/*! 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 4 A spring lies on a horizontal ta... [FREE SOLUTION] | 91Ó°ÊÓ

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A spring lies on a horizontal table, and the left end of the spring is attached to a wall. The other end is connected to a box. The box is pulled to the right, stretching the spring. Static friction exists between the box and the table, so when the spring is stretched only by a small amount and the box is released, the box does not move. The mass of the box is \(0.80 \mathrm{kg}\), and the spring has a spring constant of \(59 \mathrm{N} / \mathrm{m}\). The coefficient of static friction between the box and the table on which it rests is \(\mu_{\mathrm{s}}=0.74 .\) How far can the spring be stretched from its unstrained position without the box moving when it is released?

Short Answer

Expert verified
The spring can be stretched approximately 0.098 meters (9.8 cm) without the box moving.

Step by step solution

01

Identify the Forces

When the spring is stretched, it exerts a force on the box towards its unstretched position. This force is opposed by the static frictional force that prevents the box from moving.
02

Calculate Maximum Static Friction Force

The static friction force can be calculated using the equation: \( f_s = \mu_s \times N \), where \( \mu_s = 0.74 \) and \( N = mg \) is the normal force. Here \( m = 0.80 \ \mathrm{kg} \) and \( g = 9.8 \ \mathrm{m/s^2} \). Thus, \( f_s = 0.74 \times 0.80 \times 9.8 \approx 5.792 \ \mathrm{N} \).
03

Apply Hooke's Law

Hooke's Law states that the force exerted by the spring is \( F = kx \), where \( k = 59 \mathrm{N/m} \) is the spring constant and \( x \) is the displacement. This spring force must be equal to the maximum static friction force for the box to remain stationary.
04

Solve for Maximum Stretch x

Set the spring force equal to the maximum static friction force: \( kx = f_s \). Substitute the values, \( 59x = 5.792 \). Solve for \( x \): \( x = \frac{5.792}{59} \approx 0.098 \ \mathrm{m} \).

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Key Concepts

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

Hooke's Law
Imagine you have a spring. When you stretch or compress it, it wants to return to its original shape. Hooke's Law explains this behavior. This law states that the force required to compress or stretch a spring by a distance, which we call displacement, is directly proportional to that distance. Mathematically, it is expressed as:\[ F = kx \]where - \( F \) is the force applied by or on the spring, - \( k \) is the spring constant (a measure of the spring's stiffness), - \( x \) is the displacement of the spring from its natural or rest position. A larger spring constant means a stiffer spring, requiring more force to stretch it. Understanding this law helps in determining how far a spring can stretch under a certain force without causing movement if friction is also involved.
Spring Force
Spring force is the force exerted by a spring when it is compressed or stretched. When dealing with problems involving springs, like the one in our exercise, determining the spring force is crucial. The spring force is always directed towards returning the spring back to its equilibrium or natural position. It acts in the opposite direction of the displacement. This restorative force is what keeps the spring in tension when it is pulled. Let's recall Hooke's Law: - If the spring constant \( k = 59 \, \mathrm{N/m} \)- The displacement \( x \) could be calculated based on the static friction This tells us how the spring force acts against other forces, like friction, in keeping the box from moving until enough force is applied.
Normal Force
Normal force is a basic yet foundational concept in physics. It is the support force exerted by a surface, perpendicular to the object it is in contact with. In our exercise, the normal force is crucial for understanding friction.Consider the box on the table. The normal force, denoted as \( N \), is equal in magnitude and opposite to the gravitational force acting on the box. The equation for the normal force is:\[ N = mg \]where:- \( m \) is the mass of the box, - \( g \) is the acceleration due to gravity (approximately \( 9.8 \, \mathrm{m/s^2} \) on Earth). In our case, calculating the normal force helps us find the maximum static friction force, which keeps the box stationary on the spring-loaded table.
Coefficient of Friction
The coefficient of friction is a measure of how much frictional force exists between two surfaces. It is a dimensionless constant denoted by \( \mu \). There are two types: static and kinetic. In our case, we focus on the static coefficient of friction, \( \mu_s \), because we want to know how much force is needed to start moving the box.The frictional force between the box and the table is calculated as:\[ f_s = \mu_s \times N \]where - \( f_s \) is the static friction force,- \( \mu_s \) is the static coefficient of friction, - \( N \) is the normal force. In the context of our problem, the static friction keeps the box from moving until the spring force is strong enough. The high \( \mu_s \) value means more friction, which requires a greater force to overcome, highlighting the importance of properly understanding and calculating this parameter.

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Most popular questions from this chapter

One end of a piano wire is wrapped around a cylindrical tuning peg and the other end is fixed in place. The tuning peg is turned so as to stretch the wire. The piano wire is made from steel \(\left(Y=2.0 \times 10^{11} \mathrm{N} / \mathrm{m}^{2}\right) .\) It has a radius of \(0.80 \mathrm{mm}\) and an unstrained length of \(0.76 \mathrm{m}\). The radius of the tuning peg is \(1.8 \mathrm{mm} .\) Initially, there is no tension in the wire, but when the tuning peg is turned, tension develops. Find the tension in the wire when the tuning peg is turned through two revolutions. Ignore the radius of the wire compared to the radius of the tuning peg.

A 75 -kg diver is standing at the end of a diving board while it is vibrating up and down in simple harmonic motion, as indicated in the figure. The diving board has an effective spring constant of \(k=\) \(4100 \mathrm{N} / \mathrm{m},\) and the vertical distance between the highest and lowest points in the motion is \(0.30 \mathrm{m} .\) Concepts: (i) How is the amplitude \(A\) related to the vertical distance between the highest and lowest points of the diver's motion? (ii) Starting from the top, where is the diver located one-quarter of a period later, and what can be said about his speed at this point? (iii) If the amplitude were to double, would the period also double? Explain. Calculations: (a) What is the amplitude of the motion? (b) Starting when the diver is at the highest point, what is his speed one-quarter of a period later? (c) If the vertical distance between his highest and lowest points were changed to \(0.10 \mathrm{m},\) what would be the time required for the diver to make one complete motional cycle?

Multiple-Concept Example 6 reviews the principles that play roles in this problem. A bungee jumper, whose mass is 82 kg, jumps from a tall platform. After reaching his lowest point, he continues to oscillate up and down, reaching the low point two more times in 9.6 s. Ignoring air resistance and assuming that the bungee cord is an ideal spring, determine its spring constant.

When an object of mass \(m_{1}\) is hung on a vertical spring and set into vertical simple harmonic motion, it oscillates at a frequency of \(12.0 \mathrm{Hz} .\) When another object of mass \(m_{2}\) is hung on the spring along with the first object, the frequency of the motion is \(4.00 \mathrm{Hz} .\) Find the ratio \(m_{2} / m_{1}\) of the masses.

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?

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