/*! 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 81 A block rests on a frictionless ... [FREE SOLUTION] | 91Ó°ÊÓ

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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 \(v_{0}\).

Short Answer

Expert verified
The minimum speed \( v_0 \) needed is 0.56 m/s.

Step by step solution

01

Identify the Problem

We need to find the minimum speed, \(v_0\), required to knock over a bottle positioned at 0.080 m from the equilibrium position of a block attached to a spring. The block is set into motion by pulling it by 0.050 m and then releasing it.
02

Establish Known Values

The given angular frequency is \( \omega = 7.0 \ \text{rad/s} \), the initial displacement is \( x_0 = 0.050 \ \text{m} \), and the bottle is located at \( x_b = 0.080 \ \text{m} \). The equilibrium position is \( x = 0 \).
03

Use Energy Conservation in SHM

Since the block is in simple harmonic motion, the mechanical energy is conserved. The total energy can be expressed as the sum of kinetic and potential energies. Initially, all energy is potential: \( E = \frac{1}{2}k x_0^2 \). When the block starts moving, it will have both kinetic and potential energy.
04

Relate Angular Frequency to Spring Constant

The angular frequency \( \omega \) is related to the spring constant \( k \) and mass \( m \) by \( \omega = \sqrt{\frac{k}{m}} \). Rearrange this to find \( k = m\omega^2 \).
05

Calculate Maximum Displacement for Kinetic Energy

At the extreme right, the block must at least reach \( x=0.080 \ \text{m}\) to knock over the bottle. The maximum amplitude \( A \) of the block’s motion can be expressed as \( x_0 + v_0/\omega = A \).
06

Setup the Energy Equation

The energy conservation equation from the initial position to just before hitting the bottle is: \( \frac{1}{2} m v_0^2 + \frac{1}{2} k x_0^2 = \frac{1}{2} k A^2 \). Since \( A = 0.080 \ \text{m} \), substitute \( k = m\omega^2 \) and \( A = x_0 + v_0/\omega \) into the equation.
07

Solve for Initial Velocity \(v_0\)

Rearrange the energy equation to solve for \( v_0 \). The equation gives \( v_0^2 = \omega^2 (A^2 - x_0^2) \). Using \( A = 0.080 \ \text{m} \) and \( x_0 = 0.050 \ \text{m} \), calculate \( v_0 = \omega \sqrt{0.080^2 - 0.050^2} \).
08

Compute the Value

Substituting the values gives \( v_0 = 7.0 \sqrt{0.0064} = 7.0 \times 0.08 = 0.56 \ \text{m/s} \).

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Key Concepts

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

Angular Frequency in Simple Harmonic Motion
Angular frequency is an essential aspect of simple harmonic motion. It represents how quickly an object oscillates back and forth around the equilibrium position. In the context of a block attached to a spring, angular frequency, denoted as \( \omega \), is the rate at which the block completes one full cycle of motion. This is expressed in radians per second (rad/s). The given problem states that the block has an angular frequency of \( 7.0 \ ext{rad/s} \). This tells us that the block completes one oscillation very rapidly.

The relationship of angular frequency to the spring constant \( k \) and the mass \( m \) is given by \( \omega = \sqrt{\frac{k}{m}} \). This formula shows that the angular frequency depends on both how strong the spring is and the mass of the object in motion. A larger spring constant or a smaller mass results in a higher angular frequency, meaning the oscillations are quicker. Understanding this relationship helps in analyzing how changes in mass or spring strength affect the system's dynamics.
Understanding the Spring Constant
The spring constant, symbolized as \( k \), measures the stiffness of a spring. In the context of simple harmonic motion, it is a crucial factor that influences how the block moves back and forth. The spring constant tells us how much force is needed to stretch or compress the spring by a certain length. Mathematically, it relates to Hooke's law, \( F = -kx \), which states that the force needed to stretch or compress a spring is proportional to the displacement \( x \) from its equilibrium position.

If the spring constant is high, it means the spring is stiffer, requiring more force to create the same displacement compared to a spring with a lower constant. In simple harmonic motion, along with angular frequency, the spring constant helps determine how the energy is stored in the system. The potential energy stored in a spring is calculated using \( \frac{1}{2} k x^2 \), illustrating the energy's dependency on both the displacement and the stiffness of the spring.
Energy Conservation in Oscillating Systems
Energy conservation is a fundamental principle in physics, applying perfectly to simple harmonic motion. When an object like a block attached to a spring oscillates, its total mechanical energy remains constant if no external forces, like friction, act on the system. This principle is expressed as the sum of kinetic energy (energy due to motion) and potential energy (energy stored in the spring).

In the problem context, when the block is initially released after being pulled, it has maximum potential energy and zero kinetic energy. As it moves, this potential energy converts to kinetic energy until all the potential energy becomes kinetic energy when passing through the equilibrium position. This continuous change from potential to kinetic and back illustrates energy conservation. Eventually, the energy transferred must be enough to enable the block to reach a position where it can strike the bottle, necessitating enough initial velocity to ensure sufficient kinetic energy.
The Nature of Oscillation in Simple Harmonic Motion
Oscillation refers to the repetitive movement back and forth around an equilibrium position, like the movement of a block attached to a spring. In simple harmonic motion, oscillation is a smooth, continuous motion characterized by a sinusoidal pattern. The movement repeats over equal intervals of time, showcasing the periodic nature of oscillations.

The problem describes an initial displacement where the block is pulled from its rest position, instigating oscillation. This displacement initiates the back-and-forth motion governed by the spring's force. Important parameters such as amplitude, frequency, and the period of oscillations define how the oscillation behaves. The amplitude is the maximum distance from the equilibrium, which in the problem context is 0.050 m initially, indicating how far the block was stretched. The frequency is directly connected to the angular frequency, dictating how quickly each oscillation cycle completes. Recognizing these aspects aids in predicting the system's motion and calculating necessary conditions to achieve tasks, like knocking over the bottle.

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Most popular questions from this chapter

An archer, about to shoot an arrow, is applying a force of \(+240 \mathrm{N}\) to a drawn bowstring. The bow behaves like an ideal spring whose spring constant is \(480 \mathrm{N} / \mathrm{m} .\) What is the displacement of the bowstring?

Depending on how you fall, you can break a bone easily. The severity of the break depends on how much energy the bone absorbs in the accident, and to evaluate this let us treat the bone as an ideal spring. The maximum applied force of compression that one man's thighbone can endure without breaking is \(7.0 \times 10^{4} \mathrm{N} .\) The minimum effective cross-sectional area of the bone is \(4.0 \times 10^{-4} \mathrm{m}^{2},\) its length is \(0.55 \mathrm{m},\) and Young's modulus is \(Y=9.4 \times 10^{9} \mathrm{N} / \mathrm{m}^{2} .\) The mass of the man is \(65 \mathrm{kg} .\) He falls straight down without rotating, strikes the ground stiff-legged on one foot, and comes to a halt without rotating. To see that it is easy to break a thighbone when falling in this fashion, find the maximum distance through which his center of gravity can fall without his breaking a bone.

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?

You are running from pirates on a tropical island somewhere in the Caribbean. You have somehow become separated from the rest of your group and now find yourself on the edge of a cliff with your pursuers less than 10 minutes behind you. According to a sign posted on the guardrail at the cliff's edge, the drop to the beach below is \(h=140\) feet. Your team members (waiting for you on the beach, near your boat) have a rope, but there is no time for anyone to climb the cliff to save you. You break into a deserted cabin nearby, and rummage around for a rope. Instead, you find a brand new, still-in-package, bungee cord that must have been intended for tourists jumping from a nearby bridge. You figure you might be able to attach it to the guardrail and jump to the beach, letting go at the bottom before it reverses your motion. You read the bungee cord specifications on the package: the total length of the cord is \(L_{0}=100 \mathrm{m},\) the maximum elastic deformation is \(200 \%\) (i.e., it can safely triple its length), and the elastic constant is \(k=75.0 \mathrm{N} / \mathrm{m}\). (a) If you weigh \(170 \mathrm{lb},\) how far is the bungee cord designed to let you fall before it stops you and reverses your direction? Will this afford you a safe landing? (b) You realize that you don't have to hang from the very end of the bungee, but rather from some point in the middle. How far from the attached end should you grasp the unstretched bungee cord so that you land softly on the beach? Will you be able to perform the jump and stay under the elastic deformation limit?

You are given the task of opening an antiquated "light lock," which is unlocked by shining red light pulses of a certain frequency for a long duration of time into a light sensor on the lock. You are given a red laser pointer, a spring of unstretched length \(L=15.0 \mathrm{cm}\) and spring constant \(k=7.20 \mathrm{N} / \mathrm{m},\) a sheet of steel \(\left(\rho=7.60 \mathrm{g} / \mathrm{cm}^{3}\right)\) that is 0.125 inches thick, and some tools. You come up with the idea to take a piece of the steel sheet (of mass \(m\) ), cut a slot in it, and hang it from the spring. If you shine the laser through the slot and onto the sensor, and then stretch the spring and let it go, the steel plate will oscillate and cause the beam to pass through the slot periodically. (a) Assuming the beam is passing through the slot (and onto the lock's sensor) when the spring-mass system is in equilibrium, how is the frequency at which the light pulses hit the sensor related to the frequency of the spring/mass (i.e., steel plate) system. (b) Based on your answer for (a), what should be the frequency of the spring/mass system if the unlocking frequency is \(3.50 \mathrm{Hz} ?\) (c) What should be the mass \(m\) of the steel plate? (d) Calculate some reasonable dimensions for the steel plate (i.e., they should be consistent with the mass that is required for the spring-mass system). You may assume the material removed to make the slot in the steel plate is of negligible mass.

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