/*! 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 28 An ideal spring with spring cons... [FREE SOLUTION] | 91Ó°ÊÓ

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An ideal spring with spring constant \(k\) is hung from the ceiling and a block of mass \(M\) is attached to its lower end. The mass is released with the spring initially unstretched. Then the maximum extension in the spring is (a) \(\frac{4 \mathrm{Mg}}{k}\) (b) \(\frac{2 M g}{k}\) (c) \(\frac{\mathrm{Mg}}{\mathrm{k}}\) (d) \(\frac{M g}{2 k}\)

Short Answer

Expert verified
(b) \(\frac{2 Mg}{k}\) is the maximum extension.

Step by step solution

01

Understand the System

We have a spring with a spring constant \(k\) and a block of mass \(M\) attached to its end. The system is released from rest with the spring initially unstretched, meaning the spring's extension is initially zero.
02

Apply Conservation of Energy

As the block falls, gravitational potential energy is converted into the spring's elastic potential energy. We will use the principle of conservation of energy:\[ \text{Initial Potential Energy} = \text{Final Elastic Potential Energy} \] Initially, the gravitational potential energy is \(Mgh\), where \(h\) is the extension when the spring is fully extended. The final elastic potential energy stored in the spring is \(\frac{1}{2} k h^2\).
03

Set Up the Energy Equation

Using conservation of energy: \[ Mgx = \frac{1}{2} k x^2 \]where \(x\) is the maximum extension of the spring.
04

Solve for Maximum Spring Extension

Rearrange the energy equation from the previous step:\[ Mgx = \frac{1}{2} k x^2 \]Rearrange it to\[ k x^2 = 2Mgx \]Divide both sides by \(x\) (assuming \(x eq 0\)):\[ kx = 2Mg \]Now, solve for \(x\):\[ x = \frac{2Mg}{k} \]
05

Choose the Correct Answer

Compare the derived maximum extension \(x = \frac{2Mg}{k}\) with the provided options. The correct answer is option (b) \(\frac{2Mg}{k}\).

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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 is a fundamental concept in physics. It states that energy cannot be created or destroyed, only transformed from one form to another. In the context of a spring-mass system like our exercise, this principle is crucial.

When the block is released, it begins to fall due to gravity. Initially, all of the system's energy is in the form of gravitational potential energy, which depends on the height of the block. As the block falls, this energy doesn't disappear. Instead, it transforms into another type of energy—elastic potential energy of the spring.

This energy transformation is what allows us to set up an equation where the initial gravitational potential energy equals the final elastic potential energy. By doing this, we can solve for the maximum extension of the spring. It's like transferring money from one pocket to another without losing a dime, just changing its form or location.
Elastic Potential Energy
Elastic potential energy is the energy stored in elastic materials as the result of their stretching or compressing. For springs, it's the energy stored when the spring is either compressed or extended. The formula for elastic potential energy is:
  • \( U = \frac{1}{2} k x^2 \)
where \( U \) is the elastic potential energy, \( k \) is the spring constant, and \( x \) is the displacement from the unstretched position.

In our exercise, as the block falls and the spring stretches to its maximum extension, all of the block's gravitational potential energy is converted into elastic potential energy. The spring constant \( k \) tells us how stiff the spring is. A higher \( k \) means a stiffer spring, needing more force to stretch it. Understanding this helps explain why energy is stored efficiently in the spring as the block's kinetic energy decreases.
Gravitational Potential Energy
Gravitational potential energy is the energy possessed by an object due to its height above the ground or another reference point. This energy depends on three factors:
  • The mass of the object (\( M \)).
  • The height of the object (\( h \)).
  • The acceleration due to gravity (\( g \), approximately \( 9.8 \text{ m/s}^2 \) on Earth).
It can be calculated as:
  • \( U_{g} = Mgh \)
In the spring-mass system, this energy is what initiates the motion when the block is first released. As the block descends and the spring begins to stretch, we're observing gravitational potential energy being transformed into elastic potential energy.

The interplay of these forms of energy is crucial. It explains how the energy initially stored in the elevated block transitions into the spring once it reaches its maximum extension. Grasping this process reveals why the final extension can be calculated using the conservation principle, as shown by the solution's formula \( x = \frac{2Mg}{k} \).

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