/*! 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 33 \(\mathrm{A} 1.00 \times 10^{-2}... [FREE SOLUTION] | 91影视

91影视

Log out

Learning Materials

Features

Discover

Chapter 10: Problem 33

\(\mathrm{A} 1.00 \times 10^{-2}-\mathrm{kg}\) block is resting on a horizontal frictionless surface and is attached to a horizontal spring whose spring constant is 124 \(\mathrm{N} / \mathrm{m}\) . The block is shoved parallel to the spring axis and is given an initial speed of 8.00 \(\mathrm{m} / \mathrm{s}\) , while the spring is initially unstrained. What is the amplitude of the resulting simple harmonic motion?

Short Answer

Expert verified
The amplitude is approximately 0.0718 m (7.18 cm).

Step by step solution

01

Understand the Problem

We need to find the amplitude of the simple harmonic motion resulting from a block of mass resting on a frictionless surface attached to a spring. The block is given an initial speed along the direction of the spring axis, and the spring is initially unstrained.
02

Identify Given Values and Equations

The mass of the block is given as \(m = 1.00 \times 10^{-2}\, \text{kg}\), the spring constant as \(k = 124\, \text{N/m}\), and the initial speed of the block as \(v_0 = 8.00\, \text{m/s}\). The energy stored in the system can be equated as kinetic energy initially to potential energy when the spring is at maximum compression or extension.
03

Apply Conservation of Energy

Initially, the system has only kinetic energy, which is given by \( E_k = \frac{1}{2} m v_0^2 \). At the point of maximum amplitude, the energy is entirely potential, given by \(E_p = \frac{1}{2} k A^2\), where \(A\) is the amplitude. Set \(E_k = E_p\).
04

Solve for the Amplitude

Using the conservation of energy, set \(\frac{1}{2} m v_0^2 = \frac{1}{2} k A^2\). Solving for amplitude \(A\), we get:\[ A = \sqrt{\frac{m v_0^2}{k}} \]Substitute the given values: \( m = 1.00 \times 10^{-2} \), \( v_0 = 8.00 \), and \( k = 124 \).\[ A = \sqrt{\frac{(1.00 \times 10^{-2}) (8.00)^2}{124}} \]\[ A = \sqrt{\frac{(1.00 \times 10^{-2}) \times 64}{124}} \]\[ A = \sqrt{\frac{0.64}{124}} \]\[ A = \sqrt{0.0051613} \]\[ A \approx 0.0718 \, \text{m} \]
05

Final Calculation Check

Reevaluate the calculation by ensuring all arithmetic and algebra is correct and consistent with units. The calculations confirm that the amplitude \( A \) of the motion is approximately 0.0718 meters (or 7.18 cm).

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 91影视!

Key Concepts

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

Conservation of Energy
In simple harmonic motion, the conservation of energy principle is crucial. It states that the total mechanical energy in a closed system remains constant, provided no external forces do work.
For the block and spring system, this principle helps us understand how energy transforms. Initially, the block has kinetic energy due to its velocity. As it compresses or extends the spring, this kinetic energy is converted into potential energy.
  • Energy never disappears; it just changes form.
  • At maximum displacement (amplitude), all kinetic energy is transformed into potential energy.
  • The total energy is the sum of kinetic and potential energies.
In this exercise, the energy shifts back and forth between kinetic and potential as the block oscillates.
Spring Constant
The spring constant, often denoted by k, is a measure of a spring's stiffness.
A higher spring constant means a stiffer spring, requiring more force to stretch or compress. In our exercise, the block-spring system has a spring constant of 124 N/m.
  • This value tells us how strongly the spring resists deformation.
  • The spring constant directly affects the frequency and period of oscillation in simple harmonic motion.
  • In the equation for potential energy \(E_p = \frac{1}{2} k x^2\), k influences how quickly energy is stored or released.
Understanding the spring constant helps predict how the system will oscillate.
Amplitude
Amplitude in simple harmonic motion refers to the maximum extent of vibration measured from the equilibrium position. It represents the peak displacement of the block in our exercise.
The amplitude is determined by the system's initial energy, which stems from the block's initial speed.
  • It is symbolized by A and measured in meters.
  • A larger amplitude means greater maximum displacement from the rest position.
  • In the solution, amplitude is found using \(A = \sqrt{\frac{m v_0^2}{k}}\).
Amplitude reflects the energy introduced into the system initially.
Kinetic Energy
Kinetic energy is the energy of motion and is given by the expression \( E_k = \frac{1}{2} m v^2 \).
In this block-spring system, the initial kinetic energy arises from the block's velocity when first pushed.
  • This energy determines how far the block will move and subsequently the extent to which the spring will be compressed or stretched.
  • As the block slows down, its kinetic energy is converted to potential energy.
  • At the amplitude, the kinetic energy drops to zero temporarily.
This alternating conversion drives the block's oscillation in the spring system.
Potential Energy
Potential energy in a spring system is stored energy due to its position.
For the spring, this energy depends on how much it has been stretched or compressed, calculated using \(E_p = \frac{1}{2} k x^2\).
  • In our scenario, the potential energy is maximized when the spring is at its maximum compression or extension.
  • Potential energy reflects the capacity to do work when the spring returns to its original shape.
  • It transforms back into kinetic energy, continuing the cycle of motion.
Understanding potential energy helps us see how energy transitions maintain harmonic motion.

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

\(\mathrm{A} 1.00 \times 10^{-2}\) -kg bullet is fired horizontally into a \(2.50-\mathrm{kg}\) wooden block attached to one end of a massless horizontal spring \((k=845 \mathrm{N} / \mathrm{m})\) The other end of the spring is fixed in place, and the spring is unstrained initially. The block rests on a horizontal, frictionless surface. The bullet strikes the block perpendicularly and quickly comes to a halt within it. As a result of this completely inelastic collision, the spring is compressed along its axis and causes the block/bullet to oscillate with an amplitude of 0.200 \(\mathrm{m}\) . What is the speed of the bullet?

A block rests on a frictionless horizontal surface and is attached to a spring. When set into simple harmonic motion, the block oscillates back and forth with an angular frequency of 7.0 \(\mathrm{rad} / \mathrm{s}\) . The drawing indicates the position of the block when the spring is unstrained. This position is labeled \( x=0 \mathrm{m} .\) The drawing also shows a small bottle located 0.080 \(\mathrm{m}\) to the right of this position. The block is pulled to the right, stretching the spring by \(0.050 \mathrm{m},\) and is then thrown to the left. In order for the block to knock over the bottle, it must be thrown with a speed exceeding \(v_{0}\) . Ignoring the width of the block, find \(v_{0}\) .

The length of a simple pendulum is 0.79 \(\mathrm{m}\) and the mass of the particle (the bob) at the end of the cable is 0.24 \(\mathrm{kg}\) . The pendulum is pulled away from its equilibrium position by an angle of \(8.50^{\circ}\) and released from rest. Assume that friction can be neglected and that the resulting oscillatory motion is simple harmonic motion. (a) What is the angular frequency of the motion? (b) Using the position of the bob at its lowest point as the reference level, determine the total mechanical energy of the pendulum as it swings back and forth. (c) What is the bob's speed as it passes through the lowest point of the swing?

An object attached to a horizontal spring is oscillating back and forth along a frictionless surface. The maximum speed of the object is \(1.25 \mathrm{m} / \mathrm{s},\) and its maximum acceleration is 6.89 \(\mathrm{m} / \mathrm{s}^{2}\) . How much time elapses between an instant when the object's speed is at a maximum and the next instant when its acceleration is at a maximum?

A uniform \(1.4-\mathrm{kg}\) rod that is 0.75 \(\mathrm{m}\) long is suspended at rest from the ceiling by two springs, one at each end of the rod. Both springs hang straight down from the ceiling. The springs have identical lengths when they are unstretched. Their spring constants are 59 \(\mathrm{N} / \mathrm{m}\) and 33 \(\mathrm{N} / \mathrm{m}\) . Find the angle that the rod makes with the horizontal.
See all solutions

Recommended explanations on Physics Textbooks

Fundamentals of Physics

Read Explanation

Kinematics Physics

Read Explanation

Circular Motion and Gravitation

Read Explanation

Particle Model of Matter

Read Explanation

Rotational Dynamics

Read Explanation

Dynamics

Read Explanation
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.