/*! 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 48 A body is suspended vertically f... [FREE SOLUTION] | 91Ó°ÊÓ

91Ó°ÊÓ

A body is suspended vertically from an ideal spring of spring constant $2.5 \mathrm{N} / \mathrm{m}$. The spring is initially in its relaxed position. The body is then released and oscillates about its equilibrium position. The motion is described by $$y=(4.0 \mathrm{cm}) \sin [(0.70 \mathrm{rad} / \mathrm{s}) t]$$ What is the maximum kinetic energy of the body?

Short Answer

Expert verified
Answer: The maximum kinetic energy of the body is \(0.050\mathrm{J}\).

Step by step solution

01

Find the amplitude and angular frequency

In the equation of motion, the amplitude (A) and angular frequency (ω) can be found directly: $$y=(4.0\mathrm{cm})\sin[(0.70 \mathrm{rad}/\mathrm{s})t]$$. From this equation, we can see that the amplitude is \(A=4.0 \mathrm{cm}=0.040\mathrm{m}\) and the angular frequency is \(\omega=0.70 \mathrm{rad/s}\).
02

Find the mass of the body

We can find the mass of the body by using the relationship between the spring constant (k), the angular frequency (ω), and the mass (m): $$\omega=\sqrt{\frac{k}{m}}$$. We are given the spring constant \(k=2.5 \mathrm{N/m}\). To find the mass, rearrange the equation and solve for m: $$m=\frac{k}{\omega^2}$$. Now, plug in the values for k and ω: $$m=\frac{2.5\mathrm{N/m}}{(0.70\mathrm{rad/s})^2}$$.
03

Calculate the mass

Calculate the mass using the values provided: $$m=\frac{2.5\mathrm{N/m}}{(0.70\mathrm{rad/s})^2}=5.10\mathrm{kg}$$
04

Calculate the maximum kinetic energy

Now that we have the mass, amplitude, and angular frequency, we can find the maximum kinetic energy (K_max) using the equation for the maximum kinetic energy of a simple harmonic oscillator: $$K_\text{max}=\frac{1}{2}m\omega^2A^2$$. Plug in the values for m, ω, and A: $$K_\text{max}=\frac{1}{2}(5.10\mathrm{kg})(0.70\mathrm{rad/s})^2(0.040\mathrm{m})^2$$.
05

Find the maximum kinetic energy

Calculate the maximum kinetic energy using the values provided: $$K_\text{max}=\frac{1}{2}(5.10\mathrm{kg})(0.70\mathrm{rad/s})^2(0.040\mathrm{m})^2=0.050\mathrm{J}$$. The maximum kinetic energy of the body is \(0.050\mathrm{J}\).

Unlock Step-by-Step Solutions & Ace Your Exams!

  • Full Textbook Solutions

    Get detailed explanations and key concepts

  • Unlimited Al creation

    Al flashcards, explanations, exams and more...

  • Ads-free access

    To over 500 millions flashcards

  • Money-back guarantee

    We refund you if you fail your exam.

Over 30 million students worldwide already upgrade their learning with 91Ó°ÊÓ!

One App. One Place for Learning.

All the tools & learning materials you need for study success - in one app.

Get started for free

Most popular questions from this chapter

The maximum height of a cylindrical column is limited by the compressive strength of the material; if the compressive stress at the bottom were to exceed the compressive strength of the material, the column would be crushed under its own weight. (a) For a cylindrical column of height \(h\) and radius \(r,\) made of material of density \(\rho,\) calculate the compressive stress at the bottom of the column. (b) since the answer to part (a) is independent of the radius \(r,\) there is an absolute limit to the height of a cylindrical column, regardless of how wide it is. For marble, which has a density of $2.7 \times 10^{3} \mathrm{kg} / \mathrm{m}^{3}\( and a compressive strength of \)2.0 \times 10^{8} \mathrm{Pa},$ find the maximum height of a cylindrical column. (c) Is this limit a practical concern in the construction of marble columns? Might it limit the height of a beanstalk?
Four people sit in a car. The masses of the people are $45 \mathrm{kg}, 52 \mathrm{kg}, 67 \mathrm{kg},\( and \)61 \mathrm{kg} .\( The car's mass is \)1020 \mathrm{kg}$. When the car drives over a bump, its springs cause an oscillation with a frequency of \(2.00 \mathrm{Hz}\). What would the frequency be if only the \(45-\mathrm{kg}\) person were present?
Resilin is a rubber-like protein that helps insects to fly more efficiently. The resilin, attached from the wing to the body, is relaxed when the wing is down and is extended when the wing is up. As the wing is brought up, some elastic energy is stored in the resilin. The wing is then brought back down with little muscular energy, since the potential energy in the resilin is converted back into kinetic energy. Resilin has a Young's modulus of $1.7 \times 10^{6} \mathrm{N} / \mathrm{m}^{2}$ (a) If an insect wing has resilin with a relaxed length of \(1.0 \mathrm{cm}\) and with a cross-sectional area of \(1.0 \mathrm{mm}^{2},\) how much force must the wings exert to extend the resilin to \(4.0 \mathrm{cm} ?\) (b) How much energy is stored in the resilin?
A sphere of copper is subjected to 100 MPa of pressure. The copper has a bulk modulus of 130 GPa. By what fraction does the volume of the sphere change? By what fraction does the radius of the sphere change?
A \(230.0-\mathrm{g}\) object on a spring oscillates left to right on a frictionless surface with a frequency of \(2.00 \mathrm{Hz}\). Its position as a function of time is given by \(x=(8.00 \mathrm{cm})\) sin \(\omega t\) (a) Sketch a graph of the elastic potential energy as a function of time. (b) The object's velocity is given by \(v_{x}=\omega(8.00 \mathrm{cm}) \cos \omega t .\) Graph the system's kinetic energy as a function of time. (c) Graph the sum of the kinetic energy and the potential energy as a function of time. (d) Describe qualitatively how your answers would change if the surface weren't frictionless.
See all solutions

Recommended explanations on Physics Textbooks

View all explanations

What do you think about this solution?

We value your feedback to improve our textbook solutions.

Study anywhere. Anytime. Across all devices.