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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 m.

Step by step solution

01

Understand the problem

The problem states that an archer is using a force to draw a bowstring, which behaves like a spring. We know the force applied is 240 N, and the spring constant (k) is 480 N/m. We need to find the displacement (x) of the bowstring.
02

Use Hooke's Law

Hooke's Law is a principle of physics that states the force needed to extend or compress a spring by some distance is proportional to that distance. It is usually formulated as \( F = kx \), where \( F \) is the force applied, \( k \) is the spring constant, and \( x \) is the displacement we want to find.
03

Rearrange the formula to solve for displacement

We need to solve for \( x \), the displacement of the bowstring. Rearrange the equation from Hooke's Law: \( x = \frac{F}{k} \).
04

Substitute the known values

Substitute the known values into the equation: \( x = \frac{240 \, \mathrm{N}}{480 \, \mathrm{N/m}} \).
05

Calculate the displacement

Perform the calculation: \( x = \frac{240}{480} = 0.5 \, \mathrm{m} \). Thus, 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 Explained
Imagine a spring as a fun and bouncy gadget! The spring constant, usually designated by the letter \( k \), is a value that tells us how stiff or stretchy the spring is. It's measured in newtons per meter (N/m), and it essentially tells us how much force is needed to stretch the spring by a certain amount.

When you have a large spring constant, the spring is quite stiff, meaning it requires more force to stretch it. Conversely, a smaller spring constant indicates a more flexible spring that stretches easily with less force. In the case of our bowstring, the spring constant is \( 480 \mathrm{N/m} \). This value helps the archer understand just how much force needs to be applied to achieve a certain stretch or displacement.
  • High spring constant: Stiff spring, more force needed.
  • Low spring constant: Flexible spring, less force needed.
With the spring constant known, we can apply Hooke's Law to find out how much the bowstring will move when the archer pulls with a given force.
Understanding Displacement Calculation
Displacement is all about figuring out how much the bowstring stretches when a force is applied. In this context, displacement tells us the distance by which the bowstring is pulled back. To find this, we use the handy Formula from Hooke's Law: \[F = kx\]Where \( F \) is the force applied, \( k \) is the spring constant, and \( x \) is the displacement.

To solve for the displacement \( x \), we rearrange the formula:\[x = \frac{F}{k}\]

Using the values from the problem, where the force \( F \) is \( 240 \mathrm{N} \) and the spring constant \( k \) is \( 480 \mathrm{N/m} \), the calculation becomes:\[x = \frac{240}{480} = 0.5 \,\mathrm{m}\]

This means the bowstring is pulled back by 0.5 meters. It’s a simple division that gives us a clear understanding of how much movement occurs when a specific force is applied to an ideal spring system.
Concept of Ideal Spring Behavior
An ideal spring is a perfect simplification used in physics. It behaves in an absolutely predictable way according to Hooke's Law. But what does this actually mean? It implies that the force exerted by the spring is directly proportional to the displacement from its rest position.

For real-world applications, ideal springs provide a foundational model, even if real springs may not align perfectly due to factors like material fatigue or non-linear forces at extreme stretches. However, under normal conditions and within its elastic limit, many springs approximate this ideal behavior quite closely.

The bow in our exercise represents an ideal spring. This means it reliably follows Hooke's Law across the range of its motion, allowing straightforward calculations and predictions about how it will behave when the archer applies force. This ideality assumes no energy loss to friction or air resistance, creating a pure mathematical model that simplifies solving for displacement or force in practical situations.

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Most popular questions from this chapter

A vertical spring with a spring constant of \(450 \mathrm{N} / \mathrm{m}\) is mounted on the floor. From directly above the spring, which is unstrained, a \(0.30-\mathrm{kg}\) block is dropped from rest. It collides with and sticks to the spring, which is compressed by \(2.5 \mathrm{cm}\) in bringing the block to a momentary halt. Assuming air resistance is negligible, from what height (in \(\mathrm{cm}\) ) above the compressed spring was the block dropped?

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?

A \(0.70-\mathrm{kg}\) block is hung from and stretches a spring that is attached to the ceiling. A second block is attached to the first one, and the amount that the spring stretches from its unstrained length triples. What is the mass of the second block?

You are running from pirates on a tropical island somewhere in the Caribbean. You have somehow become separated from the rest of your group and now find yourself on the edge of a cliff with your pursuers less than 10 minutes behind you. According to a sign posted on the guardrail at the cliff's edge, the drop to the beach below is \(h=140\) feet. Your team members (waiting for you on the beach, near your boat) have a rope, but there is no time for anyone to climb the cliff to save you. You break into a deserted cabin nearby, and rummage around for a rope. Instead, you find a brand new, still-in-package, bungee cord that must have been intended for tourists jumping from a nearby bridge. You figure you might be able to attach it to the guardrail and jump to the beach, letting go at the bottom before it reverses your motion. You read the bungee cord specifications on the package: the total length of the cord is \(L_{0}=100 \mathrm{m},\) the maximum elastic deformation is \(200 \%\) (i.e., it can safely triple its length), and the elastic constant is \(k=75.0 \mathrm{N} / \mathrm{m}\). (a) If you weigh \(170 \mathrm{lb},\) how far is the bungee cord designed to let you fall before it stops you and reverses your direction? Will this afford you a safe landing? (b) You realize that you don't have to hang from the very end of the bungee, but rather from some point in the middle. How far from the attached end should you grasp the unstretched bungee cord so that you land softly on the beach? Will you be able to perform the jump and stay under the elastic deformation limit?

A heavy-duty stapling gun uses a 0.140 -kg metal rod that rams against the staple to eject it. The rod is attached to and pushed by a stiff spring called a "ram spring" \((k=32000 \mathrm{N} / \mathrm{m})\). The mass of this spring may be ignored. The ram spring is compressed by \(3.0 \times 10^{-2} \mathrm{m}\) from its unstrained length and then released from rest. Assuming that the ram spring is oriented vertically and is still compressed by \(0.8 \times 10^{-2} \mathrm{m}\) when the downward-moving ram hits the staple, find the speed of the ram at the instant of contact.

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