/*! 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-kg block is hung from and... [FREE SOLUTION] | 91Ó°ÊÓ

91Ó°ÊÓ

A 0.70-kg block is hung from and stretches a spring that is attached to the ceiling. A second block is attached to the fi rst 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 1.4 kg.

Step by step solution

01

Understand Hooke's Law

Hooke's Law states that the force exerted by a spring is proportional to its displacement from its equilibrium position, represented by the equation \( F = k \cdot x \), where \( F \) is the force, \( k \) is the spring constant, and \( x \) is the displacement.
02

Calculate Original Force

First, we calculate the force exerted by the spring when only the 0.70-kg block is hanging from it. The gravitational force acting on the block is \( F = m \cdot g \), where \( m = 0.7 \text{ kg} \) and \( g = 9.8 \text{ m/s}^2 \). Thus, \( F = 0.7 \times 9.8 = 6.86 \text{ N} \).
03

Apply the Stretch Condition

When the second block is added, the stretch of the spring triples, which means that the total force exerted by the spring is also tripled. So the new force on the spring is \( 3 \times 6.86 = 20.58 \text{ N} \).
04

Calculate Total Mass from Total Force

Now, solve for the total mass hanging from the spring using the equation \( F = (m_1 + m_2) \cdot g \), where \( F = 20.58 \text{ N} \), \( m_1 = 0.7 \text{ kg} \), and \( g = 9.8 \text{ m/s}^2 \). Substitute and solve for \( m_1 + m_2 \): \( 20.58 = (0.7 + m_2) \cdot 9.8 \), gives \( 0.7 + m_2 = 2.1 \).
05

Solve for Second Block's Mass

Subtract the mass of the first block from the total mass to find the mass of the second block: \( m_2 = 2.1 - 0.7 = 1.4 \text{ kg} \).

Unlock Step-by-Step Solutions & Ace Your Exams!

  • Full Textbook Solutions

    Get detailed explanations and key concepts

  • Unlimited Al creation

    Al flashcards, explanations, exams and more...

  • Ads-free access

    To over 500 millions flashcards

  • Money-back guarantee

    We refund you if you fail your exam.

Over 30 million students worldwide already upgrade their learning with 91Ó°ÊÓ!

Key Concepts

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

Spring Constant
The spring constant, symbolized by \( k \), is a crucial part of Hooke's Law. It tells us how stiff a spring is or how much force is needed to extend or compress it by a unit length. The equation as per Hooke's Law is \( F = k \cdot x \), where \( F \) is the force applied, and \( x \) is the displacement caused by the force. The larger the spring constant \( k \), the stiffer the spring is, requiring more force to achieve the same displacement.
Use this knowledge to compare different springs. The one with a higher spring constant will be harder to stretch or compress. This property is essential in designing systems involving springs, such as vehicle suspensions or mechanical watches. To find the spring constant in practical scenarios, you can rearrange Hooke’s Law to \( k = \frac{F}{x} \), ensuring you have both the force applied and the displacement measured accurately.
Gravitational Force
Gravitational force is the attractive force between any two masses. When it comes to objects near Earth's surface, gravity is acting on them with a force \( F \) calculated as \( F = m \cdot g \), where \( m \) is the object's mass, and \( g \) is the acceleration due to gravity, approximately \( 9.8 \text{ m/s}^2 \).
This fundamental force plays a vital role in our daily lives, keeping everything grounded. It's also critical in physics when studying the motion of objects, such as the way a block stretches a spring. Understanding gravitational force can help explain why objects fall and how they interact with each other in space.
Displacement
Displacement refers to the change in position of an object, measured in a straight line from its original position to its new position. In the context of springs, displacement \( x \) is the measure of how much a spring is stretched or compressed from its rest position.
This concept is critical in analyzing how systems involving springs respond to forces. For example, if a spring's initial displacement is tripled by an additional weight, it indicates a linear relationship as expressed in Hooke's Law. Understanding displacement helps predict how a system will behave under increased loads or altered conditions.
  • Displacement occurs in the direction of the applied force.
  • It's a vector quantity, having both magnitude and direction.
  • In Hooke's Law, displacement directly influences the force the spring exerts back.
Mass Calculation
Calculating mass, especially in physics problems, involves understanding the relationship between force, mass, and acceleration. When a force is known, such as gravitational force, you can find mass using the equation \( F = m \cdot g \). This equation can be rearranged to \( m = \frac{F}{g} \) when solving for mass.
In the exercise described, after determining the total force when the spring displacement triples, we used this rearranged equation to solve for the total mass of two blocks. By disassembling the total mass into known and unknown parts, we can calculate the unknown mass. Such problems often involve iterative problem-solving skills, requiring careful manipulation of equations to find the desired quantities. Understanding mass calculations is pivotal in engineering, physics, and related fields where precise measurements and predictions of material behavior are required.

One App. One Place for Learning.

All the tools & learning materials you need for study success - in one app.

Get started for free

Most popular questions from this chapter

A 0.60-kg metal sphere oscillates at the end of a vertical spring. As the spring stretches from 0.12 to 0.23 m (relative to its unstrained length), the speed of the sphere decreases from 5.70 to 4.80 m/s. What is the spring constant of the spring?

A spring is hung from the ceiling. A \(0.450-\mathrm{kg}\) block is then attached to the free end of the spring. When released from rest, the block drops \(0.150 \mathrm{m}\) before momentarily coming to rest, after which it moves back upward. (a) What is the spring constant of the spring? (b) Find the angular frequency of the block's vibrations.

A vertical spring with a spring constant of \(450 \mathrm{N} / \mathrm{m}\) is mounted on the floor. From directly above the spring, which is unstrained, a \(0.30-\mathrm{kg}\) block is dropped from rest. It collides with and sticks to the spring, which is compressed by \(2.5 \mathrm{cm}\) in bringing the block to a momentary halt. Assuming air resistance is negligible, from what height (in \(\mathrm{cm}\) ) above the compressed spring was the block dropped?

Two stretched cables both experience the same stress. The first cable has a radius of \(3.5 \times 10^{-3} \mathrm{m}\) and is subject to a stretching force of \(270 \mathrm{N}\). The radius of the second cable is \(5.1 \times 10^{-3} \mathrm{m} .\) Determine the stretching force acting on the second cable.

Multiple-Concept Example 6 presents a model for solving this problem. As far as vertical oscillations are concerned, a certain automobile can be considered to be mounted on four identical springs, each having a spring constant of \(1.30 \times 10^{5} \mathrm{N} / \mathrm{m} .\) Four identical passengers sit down inside the car, and it is set into a vertical oscillation that has a period of 0.370 s. If the mass of the empty car is 1560 kg, determine the mass of each passenger. Assume that the mass of the car and its passengers is distributed evenly over the springs.

See all solutions

Recommended explanations on Physics Textbooks

View all explanations

What do you think about this solution?

We value your feedback to improve our textbook solutions.

Study anywhere. Anytime. Across all devices.