanical Energy

Mechanical energy is the sum of potential and kinetic energy in a system. In the context of springs, we often focus on the potential aspect of mechanical energy since springs store energy when they are extended or compressed. In a perfect system where no energy is lost to friction or air resistance, mechanical energy is conserved. This means that the energy you put into the system, such as by compressing a spring, is entirely stored, and when released, converts back.

The mechanical energy stored in a spring during displacement is dictated by Hooke's Law and its potential energy formula \( PE = \frac{1}{2}kx^2 \). This reflects how much energy is stored depending on the stiffness of the spring (\( k \)) and its displacement (\( x \)). An understanding of mechanical energy in springs helps explain phenomena like oscillation, where a spring compresses and extends repeatedly, converting energy back and forth.

Force and Displacement

Force and displacement are key ideas in understanding Hooke's Law. The force applied to a spring leads to its displacement, changing its length. The relationship is linear, meaning that the force will produce a proportional displacement as long as the spring doesn't exceed its elastic limit. The formula \( F = kx \) simplifies this interaction.

Displacement here is the difference in length of the spring when a force is applied, compared to its natural length. It's important to convert units properly—such as converting centimeters to meters—as seen in calculations where \( x \) is required to be in meters for consistency with \( N/m \) units of the spring constant.

Understanding the concept of force and resulting displacement helps illustrate how springs respond to external loads, and how much they stretch or compress under different conditions. It's about the balance between applied forces and the spring's natural inclination to restore its shape.