/*! 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 initial speed \(v_0\) required is 0.56 m/s.

Step by step solution

01

Identify the Known Values

The problem gives several key parameters: the angular frequency \(\omega = 7.0 \, \text{rad/s}\), the initial displacement from the equilibrium \(x_0 = 0.050 \, \text{m}\), and the distance to the bottle \(d = 0.080 \, \text{m}\). The goal is to find the initial speed \(v_0\) required for the block to reach and knock over the bottle.
02

Apply Energy Conservation Principles

In simple harmonic motion, the total mechanical energy of the system is conserved. That means the sum of kinetic energy (KE) and potential energy (PE) at any point in time is constant:\[ \frac{1}{2}mv_0^2 = \frac{1}{2}k A^2 \]where \(v_0\) is the initial velocity, \(k\) is the spring constant, and \(A\) (the amplitude of the motion) is \(0.080 \, \text{m}\), the total stretch needed to knock over the bottle.
03

Relate Angular Frequency and Spring Constant

The angular frequency \(\omega\) is related to the spring constant \(k\) and the mass \(m\) by the formula:\[ \omega = \sqrt{\frac{k}{m}} \]Rearranging for \(k\):\[ k = m \omega^2 \]
04

Use Amplitude and Energy Relation

Substitute the expression for \(k\) from Step 3 into the energy conservation formula:\[ \frac{1}{2}mv_0^2 = \frac{1}{2}(m \omega^2) A^2 \]
05

Simplify and Solve for Initial Velocity

The mass \(m\) cancels out from both sides, simplifying to:\[ v_0^2 = \omega^2 A^2 \]Taking the square root gives:\[ v_0 = \omega A \]
06

Calculate the Required Initial Speed

Substitute the values for \(\omega\) and \(A\) into the equation:\[ v_0 = 7.0 \, \text{rad/s} \times 0.080 \, \text{m} = 0.56 \, \text{m/s} \]

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Key Concepts

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

Angular Frequency
Angular frequency in simple harmonic motion describes how quickly the oscillating system cycles through its motion. It is denoted by \( \omega \) and expressed in radians per second (rad/s). Angular frequency is related to the characteristics of the oscillating system, such as its mass and spring constant, through the formula:
  • \( \omega = \sqrt{\frac{k}{m}} \)
In this expression, \( k \) represents the spring constant, and \( m \) denotes the mass of the oscillating object. The angular frequency (\( 7.0 \, \text{rad/s} \) in this exercise) signifies the block's rate of oscillation when attached to the spring. By understanding angular frequency, you can gain insight into how rapidly the system oscillates and how it relates to energy conservation in simple harmonic motion. This frequency determines not only how fast the oscillation occurs but also how it interacts with other variables, such as initial speed, to influence motion outcomes.
Conservation of Energy
Conservation of Energy is a fundamental principle in physics that dictates energy within a closed system remains constant. In simple harmonic motion, this means the total mechanical energy—composed of kinetic energy (KE) and potential energy (PE)—stays the same throughout the movement.
The equation for energy conservation in this context is:
  • \( \frac{1}{2}mv_0^2 = \frac{1}{2}k A^2 \)
Here, \( v_0 \) is the initial velocity required to achieve the necessary movement, \( k \) is the spring constant, and \( A \) is the amplitude of motion. By employing this principle, the exercise determines the initial speed needed for the block to knock over the bottle.
In this scenario, understanding energy conservation allows you to calculate the energy required to perform different tasks within the system, considering both kinetic and potential energy as they interchange during oscillation. This helps identify the required conditions for an effective motion outcome, maintaining a balance between energy types.
Spring Constant
The spring constant, represented as \( k \), quantifies a spring's stiffness—defining the force required to compress or extend the spring by a unit length. It is measured in Newtons per meter (N/m).In simple harmonic motion, the spring constant \( k \) is pivotal as it influences both the frequency of oscillation and the energy dynamics. The relationship with angular frequency is:
  • \( \omega = \sqrt{\frac{k}{m}} \)
This relationship reflects how the spring constant, combined with the object's mass \( m \), affects the system. A higher spring constant means a stiffer spring, requiring more force for the same displacement, influencing not just the frequency but also the energy conservation aspects of the motion.Understanding the spring constant helps describe how the oscillating system behaves under various forces—evaluating how changes in \( k \) can impact the motion characteristics and desired outcomes within the harmonic motion setting.
Oscillation
Oscillation refers to the repetitive movement back and forth around an equilibrium position. In the context of simple harmonic motion, oscillation characterizes the energy transfer between kinetic and potential energy, pressing the object to move cyclically.Each full cycle of movement back to the starting point represents one complete oscillation. The angular frequency \( \omega \), amplitude \( A \), and object properties dictate how the oscillation unfolds.For the exercise, the block on a spring experiences oscillation as it moves due to the initial force applied. Each movement relies on the conservation of energy to shift between kinetic energy as the block speeds up and potential energy as the spring stretches further.Understanding oscillation is crucial to predict the system’s future positions over time, determining how far or quickly it moves given specific initial conditions—such as the speed \( v_0 \) needed to knock over the bottle. Recognizing these dynamics enables a holistic comprehension of periodic systems and their responses to different manipulations.

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Most popular questions from this chapter

A small object oscillates back and forth at the bottom of a frictionless hemispherical bowl, as the drawing illustrates. The radius of the bowl is \(R\), and the angle \(\theta\) is small enough that the object oscillates in simple harmonic motion. Derive an for the angular frequency \(\omega\) of the motion. Express your answer in terms of \(R\) and \(g,\) the magnitude of the acceleration due to gravity.

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?

A 1.0 \times 10 \(^{-3}\) -kg spider is hanging vertically by a thread that has a Young's modulus of \(4.5 \times 10^{9} \mathrm{N} / \mathrm{m}^{2}\) and a radius of \(13 \times 10^{-6} \mathrm{m} .\) Suppose that a \(95-\mathrm{kg}\) person is hanging vertically on an aluminum wire. What is the radius of the wire that would exhibit the same strain as the spider's thread, when the thread is stressed by the full weight of the spider?

A uniform 1.4-kg rod that is 0.75 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 N/m and 33 N/m. Find the angle that the rod makes with the horizontal.

A block of mass \(\mathrm{m}=0.750 \mathrm{kg}\) 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.

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