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When a body of mass \(M\) is attached to lower end of a wire (of length \(L\) ) whose upper end is fixed, then the elongation of the wire is \(l\). In this situation, mark out the correct statement(s). (1) Loss in gravitational potential energy of \(M\) is \(M g l .\) (2) Elastic potential energy stored in the wire is \(\frac{M g l}{2}\). (3) Elastic potential energy stored in the wire is \(\mathrm{Mgl}\). (4) Elastic potential energy stored in the wire is \(\frac{M g l}{3}\).

Short Answer

Expert verified
The correct statements are (1) and (2).

Step by step solution

01

Understanding the Problem

In this problem, a mass \( M \) is suspended from a wire, causing the wire to elongate by a length \( l \). We need to determine which statements about potential energy are correct based on the given elongation.
02

Loss in Gravitational Potential Energy

When the mass \( M \) falls a distance \( l \), the gravitational potential energy decreases by \( Mgl \). This represents the loss in gravitational potential energy as the mass moves downward due to gravity.
03

Calculating Elastic Potential Energy

When the wire stretches, the energy stored in the wire as elastic potential energy can be calculated using the formula \( \frac{1}{2} k l^2 \), where \( k \) is the spring constant of the wire. Here, we need to find this energy in terms of \( M \), \( g \), and \( l \).
04

Using Hooke’s Law to Find Spring Constant

From Hooke's Law, \( F = k l \), where \( F = Mg \) is the force exerted by the mass. Therefore, \( k = \frac{Mg}{l} \).
05

Substitute Spring Constant into Elastic Potential Energy Formula

Substitute \( k = \frac{Mg}{l} \) into the elastic potential energy formula: \[ \text{Elastic Potential Energy} = \frac{1}{2} \left(\frac{Mg}{l}\right) l^2 = \frac{Mgl}{2} \]. This shows the elastic potential energy stored in the wire.
06

Evaluating the Statements

Based on the calculations, the correct statements are: (1) Loss in gravitational potential energy of \( M \) is \( M g l \), and (2) Elastic potential energy stored in the wire is \( \frac{M g l}{2} \). The other statements are incorrect.

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Key Concepts

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

Gravitational Potential Energy
Gravitational potential energy is a type of energy that depends on an object's position relative to the ground. When an object is lifted to a height, it gains gravitational potential energy. This energy is calculated using the formula:
  • \( ext{Gravitational Potential Energy} = mgh \)
where:
  • \( m \) is the mass of the object,
  • \( g \) is the acceleration due to gravity \( (9.8 \, \text{m/s}^2) \),
  • \( h \) is the height above the reference point.
When the object in our exercise, with a mass \( M \), descends by a distance \( l \), it loses gravitational potential energy. This loss is expressed as:
  • \( Mgl \)
This indicates the energy lost to the surroundings as the object moves downward due to gravity.
Hooke's Law
Hooke's Law relates the force necessary to either compress or extend a spring to the distance it is compressed or stretched. It's a fundamental principle that explains how springs work. The formula can be expressed as:
  • \( F = kx \)
where:
  • \( F \) is the force applied to the spring,
  • \( k \) is the spring constant, a measure of the spring's stiffness,
  • \( x \) is the displacement of the spring from its original length.
In our scenario with the wire, applying Hooke's Law involves the mass pulling on the wire causing elongation. The force exerted by the mass, \( F = Mg \), results in the elongation \( l \). By rearranging Hooke’s Law, we find the spring constant \( k \) for the wire:
  • \( k = \frac{Mg}{l} \)
This equation helps us understand how the wire behaves similarly to a spring, stretching due to the force applied by the weight.
Spring Constant
The spring constant, represented by \( k \), is a crucial factor that describes how responsive a spring is to an applied force. A higher spring constant means the spring is stiffer and requires more force to stretch or compress. Conversely, a lower spring constant indicates a less stiff spring, which is easier to deform.
In the context of a wire behaving like a spring, the spring constant can be determined using Hooke's Law:
  • \( k = \frac{F}{l} \)
In our specific example:
  • \( k = \frac{Mg}{l} \)
This shows the wire's resistance to deformation. Understanding the spring constant helps predict how much the wire will stretch given a particular force, providing insight into the material properties of the wire.
Energy Conservation
Energy conservation is a fundamental principle in physics stating that energy cannot be created or destroyed, only transformed from one form to another. In the scenario with the suspended mass and the wire, energy conservation plays a crucial role. As the mass descends and stretches the wire, gravitational potential energy is converted into elastic potential energy stored in the wire.
The initial gravitational potential energy lost by the mass as it falls a distance \( l \) is:
  • \( Mgl \)
Part of this energy is stored as elastic potential energy in the wire described by the formula:
  • \( \frac{1}{2} kl^2 \)
After substituting the spring constant \( k \) we derived earlier \( \left( k = \frac{Mg}{l} \right) \), the elastic potential energy simplifies to:
  • \( \frac{Mgl}{2} \)
This demonstrates the conservation of energy where the total initial gravitational energy is shared between lost energy and stored elastic energy, maintaining the overall energy balance.

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Most popular questions from this chapter

The density of water at the surface of ocean is \(\rho\). If the bulk modulus of water is \(B\), then the density of ocean water at depth, when the pressure is \(\alpha p_{0}\) and \(p_{0}\) is the atmospheric pressure, is (1) \(\frac{\rho B}{B-(\alpha-1) p_{0}}\) (2) \(\frac{\rho B}{B+(\alpha-1) p_{0}}\) (3) \(\frac{\rho B}{B-\alpha_{0}}\) (4) \(\frac{\rho B}{B+\alpha p_{0}}\)

A steel wire of length \(4.7 \mathrm{~m}\) and cross-sectional area \(3 \times 10^{-6} \mathrm{~m}^{2}\) stretches by the same amount as a copper wire of length \(3.5 \mathrm{~m}\) and cross-sectional area of \(4 \times 10^{-6} \mathrm{~m}^{2}\) under a given load. The ratio of Young's modulus of steel to that of copper is (1) \(1.8\) (2) \(3.6\) (3) \(0.6\) (4) \(8.7\)

A small but heavy block of mass \(10 \mathrm{~kg}\) is attached to a wire \(0.3 \mathrm{~m}\) long. Its breaking stress is \(4.8 \times 10^{7} \mathrm{~N} / \mathrm{m}^{2}\). The area of the cross section of the wire is \(10^{-6} \mathrm{~m}^{2}\). The maximum angular velocity with which the block can be rotated in the horizontal circle is (1) \(4 \mathrm{rad} / \mathrm{s}\) (2) \(8 \mathrm{rad} / \mathrm{s}\) (3) \(10 \mathrm{rad} / \mathrm{s}\) (4) \(32 \mathrm{rad} / \mathrm{s}\)

The breaking stress for a metal is \(7.8 \times 10^{9} \mathrm{Nm}^{-2}\). The density of the metal is \(7800 \mathrm{~kg} \mathrm{~m}^{-3}\). If \(g=10 \mathrm{~N} \mathrm{~kg}^{-1}\), Find the maximum length of the wire made of this metal which may be suspended without breaking.

A massive stone pillar \(20 \mathrm{~m}\) high and of uniform cross section rests on a rigid base and supports a vertical load of \(5.0 \times 10^{5} \mathrm{~N}\) at its upper end. If the compressive stress in the pillar is not exceed \(16 \times 10^{6} \mathrm{~N} / \mathrm{m}^{2}\), what is the minimum cross-sectional area of the pillar? (Density of the stone \(=2.5 \times 10^{3} \mathrm{~kg} / \mathrm{m}^{3}\) take \(\left.g=10 \mathrm{~N} / \mathrm{kg} .\right)\) (1) \(0.15 \mathrm{~m}^{2}\) (2) \(0.25 \mathrm{~m}^{2}\) (3) \(0.35 \mathrm{~m}^{2}\) (4) \(0.45 \mathrm{~m}^{2}\)

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