Problem 5 - Physics

Question

A car is hauling a 92-kg trailer, to which it is connected by a spring. The spring constant is \(2300 \mathrm{~N} / \mathrm{m}\). The car accelerates with an acceleration of \(0.30 \mathrm{~m} / \mathrm{s}^{2} .\) By how much does the spring stretch?

Step-by-Step Solution

Answer

The spring stretches by 0.012 m (or 12 mm).

Step 1: Understanding the Problem

We have a car hauling a trailer using a spring. The trailer's mass is 92 kg, and the spring constant is 2300 N/m. The car and trailer are accelerating at 0.30 m/s². We need to find out how much the spring stretches.

Step 2: Calculate the Force on the Trailer

Use Newton’s second law to find the force exerted on the trailer. The force can be calculated using the formula: \[ F = ma \]where \( m = 92 \, \text{kg} \) and \( a = 0.30 \, \text{m/s}^2 \).

Key Concepts

Newton's Second Law

Newton's Second Law of Motion is a fundamental concept in physics that explains how the motion of an object changes when it is subjected to external forces. It is often summarized by the equation \( F = ma \), where \( F \) is the force applied to an object, \( m \) is the object's mass, and \( a \) is the acceleration produced. This law implies that the force acting on an object is the product of its mass and the acceleration it undergoes.

In the context of the exercise, a 92-kg trailer is connected to a spring. The trailer experiences an acceleration of \( 0.30 \, \text{m/s}^2 \). By applying Newton's second law, we can calculate the force exerted on the trailer as \( F = 92 \, \text{kg} \times 0.30 \, \text{m/s}^2 = 27.6 \, \text{N} \).

This calculated force is crucial because it is what causes the spring to stretch. Recognizing this relationship helps us understand how a change in acceleration or mass can significantly alter the force involved, affecting the spring's behavior.

Spring Constant

The spring constant is a measure of a spring's stiffness, denoted by the symbol \( k \). It indicates how much force is needed to stretch or compress a spring by a certain distance. The unit of the spring constant is Newtons per meter (N/m).

In the problem, the spring constant is given as \( 2300 \, \text{N/m} \). This means that for every meter the spring is stretched, a force of 2300 N is exerted. Hooke's Law expresses this relationship with the formula \( F = kx \), where \( F \) is the force applied to the spring, \( k \) is the spring constant, and \( x \) is the displacement of the spring from its equilibrium position.

Rearranging Hooke's Law allows us to solve for \( x \), the stretch of the spring, using the force we calculated previously: \( x = \frac{F}{k} = \frac{27.6 \, \text{N}}{2300 \, \text{N/m}} \approx 0.012 \, \text{m} \). This small stretch illustrates how a relatively large spring constant results in a small displacement for a given force.

Acceleration

Acceleration is the rate at which an object's velocity changes over time. It is a vector quantity, which means it has both a magnitude and a direction. The standard unit for measuring acceleration is meters per second squared (m/s²).

In this scenario, the trailer is accelerating at a rate of \( 0.30 \, \text{m/s}^2 \). This positive acceleration indicates that the trailer is gaining speed as it is pulled by the car. The acceleration plays a critical role in determining the force exerted on the trailer, as outlined in Newton's second law.

Understanding acceleration is vital for solving dynamics problems, as it directly influences the forces involved. It provides insight into how quickly an object can change its speed, which in turn affects how systems like springs respond to these forces. As a result, being able to calculate and conceptualize acceleration is a key skill in physics problem-solving.