/*! 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

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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).

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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.

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Most popular questions from this chapter

Between each pair of vertebrae in the spinal column is a cylindrical disc of cartilage. Typically, this disc has a radius of about \(3.0 \times 10^{-2} \mathrm{m}\) and a thickness of about \(7.0 \times 10^{-3} \mathrm{m} .\) The shear modulus of cartilage is \(1.2 \times 10^{7} \mathrm{N} / \mathrm{m}^{2}\) . Suppose that a shearing force of magnitude 11 \(\mathrm{N}\) is applied parallel to the top surface of the disc while the bottom surface remains fixed in place. How far does the top surface move relative to the bottom surface?

A hand exerciser utilizes a coiled spring. A force of 89.0 \(\mathrm{N}\) is required to compress the spring by 0.0191 \(\mathrm{m} .\) Determine the force needed to compress the spring by 0.0508 \(\mathrm{m} .\)

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.

A simple pendulum is made from a \(0.65-\mathrm{m}\) -long string and a small ball attached to its free end. The ball is pulled to one side through a small angle and then released from rest. After the ball is released, how much time elapses before it attains its greatest speed?

A 15.0 -kg block rests on a horizontal table and is attached to one end of a massless, horizontal spring. By pulling horizontally on the other end of the spring, someone causes the block to accelerate uniformly and reach a speed of 5.00 \(\mathrm{m} / \mathrm{s}\) in 0.500 \(\mathrm{s}\) . In the process, the spring is stretched by 0.200 \(\mathrm{m}\) . The block is then pulled at a constant speed of 5.00 \(\mathrm{m} / \mathrm{s}\) , during which time the spring is stretched by only 0.0500 \(\mathrm{m} .\) Find \((\mathrm{a})\) the spring constant of the spring and \((\mathrm{b})\) the coefficient of kinetic friction between the block and the table.
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