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An archer, about to shoot an arrow, is applying a force of \(+240 \mathrm{N}\) to a drawn bowstring. The bow behaves like an ideal spring whose spring constant is 480 \(\mathrm{N} / \mathrm{m}\) . What is the displacement of the bowstring?

Short Answer

Expert verified
The displacement of the bowstring is 0.5 meters.

Step by step solution

01

Identify the given values

We are given that the force applied is \(+240 \, \mathrm{N}\) and the spring constant of the bow is \(480 \, \mathrm{N/m}\). These values will be used in Hooke's Law to find the displacement.
02

Recall Hooke's Law

Hooke's Law is given by the equation: \(F = k \cdot x\), where \(F\) is the force applied, \(k\) is the spring constant, and \(x\) is the displacement. This relationship will help us determine the displacement of the bowstring.
03

Rearrange Hooke's Law for displacement

Rearrange the equation to solve for the displacement \(x\): \(x = \frac{F}{k}\). This formula will allow us to calculate how far the bowstring is displaced by the force applied.
04

Substitute the given values and calculate

Substitute \(F = 240 \, \mathrm{N}\) and \(k = 480 \, \mathrm{N/m}\) into the equation: \(x = \frac{240}{480}\) which simplifies to \(x = 0.5\). Hence, the displacement of the bowstring is \(0.5\) meters.

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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, represented by the symbol "k," is a fundamental concept in physics. It provides a measure of a spring's stiffness; essentially, how resistant it is to being compressed or stretched. The greater the spring constant, the stiffer the spring. This constant is expressed in units of Newtons per meter (N/m). In the context of our exercise, the bow has a spring constant of 480 N/m, indicating it requires a force of 480 Newtons to stretch the bowstring by one meter.
Understanding the spring constant is crucial when working with springs and other elastic systems. It influences how much work is done in stretching or compressing the spring linearly and is central to calculating energy storage in spring systems as well.
Force and Displacement
The relationship between force and displacement is beautifully captured by Hooke's Law, which is foundational in classical mechanics. Using the formula \( F = k \cdot x \), where \( F \) is the force applied, \( k \) is the spring constant, and \( x \) is the displacement, we can determine how far a spring moves under a specific force.
In our problem, we applied a force of 240 Newtons to the bow. By rearranging Hooke's Law to \( x = \frac{F}{k} \), we can calculate that the displacement is 0.5 meters.
This relationship is linear, meaning that the displacement of the spring is directly proportional to the force applied. More force results in more displacement, and vice versa. This concept is essential for solving numerous physics problems related to springs and elasticity.
Ideal Spring
The notion of an ideal spring is used to simplify complex physical systems. An ideal spring is a theoretical model that perfectly follows Hooke's Law, meaning its deformation (stretching or compressing) is directly proportional to the applied force and perfectly reversible.
Real-world springs, such as those in machinery or the bow in our problem, approximate this ideal behavior but with some discrepancies due to material imperfections or environmental factors. In the context of our exercise, we treat the bow as an ideal spring to make the problem more straightforward to solve, allowing us to apply the principles of Hooke's Law without accounting for anomalies that would deviate from the theory.
  • An ideal spring has no mass,
  • Experiences no internal friction,
  • And its deformation is fully reversible.
These assumptions simplify calculations and make it easier to predict the system's behavior.
Physics Problem Solving
Solving physics problems often involves breaking down complex ideas into simpler components. Following a systematic approach helps in understanding and tackling challenges efficiently. In the problem involving the archer's bow, we started by identifying the known values and using them in Hooke's Law to find the solution.
A systematic problem-solving strategy often involves:
  • Identifying what is given and what needs to be found,
  • Applying the relevant principles or laws,
  • Rearranging formulas to solve for unknown quantities,
  • And substituting known values to find the solution.
This step-by-step method can be applied not just to spring-related problems, but broadly across various areas of physics, helping students build a strong foundation in analytical problem solving.

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Most popular questions from this chapter

An 11.2 -kg block and a 21.7 -kg block are resting on a horizontal frictionless surface. Between the two is squeezed a spring (spring constant \(=1330 \mathrm{N} / \mathrm{m}\) ). The spring is compresed by 0.141 \(\mathrm{m}\) from its unstrained length and is not attached to either block. With what speed does each block move away after the mechanism keeping the spring squeezed is released and the spring falls away?

An object attached to a horizontal spring is oscillating back and forth along a frictionless surface. The maximum speed of the object is \(1.25 \mathrm{m} / \mathrm{s},\) and its maximum acceleration is 6.89 \(\mathrm{m} / \mathrm{s}^{2}\) . How much time elapses between an instant when the object's speed is at a maximum and the next instant when its acceleration is at a maximum?

In 0.750 s, a \(7.00-\mathrm{kg}\) block is pulled through a distance of 4.00 \(\mathrm{m}\) on a frictionless horizontal surface, starting from rest. The block has a constant acceleration and is pulled by means of a horizontal spring that is attached to the block. The spring constant of the spring is 415 \(\mathrm{N} / \mathrm{m} .\) By how much does the spring stretch?

A tray is moved horizontally back and forth in simple harmonic motion at a frequency of \(f=2.00 \mathrm{Hz}\) . On this tray is an empty cup. Obtain the coefficient of static friction between the tray and the cup, given that the cup begins slipping when the amplitude of the motion is \(5.00 \times 10^{-2} \mathrm{m} .\)

Between each pair of vertebrae in the spinal column is a cylindrical disc of cartilage. Typically, this disc has a radius of about \(3.0 \times 10^{-2} \mathrm{m}\) and a thickness of about \(7.0 \times 10^{-3} \mathrm{m} .\) The shear modulus of cartilage is \(1.2 \times 10^{7} \mathrm{N} / \mathrm{m}^{2}\) . Suppose that a shearing force of magnitude 11 \(\mathrm{N}\) is applied parallel to the top surface of the disc while the bottom surface remains fixed in place. How far does the top surface move relative to the bottom surface?

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