/*! 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 17 A block of mass \(m=0.750 \mathr... [FREE SOLUTION] | 91影视

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Chapter 10: Problem 17

A block of mass \(m=0.750 \mathrm{kg}\) is fastened to an unstrained horizontal spring whose spring constant is \(k=82.0 \mathrm{N} / \mathrm{m} .\) The block is given a displacement of \(+0.120 \mathrm{m},\) where the \(+\) sign indicates that the displacement is along the \(+x\) axis, and then released from rest. (a) What is the force (magnitude and direction) that the spring exerts on the block just before the block is released? (b) Find the angular frequency \(\omega\) of the resulting oscillatory motion. (c) What is the maximum speed of the block? (d) Determine the magnitude of the maximum acceleration of the block.

Short Answer

Expert verified
(a) 9.84 N (left), (b) \(10.46\, \text{rad/s}\), (c) \(1.255\, \text{m/s}\), (d) \(13.122\, \text{m/s}^2\)."

Step by step solution

01

Determine the Force Exerted by the Spring

To find the force exerted by the spring on the block, use Hooke's Law: \[ F = -kx \]where \(k\) is the spring constant, and \(x\) is the displacement. Here, \(x = +0.120\, \text{m}\) and \(k = 82.0\, \text{N/m}\).Substitute these values into the equation:\[ F = -(82.0)(0.120) = -9.84\, \text{N} \]The negative sign indicates that the force is directed opposite to the displacement.
02

Calculate the Angular Frequency

The angular frequency \( \omega \) for a mass-spring system is given by:\[ \omega = \sqrt{\frac{k}{m}} \]where \( m = 0.750\, \text{kg} \) and \( k = 82.0\, \text{N/m} \).Substitute the values to find \( \omega \):\[ \omega = \sqrt{\frac{82.0}{0.750}} = \sqrt{109.33} \approx 10.46\, \text{rad/s} \]
03

Find the Maximum Speed of the Block

The maximum speed \( v_{max} \) occurs as the block passes through the equilibrium position. It can be calculated using the formula:\[ v_{max} = A\omega \]where \( A = 0.120\, \text{m} \) is the amplitude and \( \omega = 10.46\, \text{rad/s} \).Calculate \( v_{max} \):\[ v_{max} = 0.120 \times 10.46 = 1.255\, \text{m/s} \]
04

Determine the Maximum Acceleration

The maximum acceleration \( a_{max} \) is given by:\[ a_{max} = A\omega^2 \]where \( A = 0.120\, \text{m} \) and \( \omega = 10.46\, \text{rad/s} \).First, calculate \( \omega^2 = (10.46)^2 = 109.35 \text{rad}^2/\text{s}^2 \).Then, calculate \( a_{max} \):\[ a_{max} = 0.120 \times 109.35 = 13.122 \text{m/s}^2 \]

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Key Concepts

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

Hooke's Law
Hooke's Law is fundamental in understanding how springs behave when they are compressed or stretched. Simply put, Hooke's Law states that the force exerted by a spring is directly proportional to the displacement from its equilibrium position and acts in the direction opposite to this displacement. The mathematical expression for Hooke's Law is:
  • \( F = -kx \)
Where:
  • \( F \) is the force exerted by the spring,
  • \( k \) is the spring constant, a measure of the spring's stiffness, and
  • \( x \) is the displacement from the equilibrium position.
The negative sign indicates that the force is a restoring force, always aiming to bring the object back to its equilibrium point. Using this law, we calculated the force exerted by the spring in the problem as \(-9.84\, \text{N},\) with the negative indicating it is directed opposite to the displacement.
Angular Frequency
Angular frequency, denoted by \( \omega, \) is a crucial concept when discussing oscillatory motion like Simple Harmonic Motion (SHM). It describes how many radians an object moves through per second, which helps us understand the rate of oscillation.
  • The formula for angular frequency in a mass-spring system is:\[ \omega = \sqrt{\frac{k}{m}} \]Where:
    • \( k \) is the spring constant,
    • \( m \) is the mass attached to the spring.
Substituting the given values, the angular frequency for our problem is calculated as approximately \( 10.46 \, \text{rad/s}.\)This tells us that the mass completes one radian of motion every \(0.096\) seconds.
Maximum Speed
The maximum speed in a Simple Harmonic Motion occurs as the oscillating object passes through its equilibrium position. This point is where all the potential energy stored in the spring, when it is compressed or extended, is converted into kinetic energy.
  • To find the maximum speed, we use:\[ v_{max} = A\omega \]Where:
    • \( A \) is the amplitude, the maximum displacement from the equilibrium, and
    • \( \omega \) is the angular frequency.
In our case, substituting \( A = 0.120 \, \text{m} \) and \( \omega = 10.46 \, \text{rad/s}, \) results in a maximum speed of \( 1.255 \, \text{m/s}. \) This speed happens at the midpoint of the oscillation, showing how fast the object moves through equilibrium.
Amplitude
Amplitude is a measure of how far an object in harmonic motion moves from its equilibrium or rest position. In this context, it's the maximum extent of displacement during the oscillation.
  • The equation for the maximum speed in SHM depends on amplitude:\[ v_{max} = A\omega \]Where:
    • \( A \) is the amplitude, and
    • \( \omega \) is the angular frequency.
In other words, the larger the amplitude, the greater the potential speed and energy since the object travels a longer distance from rest before being pulled back by the spring. In the given exercise, the amplitude of \(0.120 \text{m}\) represents the furthest point reached by the block from its original position, emphasizing its role in the oscillation process.

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Most popular questions from this chapter

A vertical ideal spring is mounted on the floor and has a spring constant of \(170 \mathrm{N} / \mathrm{m} .\) A \(0.64-\mathrm{kg}\) block is placed on the spring in two different ways. (a) In one case, the block is placed on the spring and not released until it rests stationary on the spring in its equilibrium position. Determine the amount (magnitude only) by which the spring is compressed. (b) In a second situation, the block is released from rest immediately after being placed on the spring and falls downward until it comes to a momentary halt. Determine the amount (magnitude only) by which the spring is now compressed.

The femur is a bone in the leg whose minimum cross-sectional area is about \(4.0 \times 10^{-4} \mathrm{m}^{2} .\) A compressional force in excess of \(6.8 \times 10^{4} \mathrm{N}\) will fracture this bone. (a) Find the maximum stress that this bone can with- stand. (b) What is the strain that exists under a maximum-stress condition?

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.

A tray is moved horizontally back and forth in simple harmonic motion at a frequency of \(f=2.00 \mathrm{Hz} .\) On this tray is an empty cup. Obtain the coefficient of static friction between the tray and the cup, given that the cup begins slipping when the amplitude of the motion is \(5.00 \times 10^{-2} \mathrm{m} .\)

A rifle fires a \(2.10 \times 10^{-2}-\mathrm{kg}\) pellet straight upward, because the pellet rests on a compressed spring that is released when the trigger is pulled. The spring has a negligible mass and is compressed by \(9.10 \times 10^{-2} \mathrm{m}\) from its unstrained length. The pellet rises to a maximum height of \(6.10 \mathrm{m}\) above its position on the compressed spring. Ignoring air resistance, determine the spring constant.
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