/*! 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 51 A light spring having a force co... [FREE SOLUTION] | 91Ó°ÊÓ

91Ó°ÊÓ

A light spring having a force constant of \(125 \mathrm{~N} / \mathrm{m}\) is used to pull a \(9.50 \mathrm{~kg}\) sled on a horizontal frictionless ice rink. If the sled has an acceleration of \(2.00 \mathrm{~m} / \mathrm{s}^{2},\) by how much does the spring stretch if it pulls on the sled (a) horizontally, (b) at \(30.0^{\circ}\) above the horizontal?

Short Answer

Expert verified
(a) The spring stretches 0.152 m horizontally. (b) The spring stretches 0.175 m at 30°.

Step by step solution

01

Determine the Horizontal Force Required

Given that the sled has a mass of \(9.50 \mathrm{~kg}\) and an acceleration of \(2.00 \mathrm{~m} / \mathrm{s}^{2}\), use Newton's Second Law \(F = ma\) to find the force required to accelerate the sled. \[F = 9.50 \times 2.00 = 19.00 \mathrm{~N}\]
02

Calculate the Stretch of the Spring (Horizontally)

The spring force is given by Hooke's Law \( F = kx \), where \(k\) is the spring constant and \(x\) is the stretch of the spring. Solve for \(x\) when the spring pulls horizontally:\[19.00 = 125x \quad \Rightarrow \quad x = \frac{19.00}{125} = 0.152 \mathrm{~m}\]
03

Determine the Effective Force at an Angle

When the spring pulls at a \(30.0^{\circ}\) angle, only the horizontal component of the force contributes to the acceleration. This component is given by \(F_{x} = F \cos(\theta)\). Therefore:\[F_{x} = 19.00 \div \cos(30.0^{\circ}) \approx 19.00 \div 0.866 = 21.94 \mathrm{~N}\]
04

Calculate the Stretch of the Spring (At an Angle)

Use Hooke's Law again to find the stretch of the spring when pulling at an angle:\[21.94 = 125x \quad \Rightarrow \quad x = \frac{21.94}{125} = 0.175 \mathrm{~m}\]

Unlock Step-by-Step Solutions & Ace Your Exams!

  • Full Textbook Solutions

    Get detailed explanations and key concepts

  • Unlimited Al creation

    Al flashcards, explanations, exams and more...

  • Ads-free access

    To over 500 millions flashcards

  • Money-back guarantee

    We refund you if you fail your exam.

Over 30 million students worldwide already upgrade their learning with 91Ó°ÊÓ!

Key Concepts

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

Hooke's Law
Hooke's Law is a principle that relates the force exerted by a spring to its displacement from the equilibrium position. It is expressed through the formula \( F = kx \). Here, \( F \) is the force applied by the spring, \( k \) is the force constant of the spring, and \( x \) represents the stretch or compression of the spring. Hooke's Law assumes that the spring remains within its elastic limit, where it can still return to its original shape without permanent deformation. This law is fundamental when analyzing systems using springs because it provides a clear relationship between force and displacement, helping us predict how far a spring will stretch or compress under a given force. Mainly, it's crucial for determining how the performance of a spring changes with varying conditions, such as different orientations or external forces.
Force Constant
The force constant, represented by \( k \), is a value indicating the stiffness of a spring. Measured in Newtons per meter (N/m), it dictates how much force is needed to stretch or compress the spring by a unit distance. A higher force constant means a stiffer spring, which requires more force to stretch it. In the exercise, the spring has a force constant of \( 125 \, \mathrm{N/m} \). Understanding the force constant is essential because it allows you to calculate the stretch of the spring when a specific force is applied. Knowing \( k \) and using Hooke’s Law, you can easily find how a spring behaves under loading conditions. It helps in designing mechanical systems where specific spring behaviors are needed.
Spring Stretch
Spring stretch refers to how much a spring elongates when a force is applied. Using Hooke's Law, you can determine the stretch \( x \) by rearranging the formula to \( x = \frac{F}{k} \). In the horizontal scenario of the exercise, a force of 19.00 N results in a stretch of the spring by \( 0.152 \, \mathrm{m} \). When force changes due to additional factors like angles, the stretch changes as well, requiring recalculation using the adjusted force value. My understanding is knowing precisely how much a spring stretches is important in practical applications where exact distances and forces impact the overall system performance, such as in the suspension systems of vehicles or measuring forces and distances in controlled laboratory experiments.
Horizontal Force
Horizontal force is a component of the total force acting in the horizontal plane. In the simplest scenario, this force directly affects motion with no angle of deviation, as seen in the exercise. Newton’s Second Law \( F = ma \) helps calculate the horizontal force necessary to move the sled, accounting for only the mass and acceleration of the object.
In real-world applications, it is crucial to consider horizontal forces when designing and analyzing systems subject to various loads and forces that could affect their motion. Understanding horizontal forces allows engineers to predict how objects will behave under applied loads and helps in optimizing system designs to ensure safety and effectiveness. This becomes significantly more useful when dealing with non-horizontal forces, where determining the horizontal component is crucial for analysis and understanding of real-world scenarios like inclined planes or when additional forces act at angles.

One App. One Place for Learning.

All the tools & learning materials you need for study success - in one app.

Get started for free

Most popular questions from this chapter

An average person can reach a maximum height of about \(60 \mathrm{~cm}\) when jumping straight up from a crouched position. During the jump itself, the person's body from the knees up typically rises a distance of around \(50 \mathrm{~cm}\). To keep the calculations simple and yet get a reasonable result, assume that the entire body rises this much during the jump. (a) With what initial speed does the person leave the ground to reach a height of \(60 \mathrm{~cm} ?\) (b) Make a free-body diagram of the person during the jump. (c) In terms of this jumper's weight \(\underline{W}\). what force does the ground exert on him or her during the jump?

You've attached a bungee cord to a wagon and are using it to pull your little sister while you take her for a jaunt. The bungee's unstretched length is \(1.3 \mathrm{~m}\), and you happen to know that your little sister weighs \(220 \mathrm{~N}\) and the wagon weighs \(75 \mathrm{~N}\). Crossing a street, you accelerate from rest to your normal walking speed of \(1.5 \mathrm{~m} / \mathrm{s}\) in \(2.0 \mathrm{~s},\) and you notice that while you're accelerating, the bungee's length increases to about \(2.0 \mathrm{~m}\). What's the force constant of the bungee cord, assuming it obeys Hooke's law?

At a construction site, a \(22.0 \mathrm{~kg}\) bucket of concrete is connected over a very light frictionless pulley to a \(375 \mathrm{~N}\) box on the roof of a building. (See Figure \(5.55 .)\) There is no appreciable friction on the box, since it is on roller bearings. The box starts from rest. (a) Make free- body diagrams of the bucket and the box. (b) Find the acceleration of the bucket. (c) How fast is the bucket moving after it has fallen \(1.50 \mathrm{~m}\) (assuming that the box has not yet reached the edge of the roof)?

Shoes for the sports of bouldering and rock climbing are designed to provide a great deal of friction between the foot and the surface of the ground. On smooth rock these shoes might have a coefficient of static friction of 1.2 and a coefficient of kinetic friction of 0.90 . For a person wearing these shoes, what's the maximum angle (with respect to the horizontal) of a smooth rock that can be walked on without slipping? A. \(42^{\circ}\) B. \(50^{\circ}\) C. \(64^{\circ}\) D. Greater than \(90^{\circ}\)

A \(25.0 \mathrm{~kg}\) box of textbooks rests on a loading ramp that makes an angle \(\alpha\) with the horizontal. The coefficient of kinetic friction is \(0.25,\) and the coefficient of static friction is \(0.35 .\) (a) As the angle \(\alpha\) is increased, find the minimum angle at which the box starts to slip. (b) At this angle, find the acceleration once the box has begun to move. (c) At this angle, how fast will the box be moving after it has slid \(5.0 \mathrm{~m}\) along the loading ramp?

See all solutions

Recommended explanations on Physics Textbooks

View all explanations

What do you think about this solution?

We value your feedback to improve our textbook solutions.

Study anywhere. Anytime. Across all devices.