/*! 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 75 A mass \(m\) is attached to a sp... [FREE SOLUTION] | 91Ó°ÊÓ

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A mass \(m\) is attached to a spring which is held stretched a distance \(x\) by a force F, Fig. 6-45, and then released. The spring pulls the mass to the left, towards its natural equilibrium length. Assuming there is no friction, determine the speed of the mass m when the spring returns: (a) to its normal length (\(x =\) 0); (b) to half its original extension(\(x\)/2).

Short Answer

Expert verified
(a) Speed at normal length: \( v = \sqrt{\frac{k}{m}}x \). (b) Speed at half extension: \( v = \frac{1}{2}\sqrt{\frac{k}{m}}x \).

Step by step solution

01

Identify Spring Potential Energy

The energy stored in the spring at the initial position can be expressed as spring potential energy using Hooke's law. The potential energy is given by \( U = \frac{1}{2}kx^2 \), where \( k \) is the spring constant and \( x \) is the displacement from the equilibrium position.
02

Express the Energy Conservation Principle

Since there is no friction, energy is conserved. The sum of potential energy in the spring and kinetic energy of the mass must remain constant. Initially, all energy is potential, \( U = \frac{1}{2}kx^2 \), and when the mass is at position \( x \), the energy is split between kinetic \( K = \frac{1}{2}mv^2 \) and remaining potential energy \( U = \frac{1}{2}kx'^2 \) where \( x' \) is the new displacement from equilibrium.
03

Calculate Speed at Normal Length

(a) When the spring returns to its normal length, \( x = 0 \). The potential energy is zero, so all initial potential energy becomes kinetic energy. Thus, \( \frac{1}{2}kx^2 = \frac{1}{2}mv^2 \). Solving for \( v \), we get \( v = \sqrt{\frac{k}{m}}x \).
04

Calculate Speed at Half Original Extension

(b) When the spring is at half its original extension, \( x' = \frac{x}{2} \). The energy equation becomes \( \frac{1}{2}kx^2 = \frac{1}{2}mv^2 + \frac{1}{2}k\left(\frac{x}{2}\right)^2 \). Solving for \( v \), we simplify to \( v = \frac{1}{2}\sqrt{\frac{k}{m}}x \).

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Key Concepts

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

Hooke's Law
When dealing with springs and their behavior, Hooke's Law is a fundamental concept to understand. This law explains how a spring reacts when a force is applied. It states that the force required to compress or stretch a spring is directly proportional to the distance it is stretched or compressed. Mathematically, this is expressed as:
\[ F = -kx \]where:
  • \( F \) is the force applied to the spring.
  • \( k \) is the spring constant, indicating how stiff the spring is.
  • \( x \) is the displacement from the spring's natural length.
Understanding Hooke's law is important as it forms the basis for calculating spring potential energy, which is a critical part of energy conservation in a spring-mass system.
Energy Conservation
Energy conservation is a key principle in physics that suggests the total energy in an isolated system remains constant. In the context of a spring-mass system without friction, this means the total energy is the sum of potential and kinetic energies.
Initially, the energy is stored as potential energy in the spring, expressed as:\[ U = \frac{1}{2}kx^2 \]As the spring moves and the mass is displaced, this potential energy converts into kinetic energy. The conservation law ensures that the entire original potential energy eventually transforms into kinetic energy as the spring returns to its equilibrium.
This principle helps us determine the speed of the mass when the spring reaches certain points, like when it is at its normal length or half its original extension.
Kinetic Energy
Kinetic energy is the energy possessed by an object due to its motion. When dealing with a moving mass attached to a spring, kinetic energy plays a crucial role.
The kinetic energy of the mass can be given by:\[ K = \frac{1}{2}mv^2 \]where:
  • \( m \) is the mass of the object.
  • \( v \) is its velocity.
Initially, when the spring is fully stretched or compressed, the mass has zero kinetic energy. As the spring returns to its normal length, its potential energy converts to kinetic energy, allowing us to solve for the speed of the mass using energy conservation principles.
Frictionless System
In a frictionless system, there are no external forces like friction acting to impede the motion of the objects involved. This is an idealized scenario that simplifies calculations, allowing us to focus solely on other forces at play, such as those exerted by the spring.
Without friction:
  • Energy is not lost to the surroundings.
  • The total mechanical energy is conserved, consisting only of kinetic and potential energy.
This assumption makes it easier to apply the conservation of energy principle, as we don't need to account for energy loss due to heat or other resistive forces. Consequently, when analyzing the motion of the spring-mass system, we can ensure all potential energy eventually changes into kinetic energy at points of interest.

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Most popular questions from this chapter

(II) A 1.60-m-tall person lifts a 1.65-kg book off the ground so it is 2.20 m above the ground. What is the potential energy of the book relative to (\(a\)) the ground, and (\(b\)) the top of the person's head? (\(c\)) How is the work done by the person related to the answers in parts (a) and (b)?

(III) Early test flights for the space shuttle used a "glider" (mass of 980 kg including pilot). After a horizontal launch at 480 km/h at a height of 3500 m, the glider eventually landed at a speed of 210 km/h. (\(a\)) What would its landing speed have been in the absence of air resistance? (\(b\))What was the average force of air resistance exerted on it if it came in at a constant glide angle of 12\(^\circ\) to the Earth's surface?

Water flows over a dam at the rate of 680 kg/s and falls vertically 88 m before striking the turbine blades. Calculate (\(a\)) the speed of the water just before striking the turbine blades (neglect air resistance), and (\(b\)) the rate at which mechanical energy is transferred to the turbine blades, assuming 55% efficiency.

An elevator cable breaks when a 925-kg elevator is 28.5 m above the top of a huge spring \((k = 8.00 \times 10^4 N/m)\) at the bottom of the shaft. Calculate (\(a\)) the work done by gravity on the elevator before it hits the spring; (\(b\)) the speed of the elevator just before striking the spring; (\(c\)) the amount the spring compresses (note that here work is done by both the spring and gravity).

(II) A box of mass 4.0 kg is accelerated from rest by a force across a floor at a rate of 2.0 m/s\(^2\) for 7.0 s. Find the net work done on the box.

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