/*! 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 35 A 2.00-kg frictionless block att... [FREE SOLUTION] | 91Ó°ÊÓ

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A 2.00-kg frictionless block attached to an ideal spring with force constant 315 N/m is undergoing simple harmonic motion. When the block has displacement \(+\)0.200 m, it is moving in the negative \(x\)-direction with a speed of 4.00 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
Amplitude: 0.233 m; Max Acceleration: 36.7 m/s²; Max Force: 73.4 N.

Step by step solution

01

Understand the Problem

We have a block of mass 2.00 kg attached to a spring with a spring constant of 315 N/m. The block is undergoing simple harmonic motion, meaning both potential and kinetic energy interchange but the total energy remains constant. We need to find the amplitude, maximum acceleration, and the maximum force exerted by the spring when the block is in motion with certain known displacement and velocity.
02

Use Energy Conservation to find Amplitude

Simple harmonic motion in a spring system follows the principle of conservation of energy, meaning total mechanical energy stays constant:\[ E = \frac{1}{2}kx^2 + \frac{1}{2}mv^2 = \frac{1}{2}kA^2 \]Given: displacement \(x = 0.200\) m, spring constant \(k = 315\) N/m, speed \(v = 4.00\) m/s, mass \(m = 2.00\) kg. We start by calculating the total energy at the known state and equate it to the energy expression in terms of amplitude \(A\):\[E = \frac{1}{2}(315)(0.200)^2 + \frac{1}{2}(2.00)(4.00)^2 \]Calculate these individual energies, solving for \(A\).
03

Calculate Maximum Acceleration

The acceleration is maximum when the block is at the maximum displacement, i.e., at \(x = A\). It is calculated using the formula:\[a_{max} = \frac{k}{m} imes A\]Once the amplitude \(A\) is known from the previous step, substitute \(k = 315\) N/m, and \(m = 2.00\) kg to find \(a_{max}\).
04

Determine Maximum Force Exerted by the Spring

The maximum force exerted by the spring occurs when the spring is at maximum stretch or compression, which is at amplitude \(A\). Use Hooke's Law:\[F_{max} = kA\]With the known \(k = 315\) N/m and calculated amplitude \(A\), find \(F_{max}\).

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Key Concepts

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

Spring Constant
The spring constant, often denoted as \( k \), is a fundamental property of a spring that describes how stiff the spring is. It is measured in newtons per meter (N/m) and tells us how much force is required to stretch or compress the spring by one meter.

In the case of simple harmonic motion, as we have with the block attached to the spring, the spring constant plays a crucial role. The block's interaction with the spring is what allows it to oscillate, creating the phenomena of simple harmonic motion. The spring constant here is given as 315 N/m, meaning for every meter the spring is stretched, a force of 315 newtons is exerted.
  • A higher spring constant means a stiffer spring.
  • A lower spring constant indicates a more flexible spring.
In formulas concerning springs, you often see \( k \) appearing when calculating forces or energies involved in spring motion.
Amplitude
Amplitude in simple harmonic motion is the maximal distance from the equilibrium position. It signifies the extreme points of motion. In other words, it’s the peak displacement of the block attached to the spring.

Finding amplitude involves using the conservation of energy principle. Energy in the system is conserved and oscillates between kinetic and potential energy.
  • Kinetic Energy (KE): \( \frac{1}{2}mv^2 \)
  • Potential Energy (PE): \( \frac{1}{2}kx^2 \)
Calculate the energy at a known displacement and velocity, and then solve for the maximum amplitude using these equations. This tells us the maximum extent of the block's back-and-forth motion.
When the block is at its maximum distance from the center, the speed is zero, and all energy is potential: \( \frac{1}{2}kA^2 \). Using known parameters, we solve for \( A \) to find how far the block moves from its resting point in either direction.
Maximum Acceleration
Maximum acceleration in simple harmonic motion occurs at the points of maximum displacement from the equilibrium. At these points, the spring is either most compressed or most stretched.

The formula used to find maximum acceleration, \( a_{max} \), is: \[ a_{max} = \frac{k}{m} \times A \] where \( k \) is the spring constant, \( m \) is the mass of the block, and \( A \) is the amplitude.

This formula tells us that the greater the spring constant and the displacement (amplitude combined), the higher the acceleration:
  • \( a_{max} \propto k \) means stiffness increases acceleration.
  • \( a_{max} \propto A \) means greater displacement increases acceleration.
Acceleration reaches its peak because the spring force is greatest when displacement is greatest, rapidly pulling the block back towards equilibrium.
Conservation of Energy
In simple harmonic motion, conservation of energy is a core principle. The total mechanical energy of the system remains constant, shifting between kinetic energy (movement) and potential energy (spring compression or extension).
Even though kinetic and potential energies fluctuate, their sum does not change, as shown in the equation:\[ E = \frac{1}{2}kx^2 + \frac{1}{2}mv^2 \] This essential formula helps determine various aspects of motion, such as amplitude.
  • At maximum speed, energy is purely kinetic.
  • At maximum displacement, energy is purely potential.
Understanding this energy exchange allows us to solve complex motion problems by focusing on energy parameters, ensuring accuracy without always needing direct force or positional calculations. This foundational concept simplifies the analysis of oscillating systems like our spring-block model.

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Most popular questions from this chapter

A 175-g glider on a horizontal, frictionless air track is attached to a fixed ideal spring with force constant 155 N/m. At the instant you make measurements on the glider, it is moving at 0.815 m/s and is 3.00 cm from its equilibrium point. Use \({energy\ conservation}\) to find (a) the amplitude of the motion and (b) the maximum speed of the glider. (c) What is the angular frequency of the oscillations?

A 0.500-kg glider, attached to the end of an ideal spring with force constant \(k =\) 450 N/m, undergoes SHM with an amplitude of 0.040 m. Compute (a) the maximum speed of the glider; (b) the speed of the glider when it is at \(x = -\)0.015 m; (c) the magnitude of the maximum acceleration of the glider; (d) the acceleration of the glider at \(x = -\)0.015 m; (e) the total mechanical energy of the glider at any point in its motion.

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.

An object with mass 0.200 kg is acted on by an elastic restoring force with force constant 10.0 N/m. (a) Graph elastic potential energy \(U\) as a function of displacement \(x\) over a range of \(x\) from \(-\)0.300 m to \(+\)0.300 m. On your graph, let 1 cm \(=\) 0.05 J vertically and 1 cm \(=\) 0.05 m horizontally. The object is set into oscillation with an initial potential energy of 0.140 J and an initial kinetic energy of 0.060 J. Answer the following questions by referring to the graph. (b) What is the amplitude of oscillation? (c) What is the potential energy when the displacement is one-half the amplitude? (d) At what displacement are the kinetic and potential energies equal? (e) What is the value of the phase angle \(\phi\) if the initial velocity is positive and the initial displacement is negative?

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?

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