/*! 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 44 On a horizontal surface, a crate... [FREE SOLUTION] | 91影视

91影视

Log out

Learning Materials

Features

Discover

Chapter 7: Problem 44

On a horizontal surface, a crate with mass 50.0 \(\mathrm{kg}\) is placed against a spring that stores 360 \(\mathrm{J}\) of energy. The spring is released, and the crate slides 5.60 \(\mathrm{m}\) before coming to rest. What is the speed of the crate when it is 2.00 \(\mathrm{m}\) from its initial position?

Short Answer

Expert verified
The speed of the crate at 2.00 m is approximately 3.04 m/s.

Step by step solution

01

Understand the Problem

We need to determine the speed of a crate after it has moved 2.00 meters from its initial position following the release of a spring that stored potential energy.
02

Identify Given Values and Equations

We have the mass of the crate as 50.0 kg and the spring energy as 360 J. The total distance to come to rest is 5.60 m, and we need to find the speed at 2.00 m. Use the equation for kinetic energy, \( KE = \frac{1}{2} mv^2 \), and the relationship between energy and work, \( Work = Energy_{initial} - Energy_{final} \).
03

Calculate Work Done through Friction at 2.00 m.

The work done by friction can be found using the formula \( Work = \mu mgd \), where \( \mu \) is the friction coefficient, \( m \) is the mass, \( g \) is the gravitation constant (9.8 m/s虏), and \( d \) is distance. Since the total work to stop the crate is 360 J over 5.6 m, the work done over 2.0 m can be calculated as \( Work_{2.0m} = \frac{360 J}{5.60 m} \times 2.00 m = 128.57 J \).
04

Determine Kinetic Energy at 2.00 m

The kinetic energy at 2.00 m is given by the initial spring energy minus the work done against friction: \( KE_{2.00m} = 360 J - 128.57 J = 231.43 J \).
05

Solve for Speed from Kinetic Energy

Using \( KE = \frac{1}{2} mv^2 \), we can rearrange to find speed \( v \): \( v = \sqrt{\frac{2 \times KE}{m}} \). Substitute \( KE = 231.43 J \) and \( m = 50.0 kg \): \( v = \sqrt{\frac{2 \times 231.43}{50}} \approx \sqrt{9.2572} \approx 3.04 \, \text{m/s} \).

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 91影视!

Key Concepts

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

Energy Conservation
Energy conservation is a fundamental principle in physics that states that energy cannot be created or destroyed, only transformed or transferred. This principle tells us that the initial energy from the spring will eventually convert into kinetic energy of the crate and, soon after, energy lost to friction. The total mechanical energy of the system remains constant unless work is done by an external force. In the given problem, the spring initially has 360 J of potential energy. As the spring releases, this energy begins transferring into kinetic energy as well as being used to overcome friction between the crate and the surface. Understanding energy transformations is crucial. Here, energy conservation helps us to systematically divide and calculate the different forms of energy at various points, such as the kinetic energy at 2.00 m.
Kinetic Energy
Kinetic energy is the energy an object possesses due to its motion. When the crate moves after the spring is released, it gains kinetic energy. The formula for kinetic energy is given by \[ KE = \frac{1}{2} mv^2 \]where \( m \) is the mass of the crate and \( v \) is its velocity. The problem involves calculating the kinetic energy of the crate at the 2.00 m mark to estimate how fast it is moving there. In the context of the exercise, at 2.00 m, there is a conversion of potential energy from the spring into kinetic energy of the crate. As per the solution given, after accounting for the work done against friction, the kinetic energy was determined to be 231.43 J. This is the remaining energy that has been converted into motion, helping us to calculate the speed of the crate at this distance.
Frictional Force
Frictional force acts opposite to the direction of motion and converts kinetic energy into heat, thus opposing motion. This force depends on the nature of the surfaces in contact and the normal force, which in this case is the weight of the crate.In our problem, friction plays a key role since it reduces the kinetic energy of the crate, slowing it down over time. We can represent this force mathematically by:\[ Work = \mu mgd \]Here, \( \mu \) is the coefficient of friction, \( m \) is the mass of the crate, \( g \) is the acceleration due to gravity, and \( d \) is the distance moved. Since friction reduces the energy available for kinetic motion, calculating the work done by friction helps in understanding how much energy is lost as the crate slides. In this exercise, the work done by friction at the 2.00 m point is 128.57 J, subtracted from the initial energy to find the kinetic energy.
Work-Energy Principle
The work-energy principle states that the work done on an object is equal to the change in its kinetic energy. This forms an essential bridge between the energy concepts and helps solve physics problems about motion and forces.For our crate, the principle is used to connect the spring's stored energy with the crate's motion. The equation\[ Work = Energy_{initial} - Energy_{final} \]is used to calculate how much energy transforms into other forms like friction. At 2.00 m, part of the spring's energy (360 J) has been used to do 128.57 J of work against friction, leaving 231.43 J to remain as kinetic energy.Therefore, understanding the work-energy principle helps clarify how energy conversion impacts the motion of the crate, ultimately leading to the calculation of its velocity.

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 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.

In one day, a \(75-\mathrm{kg}\) mountain climber ascends from the 1500 -m level on a vertical cliff to the top at 2400 \(\mathrm{m}\) . The next day, she descends from the top to the base of the cliff, which is at an elevation of 1350 \(\mathrm{m}\) . What is her change in gravitational potential energy (a) on the first day and \((b)\) on the second day?

A machine part of mass \(m\) is attached to a horizontal ideal spring of force constant \(k\) that is attached to the edge of a friction-free horizontal surface. The part is pushed against the spring compressing it a distance \(x_{0},\) and then released from rest. Find the maximum (a) speed and (b) acceleration of the machine part. (c) Where in the motion do the maxima in parts (a) and (b) occur? (d) What will be the maximum extension of the spring? (e) Describe the subsequent motion of this machine part. Will it ever stop permanently?

A \(0.100-\mathrm{kg}\) potato is tied to a string with length \(2.50 \mathrm{m},\) and the other end of the string is tied to a rigid support. The potato is held straight out horizontally from the point of support, with the string pulled taut, and is then released. (a) What is the speed of the potato at the lowest point of its motion? (b) What is the tension in the string at this point?

Gravity in Three Dimensions. A point mass \(m_{1}\) is held in place at the origin, and another point mass \(m_{2}\) is free to move a distance \(r\) away at a point \(P\) having coordinates \(x, y,\) and \(z\) . The gravitational potential energy of these masses is found to be \(U(r)=-G m_{1} m_{2} / r,\) where \(G\) is the gravitational constant (see Exercises 7.34 and 7.35 . (a) Show that the components of the force on \(m_{2}\) due to \(m_{1}\) are $$ F_{x}=-\frac{G m_{1} m_{2} x}{\left(x^{2}+y^{2}+z^{2}\right)^{3 / 2}} \quad F_{y}=-\frac{G m_{1} m_{2} y}{\left(x^{2}+y^{2}+z^{2}\right)^{3 / 2}} $$ $$ F_{z}=-\frac{G m_{1} m_{2} z}{\left(x^{2}+y^{2}+z^{2}\right)^{3 / 2}} $$ (Hint: First write \(r\) in terms of \(x, y,\) and \(z\) ) (b) Show that the magnitude of the force on \(m_{2}\) is \(F=G m_{1} m_{2} / r^{2} .\left(\text { c) Does } m_{1} \text { attract or }\right.\) repel \(m_{2} ?\) How do you know?
See all solutions

Recommended explanations on Physics Textbooks

Engineering Physics

Read Explanation

Dynamics

Read Explanation

Particle Model of Matter

Read Explanation

Thermodynamics

Read Explanation

Waves Physics

Read Explanation

Kinematics Physics

Read Explanation
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.