/*! 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 25 You are asked to design a spring... [FREE SOLUTION] | 91Ó°ÊÓ

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You are asked to design a spring that will give a \(1160-\mathrm{kg}\) satellite a speed of 2.50 \(\mathrm{m} / \mathrm{s}\) relative to an orbiting space shuttle. Your spring is to give the satellite a maximum acceleration of 5.00 \(\mathrm{g}\) . The spring's mass, the recoil kinetic energy of the shuttle, and changes in gravitational potential energy will be negligible. (a) What must the force constant of the spring be? (b) What distance must the spring be compressed?

Short Answer

Expert verified
(a) The force constant of the spring is approximately 446584 N/m. (b) The spring must be compressed about 0.1274 m.

Step by step solution

01

Determine the Maximum Force

To find the maximum force exerted by the spring, use the maximum acceleration the satellite can have. The maximum acceleration is given by 5.00 g, where g is the acceleration due to gravity (9.81 m/s²).First, calculate the maximum acceleration:\[ a_{ ext{max}} = 5.00 \times 9.81 \, \text{m/s}^2 = 49.05 \, \text{m/s}^2 \]Next, use Newton's second law \( F = ma \) to find the maximum force:\[ F_{ ext{max}} = 1160 \, \text{kg} \times 49.05 \, \text{m/s}^2 = 56898 \, \text{N} \]
02

Use Energy Conservation to Find Spring Constant

We need to find the spring constant \( k \). Use the energy conservation principle which involves the kinetic energy of the satellite and the potential energy stored in the spring.The kinetic energy \( KE \) of the satellite when it reaches a speed of 2.50 m/s is:\[ KE = \frac{1}{2} m v^2 = \frac{1}{2} \times 1160 \, \text{kg} \times (2.50 \, \text{m/s})^2 = 3625 \, \text{J} \]The potential energy stored in the compressed spring at maximum displacement is equal to the kinetic energy:\[ \frac{1}{2} k x^2 = 3625 \, \text{J} \]
03

Relate Force and Displacement to Solve for Spring Constant

We know from Hooke's Law that \( F = kx \), where \( F = 56898 \, \text{N} \) and \( x \) is the displacement.From the previous step, we can use the relation for energy:\[ k x^2 = 2 \times 3625 \, \text{J} \to k = \frac{2 \times 3625}{x^2} \]Substitute \( F = kx \) into the equation:\[ x = \frac{F}{k} \to x = \frac{56898}{k} \]Plug this into the energy equation:\[ k \left( \frac{56898}{k} \right)^2 = 2 \times 3625 \]Solving gives:\[ 56898^2 = 2 \times 3625 \times k \Rightarrow k = \frac{56898^2}{2 \times 3625} \Rightarrow k \approx 446584 \, \text{N/m} \]
04

Calculate Spring Compression Distance

Now we can find the distance \( x \) the spring needs to be compressed using \( F = kx \):\[ x = \frac{F}{k} = \frac{56898}{446584} \approx 0.1274 \, \text{m} \]

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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 describes how the force required to extend or compress a spring is proportional to the distance it is stretched or compressed. This is expressed in the formula:
  • \( F = kx \)
Where:
  • \( F \) is the force applied to the spring (in Newtons, N)
  • \( k \) is the spring constant (in Newtons per meter, N/m), characterizing the stiffness of the spring
  • \( x \) is the displacement of the spring from its equilibrium position (in meters, m)
A larger spring constant \( k \) indicates a stiffer spring, which requires more force for a given displacement. In the context of the exercise, by understanding Hooke's Law, we calculated the maximum force the spring could exert on the satellite to determine how much it needed to be compressed.
Energy Conservation
Energy conservation is a vital principle governing physical interactions, stating that energy cannot be created or destroyed, only transformed from one form to another. In the exercise, energy conservation helps find the spring constant by equating the kinetic energy a satellite gains to the potential energy stored by the spring.The key concept is that when the spring is compressed, it stores potential energy described by:
  • \( PE_{spring} = \frac{1}{2}kx^2 \)
When released, this potential energy converts into kinetic energy \( KE \) defined by:
  • \( KE = \frac{1}{2}mv^2 \)
Ensuring energy conservation implies \( \frac{1}{2}kx^2 = \frac{1}{2}mv^2 \). This relationship allows us to calculate the spring constant \( k \) once the kinetic energy is known, confirming that no energy is lost in the transition from potential to kinetic.
Kinetic Energy
Kinetic energy is the energy of an object in motion and is indispensable for solving various physics problems, including those involving springs and accelerations. It allows scientists to predict how objects move and interact.For a moving satellite, its kinetic energy is calculated using:
  • \( KE = \frac{1}{2}mv^2 \)
Where:
  • \( m \) is the satellite's mass
  • \( v \) is its velocity
In our exercise, knowing the satellite’s final velocity lets us determine its kinetic energy, which equates to the stored potential energy in the spring. This step is crucial to sizing the spring accurately to achieve the desired satellite speed upon release without any additional forces.
Newton's Second Law
Newton's second law of motion bridges the ideas of force, mass, and acceleration through the iconic equation:
  • \( F = ma \)
This law is pivotal in calculating the force exerted by or on an object under uniform acceleration.In the context of this satellite problem, we first compute the maximum acceleration possible for the satellite (given as 5 times the acceleration due to gravity \( g \)). Substituting this acceleration into the equation allows us to calculate the maximum possible force exerted on the satellite. It embodies the essence of Newton's second law—how an unbalanced force acting on a mass always causes acceleration. Understanding and applying this law forms the foundational step for evaluating forces in mechanics, especially for systems involving springs like ours.

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Most popular questions from this chapter

A 1.20 \(\mathrm{kg}\) 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.)

A proton with mass \(m\) moves in one dimension. The potential-energy function is \(U(x)=\alpha / x^{2}-\beta / x,\) where \(\alpha\) and \(\beta\) are positive constants. The proton is released from rest at \(x_{0}=\alpha / \beta\) (a) Show that \(U(x)\) can be written as $$ U(x)=\frac{\alpha}{x_{0}^{2}}\left[\left(\frac{x_{0}}{x}\right)^{2}-\frac{x_{0}}{x}\right] $$ Graph \(U(x)\) . Calculate \(U\left(x_{0}\right)\) and thereby locate the point \(x_{0}\) on the graph. (b) Calculate \(v(x),\) the speed of the proton as a function of position. Graph \(v(x)\) and give a qualitative description of the motion. (c) For what value of \(x\) is the speed of the proton a maximum? What is the value of that maximum speed? (d) What is the force on the proton at the point in part (c)? (e) Let the proton be released instead at \(x_{1}=3 \alpha / \beta\) . Locate the point \(x_{1}\) on the graph of \(U(x)\) . Calculate \(v(x)\) and give a qualitative description of the motion. (f) For each release point \(\left(x=x_{0} \text { and } x=x_{1}\right),\) what are the maximum and minimum values of \(x\) reached during the motion?

A 0.60\(\cdot \mathrm{kg}\) book slides on a horizontal table. The kinetic friction force on the book has magnitude 1.2 \(\mathrm{N}\) . (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 \(2.00-\mathrm{kg}\) 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. 7.30\()\) . (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?

An empty crate is given an initial push down a ramp, starting it with a speed \(v_{0},\) and reaches the bottom with speed \(v\) and kinetic energy \(K .\) Some books are now placed in the crate, so that the total mass is quadrupled. The coefficient of kinetic friction is constant and air resistance is negligible. Starting again with \(v_{0}\) at the top of the ramp, what are the speed and kinetic energy at the bottom? Explain the reasoning behind your answers.

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