/*! This file is auto-generated */ .wp-block-button__link{color:#fff;background-color:#32373c;border-radius:9999px;box-shadow:none;text-decoration:none;padding:calc(.667em + 2px) calc(1.333em + 2px);font-size:1.125em}.wp-block-file__button{background:#32373c;color:#fff;text-decoration:none} Problem 28 During spring semester at MIT, r... [FREE SOLUTION] | 91Ó°ÊÓ

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During spring semester at MIT, residents of the parallel buildings of the East Campus dorms battle one another with large catapults that are made with surgical hose mounted on a window frame. A balloon filled with dyed water is placed in a pouch attached to the hose, which is then stretched through the width of the room. Assume that the stretching of the hose obeys Hooke's law with a spring constant of \(100 \mathrm{~N} / \mathrm{m} .\) If the hose is stretched by \(5.00 \mathrm{~m}\) and then released, how much work does the force from the hose do on the balloon in the pouch by the time the hose reaches its relaxed length?

Short Answer

Expert verified
The work done by the hose on the balloon is 1250 J.

Step by step solution

01

Identify Given Variables

We are given a spring constant \(k = 100 \text{ N/m}\) and the stretch length \(x = 5.00 \text{ m}\). We need to find the work done by the force when the spring returns to its natural length.
02

Apply Hooke's Law

Hooke's Law describes the force exerted by a spring: \( F = kx \). Here, \(k\) is the spring constant and \(x\) is the displacement from the equilibrium position.
03

Formula for Work Done by a Spring

The work (\(W\)) done by the spring as it moves from its stretched position back to its equilibrium is given by the formula: \[ W = \frac{1}{2} k x^2 \] This formula arises from integrating the force over the displacement.
04

Substitute Given Values

Substitute the given values into the work formula:\[ W = \frac{1}{2} \times 100 \text{ N/m} \times (5.00 \text{ m})^2 \]
05

Calculate Work Done

Carrying out the calculation: \[ W = \frac{1}{2} \times 100 \times 25 = 1250 \text{ J} \]So, the work done by the spring is 1250 Joules.

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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 concept of the spring constant is a fundamental part of Hooke's Law. It is a measure of a spring's stiffness. In our original exercise, the spring constant is given as \(100 \text{ N/m}\). This means that for every meter the spring is stretched, a force of 100 Newtons is required to hold it in that position. The larger the spring constant, the stiffer the spring. A small spring constant indicates a softer spring that stretches more easily.

The spring constant \(k\) is crucial when calculating the force exerted by a spring, described by the formula \(F = kx\), where \(x\) is the displacement. If a spring constant is higher, it indicates that the spring is more resistant to being stretched or compressed. In real-world applications, knowing the spring constant helps engineers and scientists design systems, like suspension in vehicles or the launching mechanism in our balloon example, with the desired elasticity and force properties.
Calculating the Work Done by a Spring
Work done by a spring is an interesting phenomenon governed by Hooke's Law. When a spring is stretched, potential energy is stored within it, and when released, this energy performs work. The formula used to calculate the work done by a spring is: \[W = \frac{1}{2} k x^2\]This formula results from the integration of Hooke's Law over the displacement of the spring. In our MIT physics problem, the spring with the constant \(k = 100 \text{ N/m}\) is stretched 5 meters. Substituting these values into the formula helps us find that the work done by the spring when it returns to its unstretched position is 1250 Joules.

This concept is important as it describes how energy is transferred from potential to kinetic as the spring returns to its natural state. Understanding this transfer is crucial in fields ranging from mechanical engineering to biomechanics.
Exploring the MIT Physics Problem Context
The MIT physics problem we're dealing with is not just a fun exercise involving water balloons. It serves as a practical demonstration of Hooke's Law in real-world applications. The students at MIT use surgical hoses as springs for their catapults, stretching them to store energy before launching balloons. This problem illustrates how theoretical physics concepts are applied in everyday situations, such as games or construction.

Using a spring as a launching mechanism showcases the conversion of potential energy stored during stretching into kinetic energy as the balloon is propelled. This practical problem helps solidify learning by illustrating abstract concepts such as spring constant, work, and energy conversion with an engaging and relatable scenario. Understanding such problems enhances the comprehension of how physics governs our environment and impacts technological advancements.

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Most popular questions from this chapter

The force on a particle is directed along an \(x\) axis and given by \(F=F_{0}\left(x / x_{0}-1\right)\). Find the work done by the force in moving the particle from \(x=0\) to \(x=2 x_{0}\) by (a) plotting \(F(x)\) and measuring the work from the graph and (b) integrating \(F(x)\).

A helicopter lifts a \(72 \mathrm{~kg}\) astronaut \(15 \mathrm{~m}\) vertically from the ocean by means of a cable. The acceleration of the astronaut is \(g / 10 .\) How much work is done on the astronaut by (a) the force from the helicopter and (b) the gravitational force on her? Just before she reaches the helicopter, what are her (c) kinetic energy and (d) speed?

A \(1.5 \mathrm{~kg}\) block is initially at rest on a horizontal frictionless surface when a horizontal force along an \(x\) axis is applied to the block. The force is given by \(\vec{F}(x)=\left(2.5-x^{2}\right) \hat{\mathrm{i}} \mathrm{N},\) where \(x\) is in meters and the initial position of the block is \(x=0 .\) (a) What is the kinetic energy of the block as it passes through \(x=2.0 \mathrm{~m} ?\) (b) What is the maximum kinetic energy of the block between \(x=0\) and \(x=2.0 \mathrm{~m} ?\)

A proton (mass \(m=1.67 \times 10^{-27} \mathrm{~kg}\) ) is being accelerated along a straight line at \(3.6 \times 10^{15} \mathrm{~m} / \mathrm{s}^{2}\) in a machine. If the proton has an initial speed of \(2.4 \times 10^{7} \mathrm{~m} / \mathrm{s}\) and travels \(3.5 \mathrm{~cm},\) what then is (a) its speed and (b) the increase in its kinetic energy?

To push a \(25.0 \mathrm{~kg}\) crate up a frictionless incline, angled at \(25.0^{\circ}\) to the horizontal, a worker exerts a force of \(209 \mathrm{~N}\) parallel to the incline. As the crate slides \(1.50 \mathrm{~m},\) how much work is done on the crate by (a) the worker's applied force, (b) the gravitational force on the crate, and (c) the normal force exerted by the incline on the crate? (d) What is the total work done on the crate?

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