/*! 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 81 A 2.50 -kg textbook is forced ag... [FREE SOLUTION] | 91Ó°ÊÓ

91Ó°ÊÓ

A 2.50 -kg textbook is forced against a horizontal spring of negligible mass and force constant \(250 \mathrm{N} / \mathrm{m},\) compressing the spring a distance of 0.250 \(\mathrm{m} .\) When released, the textbook slides on a horizontal tabletop with coefficient of kinetic friction \(\mu_{\mathrm{k}}=0.30 .\) Use the work-energy theorem to find how far the textbook moves from its initial position before coming to rest.

Short Answer

Expert verified
The textbook moves approximately 1.063 meters before stopping.

Step by step solution

01

Identify Known Values

Identify and list the known values from the problem statement:- Mass of the textbook, \(m = 2.50 \, \text{kg}\).- Force constant of the spring, \(k = 250 \, \text{N/m}\).- Compression distance of the spring, \(x = 0.250 \, \text{m}\).- Coefficient of kinetic friction, \(\mu_k = 0.30\).
02

Calculate Initial Potential Energy in the Spring

The potential energy stored in a compressed spring is given by the formula:\[ U = \frac{1}{2} k x^2 \]Substitute the known values:\[ U = \frac{1}{2} \times 250 \, \text{N/m} \times (0.250 \, \text{m})^2 = 7.8125 \, \text{J} \]
03

Determine the Work Done Against Friction

The work-energy theorem states that the work done by all forces is equal to the change in kinetic energy. As the textbook comes to rest, change in kinetic energy = 0, so:\[ \text{Work done by friction} = \text{Initial spring potential energy} \]Let's denote the displacement of the textbook as \(d\). The friction force \(f_k\) is:\[ f_k = \mu_k \cdot m \cdot g = 0.30 \times 2.50 \, \text{kg} \times 9.8 \, \text{m/s}^2 = 7.35 \, \text{N} \]Thus, the work done by friction is:\[ W = f_k \cdot d = 7.35 \, \text{N} \times d \]
04

Set Up Equation for Work Done

Set the work done by friction equal to the initial potential energy of the spring:\[ 7.35 \, \text{N} \times d = 7.8125 \, \text{J} \]
05

Solve for Displacement (d)

Solve the equation from Step 4 for \(d\):\[ d = \frac{7.8125 \, \text{J}}{7.35 \, \text{N}} \approx 1.063 \, \text{m} \]
06

Conclusion

The textbook moves approximately 1.063 meters from its initial position before coming to rest.

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.

Potential Energy
When dealing with springs, understanding potential energy is crucial. Potential energy is energy stored within an object, due to its position or configuration. In the context of a spring, it's known as elastic potential energy. This energy is stored when we compress or stretch a spring. The formula to calculate the elastic potential energy in a spring is:- \[ U = \frac{1}{2} k x^2 \]- where \( U \) is the potential energy,- \( k \) is the spring force constant,- \( x \) is the compression or extension distance of the spring.In our problem, before the textbook is released, all the energy is stored as potential energy in the spring. Once released, this potential energy is what propels the textbook across the table. As the textbook moves, potential energy is transformed into kinetic energy and then into other forms like heat due to friction. This process illustrates how energy transforms and conserves within a system, following the law of conservation of energy.
Kinetic Energy
Kinetic energy is the energy an object possesses due to its motion. As the spring releases, the potential energy stored gets converted into kinetic energy. This energy is what causes the textbook to move across the table.The formula for kinetic energy is:- \[ KE = \frac{1}{2} mv^2 \]- where \( KE \) stands for kinetic energy,- \( m \) is the mass of the object,- \( v \) is the velocity.In the textbook scenario, initially, all the potential energy becomes kinetic energy as the textbook starts to move. However, as the item slides, it encounters friction, which eventually reduces the textbook's kinetic energy to zero, bringing it to rest. We are interested in how this energy transition is influenced by other forces, like friction.
Friction Force
Friction force acts against the direction of motion, opposing the movement of objects. This is a crucial concept in physics, as it explains why the motion eventually stops.In this scenario, the frictional force faced by the sliding textbook is directly related to the normal force and the coefficient of kinetic friction:- \[ f_k = \mu_k \cdot m \cdot g \]- where \( f_k \) is the frictional force,- \( \mu_k \) is the coefficient of kinetic friction,- \( m \) is mass,- \( g \) is the acceleration due to gravity.This equation tells us the friction force needed to bring the textbook to rest. As the textbook's kinetic energy is entirely transformed into heat energy via friction, the textbook halts after moving a certain distance.
Spring Force Constant
The spring force constant, denoted as \( k \), is a measure of the stiffness of a spring. A higher spring force constant implies a stiffer spring that requires more force to compress or extend.In the energy formula for springs, \( k \) defines how much potential energy is stored per unit of spring compression or extension. The spring force constant is central to determining how much energy is available to be converted into kinetic energy when the spring is released.In our problem, given \( k = 250 \) N/m, this value is used to calculate the potential energy initially stored in the spring. Knowing this energy allows us to predict how far the textbook will travel, considering other forces like friction come into play during its journey.

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

A 30.0 -kg crate is initially moving with a velocity that has magnitude 3.90 \(\mathrm{m} / \mathrm{s}\) in a direction \(37.0^{\circ}\) west of north. How much work must be done on the crate to change its velocity to 5.62 \(\mathrm{m} / \mathrm{s}\) in a direction \(63.0^{\circ}\) south of east?

A 4.00-kg block of ice is placed against a horizontal spring that has force constant \(k=200 \mathrm{N} / \mathrm{m}\) and is compressed 0.025 \(\mathrm{m}\) . The spring is released and accelerates the block along a horizontal surface. You can ignore friction and the mass of the spring. (a) Calculate the work done on the block by the spring during the motion of the block from its initial position to where the spring has returned to its uncompressed length. (b) What is the speed of the block after it leaves the spring?

An ingenious bricklayer builds a device for shooting bricks up to the top of the wall where he is working. He places a brick on a vertical compressed spring with force constant \(k=450 \mathrm{N} / \mathrm{m}\) and negligible mass. When the spring is released, the brick is propelled upward. If the brick has mass 1.80 \(\mathrm{kg}\) and is to reach a maximum height of 3.6 \(\mathrm{m}\) above its initial position on the compressed spring, what distance must the bricklayer compress the spring initially? (The brick loses contact with the spring when the spring returns to its uncompressed length. Why?

Stopping Distance. Acar is traveling on a level road with speed \(v_{0}\) at the instant when the brakes lock, so that the tires slide rather than roll. (a) Use the work-energy theorem to calculate the minimum stopping distance of the car in terms of \(v_{0}, g,\) and the coefficient of kinetic friction \(\mu_{\mathrm{k}}\) between the tires and the road. b) By what factor would the minimum stopping distance change if (i) the coefficient of kinetic friction were doubled, or (ii) the initial speed were doubled, or (iii) both the coefficient of kinetic friction and the initial speed were doubled?

A tow truck pulls a car 5.00 \(\mathrm{km}\) along a horizontal roadway using a cable having a tension of 850 \(\mathrm{N}\) . (a) How much work does the cable do on the car if it pulls horizontally? If it pulls at \(35.0^{\circ}\) above the horizontal? (b) How much work does the cable do on the tow truck in both cases of part (a)? (c) How much work does gravity do on the car in part (a)?

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.