/*! 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 A \(1.00 \times 10^{-2}\) -kg bl... [FREE SOLUTION] | 91Ó°ÊÓ

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A \(1.00 \times 10^{-2}\) -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 of the SHM is approximately 0.072 meters (7.2 cm).

Step by step solution

01

Understand the System

We have a block-spring system where the block is moved on a frictionless surface. The spring is initially unstrained, meaning no initial potential energy. The block is given some initial velocity parallel to the spring. We need to find the amplitude of the simple harmonic motion (SHM) produced.
02

Define Given Variables

Let's define the given values: - Mass, \( m = 1.00 \times 10^{-2} \) kg- Spring constant, \( k = 124 \) N/m- Initial speed, \( v = 8.00 \) m/s.
03

Apply Energy Conservation

In SHM, the total mechanical energy (sum of kinetic and potential energy) is conserved. Initially, the potential energy in the spring is zero, and the kinetic energy is maximum, given by \( KE = \frac{1}{2}mv^2 \). At the maximum amplitude, all energy will be potential \( PE = \frac{1}{2}kA^2 \).
04

Set Initial Kinetic Energy Equal to Maximum Potential Energy

Set the initial kinetic energy equal to the potential energy when the displacement is equal to the amplitude: \[ \frac{1}{2}mv^2 = \frac{1}{2}kA^2 \]Substitute the known values:\[ \frac{1}{2}(1.00 \times 10^{-2} \text{ kg})(8.00 \text{ m/s})^2 = \frac{1}{2}(124 \text{ N/m})A^2 \]
05

Solve for Amplitude, A

Simplify and solve for \( A \): \[ (1.00 \times 10^{-2})(64) = 124A^2 \]\[ 0.64 = 124A^2 \]\[ A^2 = \frac{0.64}{124} \]\[ A = \sqrt{\frac{0.64}{124}} \]\[ A \approx 0.072 \]
06

Conclude

This is the maximum displacement from the equilibrium position, hence the amplitude of the simple harmonic motion is approximately \( 0.072 \) meters, or 7.2 cm.

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Key Concepts

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

Amplitude
In the context of simple harmonic motion (SHM), amplitude refers to the maximum displacement of the object from its equilibrium position. Imagine pushing a swing and observing how far it moves to one side: that's the amplitude. Amplitude is always a positive value, representing the farthest extent the system moves from its central, rest position. It's important in describing the intensity of the oscillation. In the exercise, the amplitude is calculated using energy principles, where the maximum kinetic energy is converted into potential energy at the amplitude. By determining how far the object can move when all initial kinetic energy is stored as potential energy in the spring, we calculate the amplitude as approximately 0.072 meters.
Energy Conservation
Energy conservation is a cornerstone of physics, especially in simple harmonic motion. It asserts that when no external forces like friction act on a system, the total mechanical energy remains constant. For our block and spring system, this involves converting energy between kinetic energy (energy of motion) and potential energy (stored energy in the spring). Initially, the block has maximum kinetic energy due to its initial speed. As the block compresses the spring, its speed decreases and its kinetic energy is converted into potential energy until it momentarily stops at the maximum amplitude, where the potential energy is at its peak. This seamless conversion underscores the conservation of energy: the initial kinetic energy equals the maximum potential energy.
Spring Constant
The spring constant, often denoted by the letter \( k \), is a measure of a spring's stiffness. The larger the spring constant, the stiffer the spring is, meaning more force is required to compress or extend the spring by a certain distance. In our exercise, the spring constant is given as 124 N/m. This value is crucial for determining how much the spring can store energy when compressed or stretched. It plays a vital role in calculating the amplitude, as potential energy is given by \( \frac{1}{2}kA^2 \). Hence, both the spring constant and initial energy directly influence the amplitude and the nature of the oscillations.
Frictionless Surface
A frictionless surface in physics problems is an idealization that simplifies calculations by neglecting any forces that could cause energy dissipation. This means that an object can move without any energy loss due to friction. In this problem, as the 0.01-kg block slides on a frictionless surface, all its initial kinetic energy is conserved and can fully contribute to the oscillation, transforming into potential energy and back without any loss. This ideal condition helps us focus on the principles of energy conversion without additional complexities like frictional force, ensuring simpler and more focused problem solving.

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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.

Multiple-Concept Example 6 reviews the principles that play roles in this problem. A bungee jumper, whose mass is 82 kg, jumps from a tall platform. After reaching his lowest point, he continues to oscillate up and down, reaching the low point two more times in 9.6 s. Ignoring air resistance and assuming that the bungee cord is an ideal spring, determine its spring constant.

A 6.8-kg bowling ball is attached to the end of a nylon cord with a cross- sectional area of \(3.4 \times 10^{-5} \mathrm{m}^{2} .\) The other end of the cord is fixed to the ceiling. When the bowling ball is pulled to one side and released from rest, it swings downward in a circular arc. At the instant it reaches its lowest point, the bowling ball is \(1.4 \mathrm{m}\) lower than the point from which it was released, and the cord is stretched \(2.7 \times 10^{-3} \mathrm{m}\) from its unstrained length. What is the unstrained length of the cord? (Hint: When calculating any quantity other than the strain, ignore the increase in the length of the cord.\()\)

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?

A heavy-duty stapling gun uses a 0.140 -kg metal rod that rams against the staple to eject it. The rod is attached to and pushed by a stiff spring called a "ram spring" \((k=32000 \mathrm{N} / \mathrm{m})\). The mass of this spring may be ignored. The ram spring is compressed by \(3.0 \times 10^{-2} \mathrm{m}\) from its unstrained length and then released from rest. Assuming that the ram spring is oriented vertically and is still compressed by \(0.8 \times 10^{-2} \mathrm{m}\) when the downward-moving ram hits the staple, find the speed of the ram at the instant of contact.

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