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If a fish is attached to a vertical spring and slowly lowered to its equilibrium position, it is found to stretch the spring by an amount \(d\). If the same fish is attached to the end of the unstretched spring and then allowed to fall from rest, through what maximum distance does it stretch the spring? (Hint: Calculate the force constant of the spring in terms of the distance \(d\) and the mass \(m\) of the fish.)

Short Answer

Expert verified
The maximum distance the spring stretches is given by \(d \cdot \sqrt{2}\).

Step by step solution

01

Find Force Constant

Find the spring constant \(k\) by setting up the equation, \(mg = k \cdot d\), solving for \(k\) gives \(k=m \cdot g / d\) where \(m\) is the mass of the fish, \(g\) is acceleration due to gravity, and \(d\) is the initial stretch of the spring caused by fish's weight.
02

Equating Potential Energy to Elastic Potential Energy

The energy conservation principle states that the initial potential energy equals the elastic potential energy when the spring is stretched to its maximum. Formulate this by \(mgh = 1/2 \cdot k \cdot x^2\) where \(h\) is the height the fish falls from (the same as \(d\) in this case) and \(x\) is the maximum stretch.
03

Solve for Maximum Stretch of Spring

Substitute \(k\) from Step 1 into the equation from Step 2, to get \(m \cdot g \cdot d = 1/2 \cdot (m \cdot g / d) \cdot x^2\). This equation simplifies to \(2d^2 = x^2\). Therefore \(x = d \cdot \sqrt{2}\).

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Key Concepts

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

Hooke's Law
Imagine a spring, the kind you might find in a pen or a toy. When you apply a force to stretch or compress it, the spring reacts with a force of its own to restore its original shape. This is where Hooke's Law comes into play, which can be stated in a simple equation:
\( F = -kx \)
Here's what each symbol means:
  • \(F\) is the restoring force exerted by the spring, measured in newtons (N).
  • \(k\) is the spring constant, which tells us how stiff the spring is. It's measured in newtons per meter (N/m).
  • \(x\) is the displacement from the spring's original length, measured in meters (m).
The negative sign indicates that the force exerted by the spring is in the opposite direction of the displacement. By understanding this law, we create a foundation for exploring how springs behave and how they store energy, which is particularly useful in problems like the textbook exercise on the spring and fish.
Elastic Potential Energy
When you stretch a spring, you're doing work against the spring's natural force. This work gets stored as elastic potential energy (EPE), and it's ready to spring back into action—quite literally! In mathematical terms, this energy can be calculated using the formula:
\( EPE = \frac{1}{2} kx^2 \)
This equation links directly to Hooke's Law:
  • \(k\) still represents the spring constant—a measure of the spring's stiffness.
  • \(x\) is the displacement, meaning how far the spring has been stretched or compressed from its natural length.
The beauty of EPE is its role in many physics problems involving springs, including our textbook example. By calculating it, students can understand how energy transforms and is conserved when forces act on a spring.
Conservation of Energy
A principle that's as sure as the ground beneath our feet is the conservation of energy. It says that the total energy in an isolated system remains constant—it can neither be created nor destroyed, only transformed from one form to another. Let's put this into context with the falling fish: As the fish falls, its gravitational potential energy is converted into elastic potential energy of the spring. Using the conservation of energy, we can equate the fish's initial potential energy when it's held at height \(d\) to the spring's elastic potential energy at its maximum stretch. The idea is that the energy content of the fish-plus-spring system remains unchanged throughout the process, hence allowing us to set up an equation to find the maximum distance the spring stretches. This is a crucial concept, not just in physics puzzles, but in understanding the natural universe.
Simple Harmonic Motion
The up and down bobbing of a fish on a spring or the to-and-fro swing of a pendulum are examples of simple harmonic motion (SHM)—a type of periodic movement that is particularly predictable. A system in SHM will oscillate back and forth over the same path, within the same amount of time for each cycle, provided there’s no external force causing it to lose energy (like friction or air resistance).
The most important aspect of SHM is that it's governed by a restoring force directly proportional to the displacement and directed towards the equilibrium position—the very principle laid out by Hooke's Law. In the case of our textbook's fish, if we disregard air resistance and other non-conservative forces, it would oscillate around the equilibrium position in a simple harmonic manner once it has been dropped and stretched the spring, adding a layer of real-world complexity and application to the basic principles of physics.

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Most popular questions from this chapter

A \(0.60 \mathrm{~kg}\) book slides on a horizontal table. The kinetic friction force on the book has magnitude \(1.8 \mathrm{~N}\). (a) How much work is done on the book by friction during a displacement of \(3.0 \mathrm{~m}\) to the left? (b) The book now slides \(3.0 \mathrm{~m}\) to the right, returning to its starting point. During this second \(3.0 \mathrm{~m}\) displacement, how much work is done on the book by friction? (c) What is the total work done on the book by friction during the complete round trip? (d) On the basis of your answer to part (c), would you say that the friction force is conservative or nonconservative? Explain.

A 1.20 kg piece of cheese is placed on a vertical spring of negligible mass and force constant \(k=1800 \mathrm{~N} / \mathrm{m}\) that is compressed \(15.0 \mathrm{~cm} .\) When the spring is released, how high does the cheese rise from this initial position? (The cheese and the spring are not attached.)

\(\mathrm{A} 1500 \mathrm{~kg}\) rocket is to be launched with an initial upward speed of \(50.0 \mathrm{~m} / \mathrm{s} .\) In order to assist its engines, the engineers will start it from rest on a ramp that rises \(53^{\circ}\) above the horizontal (Fig. \(\mathbf{P 7 . 5 0}\) ). At the bottom, the ramp turns upward and launches the rocket vertically. The engines provide a constant forward thrust of \(2000 \mathrm{~N}\), and friction with the ramp surface is a constant \(500 \mathrm{~N}\). How far from the base of the ramp should the \text { rocket start, as measured along the surface of the ramp?

A \(2.00 \mathrm{~kg}\) block is pushed against a spring with negligible mass and force constant \(k=400 \mathrm{~N} / \mathrm{m}\), compressing it \(0.220 \mathrm{~m}\). When the block is released, it moves along a frictionless, horizontal surface and then up a frictionless incline with slope \(37.0^{\circ}\) (Fig. \(\mathbf{P 7 . 4 0}\) ). (a) What is the speed of the block as it slides along the horizontal surface after having left the spring? (b) How far does the block travel up the incline before starting to slide back down?

CALC A small object with mass \(m=0.0900 \mathrm{~kg}\) moves along the \(+x\) -axis. The only force on the object is a conservative force that has the potential-energy function \(U(x)=-\alpha x^{2}+\beta x^{3},\) where \(\alpha=2.00 \mathrm{~J} / \mathrm{m}^{2}\) and \(\beta=0.300 \mathrm{~J} / \mathrm{m}^{3} .\) The object is released from rest at small \(x .\) When the object is at \(x=4.00 \mathrm{~m},\) what are its (a) speed and (b) acceleration (magnitude and direction)? (c) What is the maximum value of \(x\) reached by the object during its motion?

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