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Spring constant It took 1800 \(\mathrm{J}\) of work to stretch a spring from its natural length of 2 \(\mathrm{m}\) to a length of 5 \(\mathrm{m}\) . Find the spring s force constant.

Short Answer

Expert verified
The spring force constant is 400 N/m.

Step by step solution

01

Understanding the Problem

We need to determine the spring constant, denoted by \( k \), using the work done to stretch the spring. The work done, \( W \), to stretch a spring from its natural length to a new length is given by the formula \( W = \frac{1}{2} k x^2 \) where \( x \) is the change in length from the spring's natural length.
02

Determine Change in Length

The spring's natural length is 2 meters, and it is stretched to 5 meters. Therefore, the change in length \( x \) is given as \( x = 5 \text{ m} - 2 \text{ m} = 3 \text{ m} \).
03

Set Up the Equation

We know that \( W = \frac{1}{2} k x^2 \). Substituting the given values, 1800 J of work, and \( x = 3 \text{ m} \), we have the equation: \[ 1800 = \frac{1}{2} k (3)^2. \]
04

Solve for the Spring Constant \(k\)

First, calculate \( x^2 = 3^2 = 9 \). Substitute back into the equation to get \[ 1800 = \frac{1}{2} k \cdot 9. \] Simplify to \[ 1800 = \frac{9}{2} k, \] and multiply both sides by 2 to eliminate the fraction: \[ 3600 = 9k. \] Divide both sides by 9 to solve for \( k \): \( k = \frac{3600}{9} \).
05

Calculate the Value of \(k\)

After dividing, we find that \( k = 400 \). Thus, the force constant of the spring is 400 N/m.

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Key Concepts

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

Hooke's Law
Hooke's Law is a fundamental principle in physics that illustrates how springs and other elastic objects behave under force. The law is named after Robert Hooke, a 17th-century scientist. He discovered that the force needed to extend or compress a spring by some distance is proportional to that distance. In mathematical terms, Hooke's Law is expressed as:
\[ F = kx \]
where:
  • \( F \) is the force applied to the spring (measured in Newtons),
  • \( k \) is the spring constant (a measure of the stiffness of the spring),
  • \( x \) is the displacement of the spring from its natural length (measured in meters).
Hooke's Law shows us that the spring constant \( k \) is unique for each spring, indicating how much force is needed for each unit of distance extended or compressed. Understanding this law is vital in solving problems related to spring mechanics.
Work-Energy Principle
The work-energy principle connects the concepts of work and energy. In the context of springs, this principle helps us understand how energy is stored in a spring when force is applied to stretch or compress it. The work-energy principle states that the work done on an object is equal to the change in energy of that object.
For springs, the work done \( W \) when stretching or compressing the spring is translated into potential energy stored in the spring. This formula can express that relationship:
\[ W = \frac{1}{2} k x^2 \]
where:
  • \( W \) is the work done on the spring (in Joules),
  • \( k \) is the spring constant,
  • \( x \) is the displacement from the spring's natural length.
By using the work-energy principle, one can determine how much energy is stored in a spring or calculate the work needed to stretch or compress a spring by a certain amount.
Calculus Problem Solving
Calculus provides tools that are invaluable for solving complex problems involving changing quantities such as the length of a spring. In spring problems, derivatives and integrals help us determine how changes in length result in changes in forces and energies.
In our spring constant problem, we utilize the integration aspect of calculus. The relationship between work and displacement for the spring uses integration to evaluate how the incrementally small forces over displacement accumulate to the total work done.
The equation \( W = \frac{1}{2} k x^2 \) is derived from integrating the increment (\( F = kx \)) over the distance \( x \). Calculus allows us to methodically and accurately solve such physical problems, giving precise results where algebra alone might become cumbersome.
Physics of Springs
Springs are ubiquitous in physics, offering a rich field of study into how objects store and transfer energy. They are essential components in various machines and devices, from simple toys to complex mechanical systems.
The physics of springs not only includes understanding Hooke's Law and the energy they store but also delves into oscillatory motion and harmonic waves when springs are used as part of a system in motion.
  • Springs can model elastic forces, providing a baseline understanding of potential energy storage.
  • They obey unique rules when several springs work together, such as springs in series and parallel configurations, which affect the overall spring constant.
  • Springs also help in understanding dampening where energy is lost in mechanical systems due to resistance, showcasing the real-world complexities of energy transfer.
In summary, the study of springs is fundamental to comprehending broader physical systems and predicting how they behave under various forces.

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Most popular questions from this chapter

Volume of a bowl \(A\) bowl has a shape that can be generated by revolving the graph of \(y=x^{2} / 2\) between \(y=0\) and \(y=5\) about the \(y\) -axis. a. Find the volume of the bowl. b. Related rates If we fill the bowl with water at a constant rate of 3 cubic units per second, how fast will the water level in the bowl be rising when the water is 4 units deep?

The area of the region \(R\) enclosed by the semiellipse \(y=(b / a) \sqrt{a^{2}-x^{2}}\) and the \(x\) -axis is \((1 / 2) \pi a b\) and the volume of the ellipsoid generated by revolving \(R\) about the \(x\) -axis is \((4 / 3) \pi a b^{2} .\) Find the centroid of \(R .\) Notice that the location is independent of \(a .\)

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The volume of a torus The disk \(x^{2}+y^{2} \leq a^{2}\) is revolved about the line \(x=b(b>a)\) to generate a solid shaped like a doughnut and called a torus. Find its volume. (Hint: \(\int_{-a}^{a} \sqrt{a^{2}-y^{2}} d y=\) \(\pi a^{2} / 2,\) since it is the area of a semicircle of radius \(a . )\)

Set up, but do not evaluate, an integral that represents the area of the surface obtained by rotating the curve \(x=a(t-\sin t)\) \(y=a(1-\cos t), 0 \leq t \leq 2 \pi,\) about the \(x\) -axis.

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