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Two identical massless springs \(A\) and \(B\) consist spring constant \(k_{A}\) and \(k_{B}\) respectively. Then: (a) if they are compressed by same force, work done on \(A\) is more expanded when \(k_{A}>k_{B}\) (b) if they are compressed by same amount, work done on \(A\) is more expanded when \(k_{A}k_{B}\) (d) both (a) and (b) are correct

Short Answer

Expert verified
Option (c) is correct.

Step by step solution

01

Understanding Spring Work Done Formulas

The work done on a spring when compressed or stretched is given by \( W = \frac{1}{2}kx^2 \), where \( k \) is the spring constant and \( x \) is the compression distance.
02

Analyzing the Situation for Part (a)

For both springs compressed by the same force, the compression distance \( x \) is given by \( x = \frac{F}{k} \). Substituting in the work formula \( W = \frac{1}{2}k \left(\frac{F}{k}\right)^2 = \frac{F^2}{2k} \). If \( k_A > k_B \), then \( \frac{F^2}{2k_A} < \frac{F^2}{2k_B} \). Hence work done on A is less.
03

Analyzing the Situation for Part (b)

For same compression distance \( x \), work done \( W = \frac{1}{2}kx^2 \). If \( k_A < k_B \), then \( \frac{1}{2}k_Ax^2 < \frac{1}{2}k_Bx^2 \). Hence work done on A is less.
04

Analyzing the Situation for Part (c)

Still considering same compression distance \( x \), if \( k_A > k_B \), then \( \frac{1}{2}k_Ax^2 > \frac{1}{2}k_Bx^2 \). Hence work done on A is more.
05

Conclusion Based on Analysis

For option (a) work done is more on B. For option (b), work done is less on A. Finally, for option (c), work done is more on A, which matches our formula analysis. Thus, option (c) is correct.

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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, denoted as \( k \), is a measure of a spring's stiffness. It determines how resistant the spring is to deformation when a force is applied. If the spring is very stiff, it will have a high spring constant value, meaning it doesn't deform much under an applied force. Conversely, a spring with a low spring constant is more flexible and deforms more easily.

Mathematically, the spring constant \( k \) is expressed in units of Newtons per meter (N/m). It is an integral part of the equation \( F = kx \), where \( F \) is the force applied on the spring, and \( x \) is the compression or extension length.

  • A high spring constant means the spring is hard to stretch or compress, requiring more force for the same change in length.
  • A low spring constant signifies a more easily deformable spring, requiring less force for the same change in length.
Understanding the spring constant is crucial in determining how much work is done when a spring is compressed or stretched.
Hooke's Law
Hooke's Law is a fundamental principle in physics that describes the behavior of springs. It states that the force needed to extend or compress a spring by a certain distance is proportional to that distance.

The law is mathematically expressed as \( F = kx \), where:
  • \( F \) is the force applied to the spring,
  • \( k \) is the spring constant,
  • \( x \) is the displacement from the spring's equilibrium position.
Hooke's Law tells us that the amount of force required to compress or stretch a spring is directly related to how far it is being compressed or stretched. This principle is crucial for determining the work done on or by a spring when it is subject to external forces.

Applications of Hooke's Law are found in various fields such as mechanical engineering and biomechanics, where understanding the elastic properties of materials and systems is essential.
Physics Problem Solving
To effectively tackle physics problems, such as those involving springs, a systematic approach is essential. Physics problem solving often involves understanding the underlying principles and applying relevant formulas.

When faced with a spring-related problem, consider the following steps:

  • Identify known quantities, such as the spring constant, force applied, or distance compressed or stretched.
  • Apply Hooke's Law, \( F = kx \), to relate force and displacement.
  • Use the work done formula on springs, \( W = \frac{1}{2}kx^2 \), to calculate the energy stored or released.
  • Analyze the problem conditions, such as whether the springs are compressed by the same amount or by the same force, to know how to compare quantities.
By breaking down the problem and applying the correct concepts and equations, you can solve problems efficiently and with confidence. The systematic approach helps you understand the physical principles rather than just memorizing formulas, which is crucial in any physics problem solving.

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Most popular questions from this chapter

A block of mass \(m\) is hanging over a smooth and light pulley through a light string. The other end of the string is pulled by a constant force \(F\). The kinetic energy of the block increases by \(20 \mathrm{~J}\) in \(1 \mathrm{~s}\) is : (a) the tension in the string is \(m g\) (b) the tension in the string is \(F\) (c) the work done by the tension on the block is \(20 \mathrm{~J}\) in the above \(1 \mathrm{~s}\) (d) the work done by the force of gravity is \(20 \mathrm{~J}\) in the above 1 s

A flywheel of mass \(60 \mathrm{~kg}\), radius \(40 \mathrm{~cm}\) is revolving 300 revolutions per min. Its kinetic energy will be : (a) \(480 \pi^{2} \mathrm{~J}\) (b) \(48 \mathrm{~J}\) (c) \(48 \pi J\) (d) \(\frac{4}{\pi} J\)

In a certain situation, \(\mathbf{F}\) and \(\vec{s}\) are not equal to zero but the work done is zero. From this, we conclude that: \(\rightarrow \quad \rightarrow\) (a) \(\mathbf{F}\) and \(\vec{s}\) are in same direction (b) \(\overrightarrow{\mathrm{F}}\) and \(\overrightarrow{\mathrm{s}}\) are perpendicular to each other \(\Rightarrow\) (c) \(\mathrm{F}\) and \(\overrightarrow{\mathrm{s}}\) are in opposite direction (d) none of the above

An object of mass \(M\), initially at rest under the action of a constant force \(F\) attains a velocity \(v\) in time \(t\). Then the average power supplied to the mass is : (a) \(\mathrm{Fv}\) (b) \(\frac{1}{2} F v\) (c) zero (d) \(\frac{m v^{2}}{2 t}\)

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