Problem 5 - Physics

Question

A car is hauling a \(92-\mathrm{kg}\) 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 meters, or 12 mm.

Step 1: Understand the Force Applied

The force applied on the trailer by the car is determined by using Newton's second law of motion. This is given by the formula \( F = ma \), where \( m \) is the mass of the trailer and \( a \) is its acceleration. Here, \( m = 92\, \mathrm{kg} \) and \( a = 0.30\, \mathrm{m/s^2} \). Substitute these values to find the force.

Step 2: Calculate the Force Applied

Substitute the known values into the formula: \[ F = 92\,\mathrm{kg} \times 0.30\,\mathrm{m/s^2} = 27.6\,\mathrm{N} \]This is the force applied by the car on the trailer.

Step 3: Use Hooke's Law to Find Spring Stretch

Hooke's Law relates the force on a spring to the displacement or stretch (\( x \)) of the spring: \( F = kx \), where \( k \) is the spring constant. Here, \( F = 27.6\, \mathrm{N} \) and \( k = 2300\, \mathrm{N/m} \). Rearrange the formula to solve for \( x \):\[ x = \frac{F}{k} = \frac{27.6\,\mathrm{N}}{2300\,\mathrm{N/m}} \]

Step 4: Calculate the Spring Stretch

Substitute the values from the previous step:\[ x = \frac{27.6}{2300} \approx 0.012\, \mathrm{m} \]So, the spring stretches by \( 0.012 \mathrm{~m} \), or \( 12 \mathrm{~mm} \).

Key Concepts

Newton's Second Law of Motion

Newton's Second Law of Motion is the key to understanding how forces work in the world around us. It tells us that the force acting on an object is equal to the mass of that object multiplied by its acceleration. In simple terms, this law explains how much an object will speed up or slow down when a force is applied to it. Here are the important bits to remember:

• The formula for Newton's Second Law is given by: \( F = ma \), where \( F \) is the force, \( m \) is the mass, and \( a \) is the acceleration.
• This explains why heavier objects (having more mass) need more force to accelerate at the same rate as lighter objects.
• In the case of the car and trailer, if the car accelerates the trailer, the force can be calculated using this law.

By understanding this fundamental principle, we can calculate the force between any objects, like the car and its trailer in the problem.

Hooke's Law

Hooke's Law tells us how springs work and is fundamental to many engineered systems. Imagine you have a spring; if you pull or push it, it moves. How much it stretches or compresses depends on the force applied. Hooke's Law helps us figure that out with the equation \( F = kx \):

• \( F \) represents the force applied to the spring.
• \( k \) is the spring constant, a measure of the spring's stiffness.
• \( x \) is the displacement or stretch of the spring from its original position.

If we know how stiff the spring is (the spring constant) and the force applied, we can find out how much the spring will stretch. For the car and trailer exercise in question, we used Hooke's Law to determine how much the spring connecting them stretches due to the force applied by the car.

Spring Constant

The spring constant, represented by \( k \), is an essential part of understanding how much a spring will stretch or compress under a force. It tells you how "stiff" or "soft" a spring is. Here's what you need to know about the spring constant:

• A larger spring constant means the spring is stiffer and doesn't stretch as much for a given force.
• A smaller spring constant indicates a softer spring that stretches more easily.
• In the exercise, the spring constant was given as \( 2300 \, \mathrm{N/m} \).

By understanding the spring constant, you can predict how different types of springs will behave when forces are applied. This concept is not only useful in physics problems but also in real-world applications like vehicle suspension systems and measuring scales.