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Two springs \(P\) and \(Q\) having stiffness constants \(k_{1}\) and \(k_{2}\left(

Short Answer

Expert verified
More work is done on spring P (option 2).

Step by step solution

01

Understanding the Problem

Two springs, P and Q, have stiffness constants denoted by \(k_1\) and \(k_2\), where \(k_2 < k_1\). Both springs are stretched equally, meaning they are extended by the same amount \(x\). We need to determine how the work done on each spring compares.
02

Expression for Work in a Spring

The work done on a spring when it is stretched by a distance \(x\) is given by the formula \(W = \frac{1}{2} k x^2\), where \(k\) is the stiffness constant of the spring. This formula highlights that the work done is proportional to both the stiffness constant \(k\) and the square of the stretch \(x^2\).
03

Calculate Work Done on Each Spring

Since both springs are stretched by equal amounts, the stretching distance \(x\) is the same for both. Therefore, we calculate the work done on each spring as follows: - Work done on spring P: \(W_P = \frac{1}{2} k_1 x^2\)- Work done on spring Q: \(W_Q = \frac{1}{2} k_2 x^2\). Since \(k_2 < k_1\), it's clear that \(W_Q < W_P\).
04

Conclusion

Given that \(k_2 < k_1\), and using the formula \(W = \frac{1}{2} k x^2\), the work done on spring Q is less than the work done on spring P. Therefore, more work is done on spring P when both springs are stretched by the same amount.

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Key Concepts

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

Stiffness Constant
The stiffness constant, often denoted as "k," is a measure of how resistant a spring is to being deformed or stretched. Think of it like the spring's own "strength rating". The higher the stiffness constant, the more force you'd need to stretch or compress the spring by a certain amount.
Understanding this constant is crucial in mechanics since it defines how a spring will react to forces. The formula for the work done on a spring is given by \[ W = \frac{1}{2} k x^2 \]where:
  • \(W\) is the work done,
  • \(k\) is the stiffness constant,
  • \(x\) is the amount of stretch or compression.
If two springs are stretched equally, the spring with a higher stiffness constant will have more work done on it because the formula indicates a direct proportionality between work and the stiffness constant. Thus, when comparing springs through this lens, it's evident that knowing the stiffness constant is fundamental to predicting a spring's behavior.
Spring Force
Springs exert a force based on how much they are stretched or compressed. This force is described by Hooke's Law, which states that the force \( F \) exerted by a spring is \( F = - k x \),where:
  • \(F\) is the force the spring exerts,
  • \(k\) is the stiffness constant,
  • \(x\) is the displacement from the spring's equilibrium position.
The negative sign in the formula represents that the direction of the force is opposite to the direction of displacement (the spring wants to return to its original length).
This formula highlights that stronger springs (higher \(k\)) exert more force for the same displacement compared to less stiff springs. This means in our problem, spring P, with a larger \(k_1\), exerts a greater counteracting force during the stretching process than spring Q.
Mechanics Problem
In mechanics, problems involving springs generally require understanding how forces and energy interact during motion. For example, the scenario in the exercise involves two springs, each with a different stiffness constant, being stretched equally.
Given that a spring's stiffness constant directly affects the work done on it, the mechanics problem can be broken down into analyzing the forces at play and the energy consumed.When solving such problems, always:
  • Identify the given quantities, such as the stiffness constant \(k\) and displacement \(x\),
  • Use the correct formula for work done on springs, \( W = \frac{1}{2} k x^2 \),
  • Understand the relation between the stiffness constant and work done, where more work is done on a stiffer spring for the same amount of stretch.
Through these steps, we determine that the work done on a spring is proportional to its stiffness constant, leading to the conclusion that spring P, with the higher stiffness constant \(k_1\), requires more work than spring Q. This analysis helps reinforce the principle that in mechanical systems, stiffness plays a critical role in determining both force interactions and energy expenditures.

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Most popular questions from this chapter

Which of the following statements is correct? (1) Kinetic energy of a system can be changed without changing its momentum. (2) Kinetic energy of a system cannot be changed without changing its momentum. (3) Momentum of a system cannot be changed without changing its kinetic energy. (4) A system cannot have energy without having momentum.

A block of mass \(M\) rests on a table. It is fastened to the lower end of a light, vertical spring. The upper end of the spring is fastened to a block of mass \(m\). The upper block is pushed down by an additional force \(3 \mathrm{mg}\), so the spring compression is \(4 \mathrm{mg} / \mathrm{k}\). In this configuration, the upper block is released from rest. The spring lifts the lower block off the table. What is the greatest possible value for \(\frac{M}{m}\) ?

The \(\mathrm{PE}\) of a certain spring when stretched from natural length through a distance \(0.3 \mathrm{~m}\) is \(5.6 \mathrm{~J}\). Find the amount of work in joule that must be done on this spring to stretch it through an additional distance \(0.15 \mathrm{~m}\).

A block of mass \(2 \mathrm{~kg}\) is hanging over a smooth and light pulley through a light string. The other end of the string is pulled by a constant force \(F=40\) \(\mathrm{N}\). The kinetic energy of the particle increase \(40 \mathrm{~J}\) in a given interval of time. Then: \(\left(g=10 \mathrm{~m} / \mathrm{s}^{2}\right)\) (1) tension in the string is \(40 \mathrm{~N}\) (2) displacement of the block in the given interval of time is \(2 \mathrm{~m}\) (3) work done by gravity is \(-20 \mathrm{~J}\) (4) work done by tension is \(80 \mathrm{~J}\)

A man pushes a block of \(30 \mathrm{~kg}\) along a level floor at a constant speed with a force directed at \(45^{\circ}\) below the horizontal. If the coefficient of friction is \(0.20\), then match the following. $$ \begin{array}{|c|c|} \hline {\begin{array}{c} \text { Column I } \\ \end{array}} & {\begin{array}{c} \text { Column II } \\ \end{array}} \\ \hline \text { i. } \text {Work done by all forces exerted by the surface on the block in \(20 \mathrm{~m}\)} & \text { a. } \text{Zero} \\ \hline \text { ii. } \text{Work done by the force of gravity} & \text { b. } \text{\(-1500 \mathrm{~J}\)} \\ \hline \text { iii. } \text{Work done by the man on the block in pushing it through \(10 \mathrm{~m}\)} & \text { c. } \text{\(750 \mathrm{~J}\)}\\\ \hline \text { iv. } \text{Net force on the block} & \text { d. } \text{\(30 \mathrm{~J}\)} \\ \hline \end{array} $$

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