/*! 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 27 A \(2.50 \mathrm{~kg}\) mass is ... [FREE SOLUTION] | 91Ó°ÊÓ

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A \(2.50 \mathrm{~kg}\) mass is pushed against a horizontal spring of force constant \(25.0 \mathrm{~N} / \mathrm{cm}\) on a frictionless air table. The spring is attached to the tabletop, and the mass is not attached to the spring in any way. When the spring has been compressed enough to store \(11.5 \mathrm{~J}\) of potential energy, the mass is suddenly released from rest. (a) Find the greatest speed the mass reaches. When does this occur? (b) What is the greatest acceleration of the mass, and when does it occur?

Short Answer

Expert verified
(a) Greatest speed: 3.03 m/s at full spring extension. (b) Greatest acceleration: 96 m/s² at maximum spring compression.

Step by step solution

01

Convert Spring Constant to Proper Units

Given the spring constant is 25.0 N/cm, convert it to N/m: \[ k = 25.0 \, \text{N/cm} \times 100 \, \text{cm/m} = 2500 \, \text{N/m} \]
02

Use Energy Conservation to Find Maximum Speed

The potential energy stored in the spring is equal to the kinetic energy of the mass at maximum speed. Set them equal:\[ \frac{1}{2} k x^2 = \frac{1}{2} m v^2 \]Substitute the known values:\[ 11.5 = \frac{1}{2} \times 2.5 \, v^2 \]**Solve for \( v \):**\[ v^2 = \frac{11.5 \times 2}{2.5} \]\[ v^2 = \frac{23}{2.5} \approx 9.2 \]\[ v \approx 3.03 \, \text{m/s} \]The greatest speed, 3.03 m/s, occurs when all the potential energy is converted to kinetic energy.
03

Determine Greatest Acceleration

The greatest force occurs when the spring is maximally compressed, which is when the acceleration is greatest. Hooke's Law gives us:\[ F = kx \]And using Newton's Second Law, \( F = ma \), we have:\[ a = \frac{kx}{m} \]First find \( x \) using potential energy:\[ \frac{1}{2} kx^2 = 11.5 \]\[ x^2 = \frac{11.5 \times 2}{2500} \]\[ x^2 = 0.0092 \]\[ x \approx 0.096 \, \text{m} \]Substitute \( x \) back into the acceleration formula:\[ a = \frac{2500 \times 0.096}{2.5} \]\[ a \approx 96 \, \text{m/s}^2 \]The greatest acceleration, 96 m/s², occurs at maximum compression of the spring.

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Key Concepts

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

Conservation of Energy
The principle of conservation of energy states that in a closed system, the total energy remains constant. Energy can be changed from one form to another, but it cannot be created or destroyed. In our exercise, this principle is crucial in determining the speed of the mass. When the spring is compressed, it stores energy in the form of potential energy.

As the mass is released, this potential energy is converted into kinetic energy. The transformation from potential to kinetic energy allows us to calculate the maximum speed of the mass. At the point of maximum speed, all the potential energy has been converted into kinetic energy. This is when we can set the equations for potential and kinetic energy equal to each other to solve for the desired unknowns.
  • Potential Energy (initially stored): \( U = \frac{1}{2}kx^2 \)
  • Kinetic Energy (at maximum speed): \( K = \frac{1}{2}mv^2 \)
This transformation is a perfect demonstration of energy conservation, showing us how forces and motion are deeply related through energy changes.
Hooke's Law
Hooke's Law is a fundamental principle in physics that describes how springs work. It states that the force needed to either compress or extend a spring by some distance scales linearly. This relationship is captured with the formula:
  • \( F = -kx \)
The constant \( k \) represents the spring's stiffness, and \( x \) describes how far the spring is stretched or compressed from its equilibrium position.

In the context of the given exercise, we use Hooke's Law to express the force exerted by the spring on the mass. When the spring is compressed, according to Hooke's Law, it exerts a force proportional to the compression distance. This force is responsible for providing the acceleration to the mass once it is released. Understanding Hooke's Law enables us to calculate both the force the spring exerts and the maximum acceleration of the mass.
Newton's Second Law
Newton's Second Law of Motion is an essential concept for understanding movement and forces. It is expressed with the formula:
  • \( F = ma \)
This formula tells us that the acceleration \( a \) of an object is directly proportional to the net force \( F \) acting on it and inversely proportional to its mass \( m \).

In our exercise, once the spring is compressed and the mass is released, the force generated by the spring (calculated using Hooke's Law) results in the acceleration of the mass. Combining Newton's Second Law and Hooke's Law enables us to solve for the mass's acceleration. The greatest acceleration occurs at the moment of maximum spring compression, when the spring is exerting its maximum force. This law is a key backdrop for interpreting how forces cause changes in motion, particularly in systems involving springs.
Spring Potential Energy
The potential energy stored in a spring due to its compression or extension is referred to as spring potential energy. It's calculated using:
  • \( U = \frac{1}{2}kx^2 \)
This equation shows that spring potential energy is dependent on both the stiffness of the spring (\( k \)) and the square of the distance (\( x \)) that it is compressed or stretched from its natural position.

For our exercise, understanding spring potential energy is crucial for determining the initial energy transfer to the mass. The stored energy in the spring is released as kinetic energy of the mass, propelling it at its maximum speed. This is a clear example of how potential energy is harnessed to do work, illustrating the seamless interaction between stored energy and motion. By mastering the concept of spring potential energy, you can analyze and solve various mechanical problems involving springs.

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Most popular questions from this chapter

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