/*! 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 16 In Concept Simulation 10.3 at yo... [FREE SOLUTION] | 91Ó°ÊÓ

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In Concept Simulation 10.3 at you can explore the concepts that are important in this problem. 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) The spring exerts a force of 9.84 N in the negative x-direction. (b) The angular frequency is 10.46 rad/s. (c) The maximum speed is 1.26 m/s. (d) The maximum acceleration is 13.1 m/s².

Step by step solution

01

Calculate the Force Exerted by the Spring

We use Hooke's Law to find the force that the spring exerts on the block just before release. Hooke's Law is given by \[ F = -kx \]where \( F \) is the force exerted by the spring, \( k \) is the spring constant, and \( x \) is the displacement from the equilibrium position. Given, \( k = 82.0 \, \mathrm{N/m} \) and \( x = +0.120 \, \mathrm{m} \), substitute these values into the formula:\[ F = -(82.0) \times (0.120) = -9.84 \, \mathrm{N} \]Thus, the magnitude of the force is 9.84 N, and it acts in the negative \( x \)-direction because the spring force opposes the displacement.
02

Calculate the Angular Frequency

The angular frequency \( \omega \) of oscillatory motion is given by the formula:\[ \omega = \sqrt{\frac{k}{m}} \]where \( k \) is the spring constant, and \( m \) is the mass. Substitute the given values, \( k = 82.0 \, \mathrm{N/m} \) and \( m = 0.750 \, \mathrm{kg} \):\[ \omega = \sqrt{\frac{82.0}{0.750}} = \sqrt{109.333} \approx 10.46 \, \mathrm{rad/s} \]
03

Find the Maximum Speed

The maximum speed \( v_{max} \) of the block occurs at the equilibrium position and is given by:\[ v_{max} = \omega x_{max} \]where \( \omega \) is the angular frequency and \( x_{max} \) is the maximum displacement. Substitute \( \omega = 10.46 \, \mathrm{rad/s} \) and \( x_{max} = 0.120 \, \mathrm{m} \):\[ v_{max} = 10.46 \times 0.120 = 1.2552 \, \mathrm{m/s} \]
04

Calculate the Maximum Acceleration

The maximum acceleration \( a_{max} \) is given by:\[ a_{max} = \omega^2 x_{max} \]Substitute \( \omega = 10.46 \, \mathrm{rad/s} \) and \( x_{max} = 0.120 \, \mathrm{m} \):\[ a_{max} = (10.46)^2 \times 0.120 = 13.1 \, \mathrm{m/s^2} \]

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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 fundamentally describes the relationship between the force exerted by a spring and its displacement from the equilibrium position. It is mathematically expressed as \( F = -kx \). Here, \( F \) represents the force by the spring, \( k \) is the spring constant, and \( x \) is the displacement. This negative sign signifies that the spring force acts in the opposite direction of displacement, aiming to restore equilibrium.

When a spring is stretched or compressed, the force it applies is directly proportional to the amount of stretch or compression, characterized by the spring constant \( k \). In our scenario, with \( k = 82.0 \, \mathrm{N/m} \), a displacement of \( +0.120 \, \mathrm{m} \) leads to a force of \( -9.84 \, \mathrm{N} \), indicating that the spring pulls back towards equilibrium in the negative direction.
Angular Frequency
Angular frequency, a key element in simple harmonic motion, tells us how quickly an object oscillates back and forth. It is represented by \( \omega \) and given by:\[ \omega = \sqrt{\frac{k}{m}} \]

In simple harmonic oscillators like springs, \( \omega \) (in radians per second) helps determine the motion's speed and periodicity. For our block-spring system, substituting the given spring constant \( k = 82.0 \, \mathrm{N/m} \) and mass \( m = 0.750 \, \mathrm{kg} \), results in \( \omega \approx 10.46 \, \mathrm{rad/s} \).

This value of angular frequency implies the block completes its oscillatory movement fairly quickly, making it crucial for evaluating the system's dynamic properties.
Maximum Speed
Maximum speed of an object in simple harmonic motion is when it passes through its equilibrium position. This maximum speed is captured by the formula:

\[ v_{max} = \omega x_{max} \]

Here, \( v_{max} \) is the maximum speed, \( \omega \) is the angular frequency, and \( x_{max} \) is the maximum displacement. For the spring system, using \( \omega = 10.46 \, \mathrm{rad/s} \) and \( x_{max} = 0.120 \, \mathrm{m} \), we calculate a maximum speed of \( 1.2552 \, \mathrm{m/s} \).

The maximum speed occurs as energy in the spring transfers from potential to kinetic, reaching its peak at this point. This is crucial, as it links to how energy fluctuates in oscillatory systems.
Maximum Acceleration
In simple harmonic motion, maximum acceleration occurs where displacement is maximal—at the endpoints of the motion. The formula for maximum acceleration is given as:

\[ a_{max} = \omega^2 x_{max} \]

This relates acceleration not only to the maximum displacement \( x_{max} \), but also to the square of the angular frequency \( \omega \). For our case, with \( \omega = 10.46 \, \mathrm{rad/s} \) and \( x_{max} = 0.120 \, \mathrm{m} \), the maximum acceleration is \( 13.1 \, \mathrm{m/s^2} \).

Understanding maximum acceleration helps us comprehend how forces change as the object undergoes oscillations. It provides insights into the potential stresses and strains exerted within the system at regular intervals.

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Most popular questions from this chapter

A spring is hung from the ceiling. A \(0.450-\mathrm{kg}\) block is then attached to the free end of the spring. When released from rest, the block drops \(0.150 \mathrm{~m}\) before momentarily coming to rest. (a) What is the spring constant of the spring? (b) Find the angular frequency of the block's vibrations.

A horizontal spring is lying on a frictionless surface. One end of the spring is attached to a wall while the other end is connected to a movable object. The spring and object are compressed by \(0.065 \mathrm{~m}\), released from rest, and subsequently oscillate back and forth with an angular frequency of \(11.3 \mathrm{rad} / \mathrm{s}\). What is the speed of the object at the instant when the spring is stretched by \(0.048 \mathrm{~m}\) relative to its unstrained length?

A \(0.70-\mathrm{kg}\) block is hung from and stretches a spring that is attached to the ceiling. A second block is attached to the first one, and the amount that the spring stretches from its unstrained length triples. What is the mass of the second block?

A person who weighs \(670 \mathrm{~N}\) steps onto a spring scale in the bathroom, and the spring compresses by \(0.79 \mathrm{~cm}\). (a) What is the spring constant? (b) What is the weight of another person who compresses the spring by \(0.34 \mathrm{~cm}\) ?

A rifle fires a \(2.10 \times 10^{-2}\) 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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