Problem 3 - Physics for Scientists and Engineers with Modern Physics

Question

(II) A spring with \(k=63 \mathrm{~N} / \mathrm{m}\) hangs vertically next to a ruler. The end of the spring is next to the \(15-\mathrm{cm}\) mark on the ruler. If a \(2.5-\mathrm{kg}\) mass is now attached to the end of the spring, where will the end of the spring line up with the ruler marks?

Step-by-Step Solution

Answer

The spring will line up with the 53.97 cm mark on the ruler.

Step 1: Understand Hooke's Law

Hooke's Law states that the force exerted by a spring is directly proportional to its extension, given by the equation \( F = kx \), where \( F \) is the force applied, \( k \) is the spring constant, and \( x \) is the spring's extension from its equilibrium position.

Step 2: Calculate the Force Due to Mass

The force exerted by the mass on the spring is due to gravity, which is \( F = mg \), where \( m = 2.5 \text{ kg} \) is the mass, and \( g = 9.8 \text{ m/s}^2 \) is the acceleration due to gravity. Calculate \( F \). \[ F = 2.5 \times 9.8 = 24.5 \text{ N} \]

Step 3: Solve for Spring Extension

Using Hooke's Law \( F = kx \), solve for the spring extension \( x \). Given that \( F = 24.5 \text{ N} \) and \( k = 63 \text{ N/m} \): \[ 24.5 = 63x \] Solve for \( x \): \[ x = \frac{24.5}{63} \approx 0.3897 \text{ m} \] or 38.97 cm.

Step 4: Determine Final Position

The initial position of the spring was at the 15 cm mark on the ruler. Add the spring's extension to this initial position to find the final position: \( 15 \text{ cm} + 38.97 \text{ cm} = 53.97 \text{ cm} \).

Key Concepts

Spring Constant

In the fascinating realm of mechanics, the "spring constant" is a fundamental concept related to how a spring reacts to external forces. Mathematically, the spring constant is denoted by the symbol \( k \). It's measured in newtons per meter (N/m). The spring constant defines the stiffness of a spring. The higher the value of \( k \), the stiffer the spring, indicating that more force is required to stretch or compress it by a given amount.

• The spring constant \( k \) is a property intrinsic to each spring. It does not change unless the physical composition or structure of the spring is altered.
• According to Hooke's Law, the spring constant is used to relate the force exerted by the spring with its displacement \( x \). Thus, \( F = kx \).
• In our exercise, the spring constant \( k \) is given as \( 63 \; \text{N/m} \), which will help us determine how much the spring will extend when the mass is added.

Understanding the spring constant helps us predict the behavior of the spring when external forces are applied, making it a crucial part of solving problems related to springs.

Force Due to Gravity

The force due to gravity is a key player when it comes to understanding how objects interact with springs. This force, often called weight, is calculated using the formula \( F = mg \), where \( m \) is the mass in kilograms and \( g \) is the acceleration due to gravity. On Earth, \( g \) is approximately \( 9.8 \; \text{m/s}^2 \).

• Gravity pulls objects towards the center of the Earth, which is why it contributes to the extension of a spring when a mass is attached.
• In the given exercise, the mass of \( 2.5 \; \text{kg} \) results in a gravitational force: \( F = 2.5 \times 9.8 = 24.5 \; \text{N} \).
• This calculated force acts directly downward on the spring, causing it to extend.

This concept helps in understanding how the force exerted by an object due to its weight influences the stretching of a spring, which is a common application in physics and engineering.

Spring Extension

Spring extension describes how much a spring becomes longer when a force is applied to it. This physical change in length is calculated by rearranging Hooke's Law \( F = kx \) to solve for \( x \), the extension. It's a direct measure of how effective the external force is at changing the spring's shape.

• Given the force exerted by the attached mass \( F \), and the known spring constant \( k \), you can find \( x = \frac{F}{k} \).
• In the problem example, with \( F = 24.5 \; \text{N} \) and \( k = 63 \; \text{N/m} \), the extension \( x \) comes out to be approximately \( 0.3897 \; \text{m} \) or \( 38.97 \; \text{cm} \).
• This extension shows how far the spring will stretch beyond its equilibrium position when the mass is added.

Spring extension is a practical measurement that allows us to determine how systems involving springs will behave under various forces. It is crucial for designing and understanding systems where springs play a critical role, such as in vehicles' suspensions and various mechanical devices.