/*! This file is auto-generated */ .wp-block-button__link{color:#fff;background-color:#32373c;border-radius:9999px;box-shadow:none;text-decoration:none;padding:calc(.667em + 2px) calc(1.333em + 2px);font-size:1.125em}.wp-block-file__button{background:#32373c;color:#fff;text-decoration:none} Problem 7 A \(0.70-\mathrm{kg}\) block is ... [FREE SOLUTION] | 91Ó°ÊÓ

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A \(0.70-\mathrm{kg}\) block is hung from and stretches a spring that is attached to the ceiling. A second block is attached to the first one, and the amount that the spring stretches from its unstrained length triples. What is the mass of the second block?

Short Answer

Expert verified
The mass of the second block is 1.40 kg.

Step by step solution

01

Understanding the Problem

We are given that a spring is stretched by a mass of 0.70 kg, and after attaching an additional mass, the stretch of the spring becomes three times the initial stretch. We need to find the mass of this second block.
02

Initial Setup and Hooke's Law

According to Hooke's Law, the force stretching the spring is proportional to the displacement: \( F = kx \), where \( k \) is the spring constant and \( x \) is the extension. Initially, the force acting on the spring is due to the 0.70 kg block, so \( mg = kx_0 \), where \( x_0 \) is the initial stretch and \( m = 0.70 \) kg.
03

Considering the Additional Mass

When a second block is attached, the total stretch becomes \( 3x_0 \). Now, the spring force is the weight of both blocks: \( (m + M)g = k(3x_0) \). Here, \( M \) is the mass of the second block.
04

Equating Forces and Solving

We equate the forces from both situations: \[ mg = kx_0 \quad \text{and} \quad (m + M)g = 3kx_0 \]. By dividing the second equation by the first, we get: \[ (m + M) = 3m \]. Therefore, \( M = 2m \).
05

Final Calculation

Substitute \( m = 0.70 \) kg into the equation \( M = 2m \): \( M = 2 \times 0.70 = 1.40 \). Thus, the mass of the second block is 1.40 kg.

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Key Concepts

These are the key concepts you need to understand to accurately answer the question.

Understanding Spring Constant
In the world of physics, the term "spring constant" is a significant factor when it comes to understanding the behavior of springs. The spring constant, often denoted as \( k \), is a measure of the stiffness of a spring. It is defined through Hooke's Law, which states that the force \( F \) exerted by a spring is proportional to the distance \( x \) it is stretched or compressed from its original length. Mathematically, this relationship is expressed as \( F = kx \).
The spring constant is determined by the material of the spring, its coil diameter, and the number of coils it possesses. A higher spring constant means the spring is stiffer and requires more force to stretch or compress a given distance. Conversely, a lower spring constant indicates a more flexible spring. Understanding the value of \( k \) helps predict how a spring will behave under different forces, making it essential for solving problems that involve spring mechanics.
The Role of Mass Calculation
Calculating mass in physics problems involving springs can sometimes be a bit of a challenge, especially when dealing with more than one object or mass. In the given exercise, the process involved understanding how different masses alter the behavior of a spring. Initially, the spring is stretched by a mass of 0.70 kg, leading to a known extension length \( x_0 \).
When another mass is added, this changes the dynamics. The new mass together with the original one causes the spring to stretch three times the original length \( 3x_0 \).
The key to determining the mass of the second block (denoted as \( M \)) is recognizing that the forces due to both masses are acting on the spring. By using the established equations for forces \( mg = kx_0 \) and \( (m + M)g = 3kx_0 \), we can derive \( M = 2m \), showcasing how these calculations can be applied to find unknown quantities.
Exploring Displacement in Springs
In understanding spring mechanics, displacement is an important concept, referring to the change in length of the spring when a force is applied. This change, whether it's stretching or compressing, is what we refer to as displacement. In the context of the problem, we were given an initial displacement \( x_0 \) caused by hanging a 0.70 kg block.
When another block is added, the displacement becomes three times \( 3x_0 \). This introduction of additional mass causing increased displacement is a direct application of Hooke's Law. According to the law, the relationship is linear unless the spring is stretched beyond its elastic limit.
Understanding displacement allows us to analyze how different forces affect a spring's behavior, by altering the length it holds. This knowledge is especially useful in designing systems where precise control of the spring’s movement is needed, ensuring structures remain safe and stable.

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Most popular questions from this chapter

A spring lies on a horizontal table, and the left end of the spring is attached to a wall. The other end is connected to a box. The box is pulled to the right, stretching the spring. Static friction exists between the box and the table, so when the spring is stretched only by a small amount and the box is released, the box does not move. The mass of the box is \(0.80 \mathrm{kg}\), and the spring has a spring constant of \(59 \mathrm{N} / \mathrm{m}\). The coefficient of static friction between the box and the table on which it rests is \(\mu_{\mathrm{s}}=0.74 .\) How far can the spring be stretched from its unstrained position without the box moving when it is released?

A hand exerciser utilizes a coiled spring. A force of \(89.0 \mathrm{N}\) is required to compress the spring by \(0.0191 \mathrm{m} .\) Determine the force needed to compress the spring by \(0.0508 \mathrm{m}\).

A piece of mohair taken from an Angora goat has a radius of \(31 \times 10^{-6} \mathrm{m} .\) What is the least number of identical pieces of mohair needed to suspend a \(75-\mathrm{kg}\) person, so the strain experienced by each piece is less than \(0.010 ?\) Assume that the tension is the same in all the pieces.

A small object oscillates back and forth at the bottom of a frictionless hemispherical bowl, as the drawing illustrates. The radius of the bowl is \(R\), and the angle \(\theta\) is small enough that the object oscillates in simple harmonic motion. Derive an expression for the angular frequency \(\omega\) of the motion. Express your answer in terms of \(R\) and \(g,\) the magnitude of the acceleration due to gravity.

Multiple-Concept Example 6 reviews the principles that play roles in this problem. A bungee jumper, whose mass is 82 kg, jumps from a tall platform. After reaching his lowest point, he continues to oscillate up and down, reaching the low point two more times in 9.6 s. Ignoring air resistance and assuming that the bungee cord is an ideal spring, determine its spring constant.

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