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Chapter 4: Problem 130

This question has Statement 1 and Statement \(2 .\) Of the four choices given after the statements, choose the one that best describes the two statements. If two springs \(S_{1}\) and \(S_{2}\) of force constants \(k_{1}\) and \(k_{2}\), respectively, are stretched by the same force, it is found that more work is done on spring \(S_{1}\) than on spring \(S_{2}\) Statement 1: If stretched by the same amount, work done on \(S_{1}\) will be more than that on \(S_{2}\) Statement \(2: k_{1}

Short Answer

Expert verified
(B) Statement 1 is true, and Statement 2 is false.

Step by step solution

01

Recall Hooke's Law

Hooke's Law states that the force required to stretch or compress a spring is proportional to the displacement from its equilibrium position. Mathematically, this can be represented as: \(F = -kx\) where F is the force applied, k is the spring constant, and x is the displacement from the equilibrium position.
02

Calculate the work done on each spring

The work done (W) on the springs can be calculated by integrating the force over the distance of displacement. Given that the force applied to both springs is the same, we can represent the work done on S_1 and S_2 as follows: \(W_1 = \int_0^{x_1} Fdx = \frac{1}{2}k_1x_1^2\) \(W_2 = \int_0^{x_2} Fdx = \frac{1}{2}k_2x_2^2\) where \(W_1\) and \(W_2\) are the work done on springs S_1 and S_2, respectively, and \(x_1\) and \(x_2\) are their respective displacements.
03

Compare the work done on the springs

Given that the same force is applied to both springs and more work is done on spring S_1, we can write: \(\frac{1}{2}k_1x_1^2 > \frac{1}{2}k_2x_2^2\) Now, we will analyze the 2 statements and the given choices.
04

Statement 1

Since the work done on the springs when they are stretched by the same force is greater for S_1, let's find out if the work done on S_1 will also be more when stretched by the same amount x: \(W_{S_1} = \frac{1}{2}k_1x^2\) \(W_{S_2} = \frac{1}{2}k_2x^2\) We can write: \(\frac{W_{S_1}}{W_{S_2}} = \frac{k_1}{k_2}\) Since \(W_1 > W_2\), we can infer that \(k_1 > k_2\) if the springs are stretched by the same amount. Thus, Statement 1 is true.
05

Statement 2

This statement proposes that \(k_1 < k_2\), which is the opposite of what we inferred from Statement 1. Therefore, Statement 2 is false. Based on our analysis, we can conclude that: (A) Statement 1 is false, Statement 2 is true. (Incorrect) (B) Statement 1 is true, Statement 2 is false (Correct) (C) Statement 1 is true, Statement 2 is true and Statement 2 is the correct explanation for Statement 1 (Incorrect) (D) Statement 1 is true, Statement 2 is true and Statement 2 is not the correct explanation of Statement 1 (Incorrect) The correct answer is (B): Statement 1 is true, and Statement 2 is false.

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Key Concepts

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

Understanding the Spring Constant
The spring constant, denoted by the variable \(k\), is a fundamental aspect of Hooke's Law. It represents the stiffness of a spring. The larger the spring constant, the stiffer the spring. This implies that a high \(k\) value requires more force to achieve the same displacement compared to a spring with a lower \(k\). In Hooke's Law, the relationship between force \(F\) and displacement \(x\) is given by the equation:\[ F = -kx \]This equation tells us that the force applied to the spring is directly proportional to the amount it stretches or compresses from its equilibrium position. The negative sign indicates that the force is a restoring force, acting in the opposite direction of the displacement.
  • Springs with a high \(k\) value are more rigid.
  • Springs with a low \(k\) value are more flexible.
Therefore, understanding the spring constant allows us to predict how a spring will behave when force is applied.
Calculating Work Done on Springs
When you apply force to stretch or compress a spring, you are doing work on the spring. The work done is the energy transferred to the spring, and it can be calculated using the formula:\[ W = \frac{1}{2} k x^2 \]where \(W\) is the work done, \(k\) is the spring constant, and \(x\) is the displacement from the equilibrium position.
  • The work done increases with both a larger displacement \(x\) and a larger spring constant \(k\).
  • More work is needed to stretch a stiffer spring (with a larger \(k\)) by the same amount as a more flexible spring (with a smaller \(k\)).
Thus, the work done on a spring depends not only on how far the spring is stretched or compressed but also on its rigidity.
Displacement from Equilibrium and Its Role
Displacement from equilibrium is a crucial concept when dealing with springs. It refers to how far the spring is stretched or compressed from its natural, unstressed length. In the context of Hooke's Law and spring motion, this displacement is represented by \(x\) in the equation \(F = -kx\).
  • The direction and amount of displacement affect the restoring force exerted by the spring.
  • Greater displacement results in a stronger restoring force, as the spring tries to return to its original position.
When considering the work done on a spring, the displacement \(x\) plays a key role. The more you displace the spring from equilibrium, the more work is needed, hence increasing its potential energy. Understanding displacement helps in analyzing how springs behave under various forces.

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Most popular questions from this chapter

Two identical beads of \(m=100 \mathrm{~g}\) are connected by an inextensible massless string that can slide along the two arms \(A C\) and \(B C\) of a rigid smooth wireframe in a vertical plane. If the system is released from rest, the kinetic energy of the first particle when they have moved by a distance of \(0.1 \mathrm{~m}\) is \(16 x \times 10^{-3} \mathrm{~J}\). Find the value of \(x .\left(g=10 \mathrm{~m} / \mathrm{s}^{2}\right)\)

A proton is kept at rest. A positively charged particle is released from rest at a distance \(d\) in its field. Consider two experiments; one in which the charged particle is also a proton and in another, a position. In the same time \(t\), the work done on the two moving charged particles is (A) Same as the force law is involved in the two experiments. (B) Less for the case of a positron, as the positron moves away more rapidly and the force on it weakens. (C) More in the case of positron, as the positron moves away a larger distance. (D) Same as the work done by charged particle on the stationary proton.

A body of mass \(m\) accelerates uniformly from rest to \(v_{1}\) in time \(t_{1} .\) The instantaneous power delivered to the body as a function of time \(t\) is [2004] (A) \(\frac{m v_{1} t^{2}}{t_{1}}\) (B) \(\frac{m v_{1}^{2} t}{t_{1}^{2}}\) (C) \(\frac{m v_{1} t}{t_{1}}\) (D) \(\frac{m v_{1}^{2} t}{t_{1}}\)

A homogeneous rope of mass per unit length \(\lambda\) and length \(l\) kept on ground and one end of the rope is fixed to ground at \(O .\) The left end of the rope (with respect to fixed end) is pulled by an external agent which imparts constant velocity \(v\) to it. Find the work done by the external agent (in joule) to place the moving end extremely right with respect to fixed end. Take \(\lambda=1 \mathrm{~kg} / \mathrm{m}, v=1 \mathrm{~ms}^{-1}\) and \(l=1 \mathrm{~m}\)

A system consists of two identical cubes, each of mass \(3 \mathrm{~kg}\), linked together by a compressed weightless spring of force constant \(1000 \mathrm{~N} / \mathrm{m}\). The cubes are also connected by a thread which is burnt at a certain moment. At what minimum value of initial compression, \(x_{0}\) (in \(\mathrm{cm}\) ) of the spring will the lower cube bounce up after the thread is burnt through?
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