Problem 10 - College Physics

Question

\(\bullet\bullet\) (a) At what position is the magnitude of the force on a mass in a mass-spring system minimum: \((1) x=0\), (2) \(x=-A,\) or \((3) x=+A ?\) Why? \((b)\) If \(m=0.500 \mathrm{~kg}\) \(k=150 \mathrm{~N} / \mathrm{m},\) and \(A=0.150 \mathrm{~m},\) what are the magnitude of the force on the mass and the acceleration of the mass at \(x=0,0.050 \mathrm{~m},\) and \(0.150 \mathrm{~m} ?\)

Step-by-Step Solution

Answer

At \( x=0 \), the force and acceleration are minimum: force is 0 N and acceleration is 0 m/s².

Step 1: Understanding Force in a Mass-Spring System

In a mass-spring system, Hooke's Law describes the force exerted by the spring as \( F = -kx \), where \( k \) is the spring constant and \( x \) is the displacement from the equilibrium position. The force is minimum when the displacement \( x \) is zero, which means the force is zero at the equilibrium position. Therefore, the minimum magnitude of force occurs at \( x = 0 \).

Step 2: Force Magnitude Calculations

(a) The force at any position is given by \( F = k|x| \).- At \( x = 0 \), \( F = 150 \times 0 = 0 \, \text{N} \).- At \( x = 0.050 \, \text{m} \), \( F = 150 \times 0.050 = 7.5 \, \text{N} \).- At \( x = 0.150 \, \text{m} \), \( F = 150 \times 0.150 = 22.5 \, \text{N} \).Hence, the minimum force occurs at \( x = 0 \).

Step 3: Calculating Acceleration at Different Positions

(b) Using Newton's second law \( F = ma \), acceleration \( a \) can be calculated as \( a = \frac{F}{m} \).- At \( x = 0 \), since \( F = 0 \), then \( a = \frac{0}{0.500} = 0 \, \text{m/s}^2 \).- At \( x = 0.050 \, \text{m} \), \( a = \frac{7.5}{0.500} = 15 \, \text{m/s}^2 \).- At \( x = 0.150 \, \text{m} \), \( a = \frac{22.5}{0.500} = 45 \, \text{m/s}^2 \).

Step 4: Summarizing Results and Final Check

At \( x=0 \), both force and acceleration are zero indicating equilibrium with no net force acting on the mass. As displacement from equilibrium increases, both force and acceleration increase, reaching maximum at \( x=\pm A \). Thus, minimum force and acceleration occur at \( x = 0 \).

Key Concepts

Hooke's Law

Understanding Hooke's Law is crucial when dealing with a mass-spring system. Hooke's Law states that the force exerted by a spring is directly proportional to the displacement of the spring from its equilibrium position. Mathematically, it is represented as:

• \( F = -kx \)

Here, \( F \) is the force the spring exerts, \( k \) is the spring constant (a measure of the stiffness of the spring), and \( x \) is the displacement from the equilibrium position. The negative sign indicates that the direction of the force is always opposite to the direction of displacement, pulling the object back towards equilibrium.
This means whenever you stretch or compress a spring, it will always want to return to its original length. In a practical sense, this law helps us understand that the force exerted by a spring grows linearly with displacement. This makes calculations in a mass-spring system predictable and manageable, especially in oscillating systems.

Equilibrium Position

The equilibrium position in a mass-spring system is the point where the net force acting on the mass is zero. This is the natural resting position of the spring where it is neither stretched nor compressed.
The concept of equilibrium is essential, as it defines the reference point for measuring displacement. At the equilibrium position, since displacement \( x = 0 \), Hooke's Law tells us that the force \( F = -kx \) is zero.

• Therefore, when \( x = 0 \), the system experiences its minimum force magnitude, which is zero.

While in equilibrium, a mass will not accelerate unless an external force is applied to change its state. Understanding equilibrium helps us determine when potential energy is minimized in the system, contributing to solving many dynamics problems effectively.
Moreover, understanding equilibrium allows us to predict the behavior of the system under various conditions and apply principles such as conservation of energy in oscillatory motion.

Newton's Second Law

Newton's Second Law of Motion plays a fundamental role in analyzing mass-spring systems. It states that the force acting on an object is equal to the mass of the object multiplied by its acceleration, which can be expressed as:

• \( F = ma \)

In the context of a spring-mass system, the forces acting on the mass are those described by Hooke's Law. Given any force calculated using \( F = kx \), we can determine the acceleration by rearranging the formula: \( a = \frac{F}{m} \).
This relationship indicates that as the spring exerts a force on the mass, the mass subsequently accelerates. By knowing both the force and the mass, you can calculate the acceleration at any point:

• If the spring is neither compressed nor stretched, at the equilibrium, the force is zero leading thus to zero acceleration.
• As displacement from equilibrium increases, both the force and acceleration increase, becoming maximal at the extreme points (\( x = \pm A \)).

This concept helps students understand how mass, force, and acceleration are interconnected, providing the foundation for predicting and analyzing the motion within the system.