/*! 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 34 A 2.00-kg frictionless block is ... [FREE SOLUTION] | 91Ó°ÊÓ

91Ó°ÊÓ

A 2.00-kg frictionless block is attached to an ideal spring with force constant 315 N/m. Initially the spring is neither stretched nor compressed, but the block is moving in the negative direction at 12.0 m/s. Find (a) the amplitude of the motion, (b) the block's maximum acceleration, and (c) the maximum force the spring exerts on the block.

Short Answer

Expert verified
(a) Amplitude: 0.957 m, (b) Maximum acceleration: 150.5 m/s², (c) Maximum force: 301.4 N.

Step by step solution

01

Understand the problem

We have a 2.00-kg block attached to a spring with a spring constant of 315 N/m. Initially, the spring is at equilibrium, but the block is moving at 12.0 m/s. We need to find the amplitude of the motion, the block's maximum acceleration, and the maximum force exerted by the spring.
02

Calculate the Total Mechanical Energy

The total mechanical energy in a spring-block system is given by the sum of kinetic and potential energy. Initially, all energy is kinetic: \( E = \frac{1}{2}mv^2 \). Substituting the given values, we have \( E = \frac{1}{2} \times 2.00 \times (12.0)^2 = 144 \text{ J} \).
03

Relate Mechanical Energy to Amplitude

At maximum displacement (amplitude), all the energy is potential: \( E = \frac{1}{2}kA^2 \). Equate this to the total mechanical energy to solve for amplitude: \( 144 = \frac{1}{2} \times 315 \times A^2 \). Solving for \( A \), we get \( A = \sqrt{\frac{2 \times 144}{315}} \approx 0.957 \text{ m} \).
04

Calculate Maximum Acceleration

Using Hooke's law, the maximum force occurs at maximum displacement: \( F_{max} = kA \). Then, using \( F = ma \), the maximum acceleration is \( a_{max} = \frac{F_{max}}{m} = \frac{kA}{m} \). Substituting values: \( a_{max} = \frac{315 \times 0.957}{2.00} \approx 150.5 \, \text{m/s}^2 \).
05

Calculate Maximum Force Therefore, the maximum force is given by Hooke's Law, which is the product of the spring constant and the amplitude: \( F_{max} = kA \). Using the calculated amplitude, \( A = 0.957 \text{ m} \), we have \( F_{max} = 315 \times 0.957 \approx 301.4 \text{ N} \).

Verify by recalculating Hooke's Law application: \( F_{max} = k \times A = 315 \times 0.957 \approx 301.4 \text{ N} \).

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!

Key Concepts

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

Spring Constant
The spring constant, often denoted by the symbol \( k \), is a fundamental concept in harmonic motion. It measures the stiffness of a spring. Think of \( k \) as a sort of "springiness" factor. More robust springs have higher values of \( k \), meaning they require more force to be stretched or compressed. In our problem, the spring constant is given as 315 N/m. This means that for every meter the spring is stretched or compressed, 315 newtons of force are needed. Understanding the spring constant is crucial because it directly affects the behavior of the spring in motion. A higher spring constant results in a stiffer spring, affecting the oscillation's frequency and potential energy stored.
Hooke's Law
Hooke's Law is a simple yet powerful principle that relates the force exerted by a spring to its displacement from equilibrium. Mathematically, it's expressed as:\[ F = -kx \]where
  • \( F \) is the force exerted by the spring,
  • \( k \) is the spring constant,
  • \( x \) is the displacement from the spring's natural length.
The negative sign indicates that the force exerted by the spring is restorative, meaning it acts in the opposite direction of displacement, trying to bring the spring back to its equilibrium position. In the context of the problem, Hooke's Law helps us determine the maximum force when the spring is at its maximum displacement (amplitude). This force is crucial for finding how the spring affects the block's motion and its acceleration.
Kinetic and Potential Energy
Understanding kinetic and potential energy is essential for solving problems involving harmonic motion. These two types of energy continually transform into each other in a spring-system. - **Kinetic Energy (KE):** This is the energy due to motion. For an object of mass \( m \) moving at velocity \( v \), kinetic energy is given by:\[ KE = \frac{1}{2}mv^2 \]In our scenario, the block initially has complete kinetic energy since it moves at 12.0 m/s while the spring is at equilibrium.- **Potential Energy (PE):** In springs, potential energy is stored as elastic potential energy. When the spring is compressed or stretched, potential energy is at its maximum. The equation for potential energy in a spring is:\[ PE = \frac{1}{2}kx^2 \]At the amplitude, the spring's potential energy equals the system's total initial mechanical energy.The total mechanical energy in a spring-block system remains constant, allowing us to use this relationship to find unknown quantities like amplitude, as energy transitions between kinetic and potential forms during the oscillation.

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

A machine part is undergoing SHM with a frequency of 4.00 Hz and amplitude 1.80 cm. How long does it take the part to go from \(\chi =\) 0 to \(\chi = -1.80\) cm?

After landing on an unfamiliar planet, a space explorer constructs a simple pendulum of length 50.0 cm. She finds that the pendulum makes 100 complete swings in 136 s. What is the value of \(g\) on this planet?

The point of the needle of a sewing machine moves in SHM along the \(x\)-axis with a frequency of 2.5 Hz. At \(t =\) 0 its position and velocity components are \(+\)1.1 cm and \(-\)15 cm/s, respectively. (a) Find the acceleration component of the needle at \(t =\) 0. (b) Write equations giving the position, velocity, and acceleration components of the point as a function of time.

Two pendulums have the same dimensions (length \(L\)) and total mass ( \(m\) ). Pendulum \(A\) is a very small ball swinging at the end of a uniform massless bar. In pendulum \(B\), half the mass is in the ball and half is in the uniform bar. Find the period of each pendulum for small oscillations. Which one takes longer for a swing?

A spring of negligible mass and force constant \(k =\) 400 N/m is hung vertically, and a 0.200-kg pan is suspended from its lower end. A butcher drops a 2.2-kg steak onto the pan from a height of 0.40 m. The steak makes a totally inelastic collision with the pan and sets the system into vertical SHM. What are (a) the speed of the pan and steak immediately after the collision; (b) the amplitude of the subsequent motion; (c) the period of that motion?

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.