Problem 45 - College Physics

Question

You find that if you hang a \(1.25 \mathrm{~kg}\) weight from a vertical spring, it stretches \(3.75 \mathrm{~cm}\). (a) What is the force constant of this spring in \(\mathrm{N} / \mathrm{m} ?\) (b) How much mass should you hang from the spring so it will stretch by \(8.13 \mathrm{~cm}\) from its original, unstretched length?

Step-by-Step Solution

Answer

(a) 327 N/m; (b) 2.71 kg

Step 1: Understand the Problem

We are dealing with a spring that stretches according to Hooke's Law, which states that the force needed to stretch or compress a spring by some distance is proportional to that distance. This law is defined by the equation \( F = k x \), where \( F \) is the force applied, \( k \) is the spring constant, and \( x \) is the displacement from the equilibrium position.

Step 2: Calculate the Force Applied

First, calculate the force due to the weight of the mass that stretches the spring. Use the formula for gravitational force: \( F = mg \), where \( m = 1.25 \, \mathrm{kg} \) and \( g = 9.81 \, \mathrm{m/s^2} \), so \( F = 1.25 \, \mathrm{kg} \times 9.81 \, \mathrm{m/s^2} = 12.2625 \, \mathrm{N} \).

Step 3: Find the Spring Constant

Use the formula \( F = k x \) to solve for \( k \). We know \( F = 12.2625 \, \mathrm{N} \) and the displacement \( x = 3.75 \, \mathrm{cm} = 0.0375 \, \mathrm{m} \). Solve for \( k \): \( k = \frac{12.2625}{0.0375} = 327 \, \mathrm{N/m} \).

Step 4: Calculate New Force for 8.13 cm Displacement

We want to find the mass that will cause the spring to stretch \( x = 8.13 \, \mathrm{cm} = 0.0813 \, \mathrm{m} \). First, find the force required for this displacement using \( F = k x \). We have \( k = 327 \, \mathrm{N/m} \), so \( F = 327 \, \mathrm{N/m} \times 0.0813 \, \mathrm{m} = 26.5811 \, \mathrm{N} \).

Step 5: Calculate the Mass for 8.13 cm Stretch

We need to determine the mass \( m \) that corresponds to the force \( F = 26.5811 \, \mathrm{N} \). Use \( F = mg \) and solve for \( m \): \( m = \frac{26.5811 \, \mathrm{N}}{9.81 \, \mathrm{m/s^2}} = 2.71 \, \mathrm{kg} \).

Key Concepts

Spring Constant

The spring constant, often symbolized by the letter \( k \), is a measure of a spring's stiffness. In other words, it tells us how much force is needed to stretch or compress the spring by a certain amount. The unit of the spring constant is \( \text{N/m} \) (Newtons per meter). This concept is rooted in Hooke's Law, which is expressed as \( F = kx \), where:

• \( F \) is the force applied to the spring.
• \( k \) is the spring constant.
• \( x \) is the displacement of the spring from its equilibrium position.

Understanding the spring constant is essential because it helps predict how a spring will behave under different loads. A larger spring constant means a stiffer spring, while a smaller value indicates a more flexible one. When working with springs, determining \( k \) allows you to calculate the force or displacement needed in various applications, such as in our example where a 1.25 kg mass stretches a spring by 3.75 cm.

Force Calculation

The process of force calculation in the context of springs is fundamental to applying Hooke's Law. To calculate the force exerted by a spring or on a spring, you typically use the formula \( F = kx \). However, in problems involving weights, another formula is often used: \( F = mg \), where \( m \) is the mass and \( g \) is the acceleration due to gravity (approximately \( 9.81 \, \text{m/s}^2 \)).

In the given example, the force needed to stretch the spring by 3.75 cm with a mass of 1.25 kg is calculated using \( F = mg \), resulting in a force of \( 12.2625 \, \text{N} \). Once this force is known, it enables the calculation of the spring constant or the deformation that the spring undergoes under specified forces. This dual use of \( F = kx \) and \( F = mg \) is pivotal in solving mechanics problems involving springs and weights.

Mass Displacement

Mass displacement refers to the distance a spring is stretched or compressed when an object with mass is attached to it. According to Hooke's Law, this relationship between forces and displacement is predictable and directly proportional. In our example, hanging a mass of 1.25 kg causes the spring to elongate by 3.75 cm.

Understanding mass displacement is crucial because it allows for the computation of important characteristics of the spring, such as its spring constant. Furthermore, by determining how much a spring elongates for a known mass, one can also extrapolate the required mass for a desired displacement. For instance, we calculated that to achieve a displacement of 8.13 cm, a mass of 2.71 kg must be used. This helps in designing systems where specific spring displacements are needed, such as in suspensions or measuring devices.

Gravitational Force

Gravitational force is a fundamental concept in physics that describes the attraction between two masses. Near the Earth's surface, it is often simplified to the force experienced by an object due to Earth's gravity, calculated with \( F = mg \). Here, \( m \) stands for the mass of the object, and \( g \) is the acceleration due to gravity, which is approximately \( 9.81 \, \text{m/s}^2 \).

In the context of our spring example, the gravitational force acting on the mass is crucial for calculating how much force is exerted on the spring. This information is used to determine the spring's constant and how far the spring will stretch under specific weights. By understanding gravitational force, students can predict how objects will interact with springs or other systems under the influence of Earth's gravity. This is vital for correctly setting up and solving mechanics problems, ensuring systems are safe and functional within the expected parameters.