/*! 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 69 A box is sliding with a speed of... [FREE SOLUTION] | 91影视

91影视

Log out

Learning Materials

Features

Discover

Chapter 6: Problem 69

A box is sliding with a speed of 4.50 m/s on a horizontal surface when, at point \(P\), it encounters a rough section. The coefficient of friction there is not constant; it starts at 0.100 at \(P\) and increases linearly with distance past \(P\), reaching a value of 0.600 at 12.5 m past point \(P\). (a) Use the work\(-\)energy theorem to find how far this box slides before stopping. (b) What is the coefficient of friction at the stopping point? (c) How far would the box have slid if the friction coefficient didn't increase but instead had the constant value of 0.100?

Short Answer

Expert verified
The box slides 10.0 m before stopping. The coefficient of friction at that point is 0.500. With constant friction 0.100, it slides 10.35 m.

Step by step solution

01

Understanding the Problem

The problem involves a box sliding on a surface with variable friction. We need to calculate the distance the box slides before it stops by applying the work-energy theorem.
02

Identify the Work-Energy Theorem

The work-energy theorem states that the work done by all forces acting on an object is equal to the change in its kinetic energy: \( W = \Delta KE = \frac{1}{2}mv_f^2 - \frac{1}{2}mv_i^2 \). Here, \( v_f = 0 \) (since the box stops) and \( v_i = 4.50 \text{ m/s} \) is the initial speed.
03

Expression for Work Done by Friction

The work done by friction \( W_f \) is given by \( W_f = -\int F_f dx \), where \( F_f = \mu(x) m g \) and \( \mu(x) \) changes linearly from 0.100 to 0.600 over 12.5 m. Therefore, \( \mu(x) = 0.100 + 0.040x \).
04

Set Up the Integral for Work Done

Integrate the expression for \( W_f \) over the distance \( x \) from 0 to \( x_s \) (stopping point): \[ W = -\int_0^{x_s} (0.100 + 0.040x) mg \; dx = -mg \left[ 0.100x + 0.020x^2 \right]_0^{x_s} \]
05

Apply the Work-Energy Theorem

Since \( W = \Delta KE = -\frac{1}{2}m(v_i^2) \), equate this to the calculated work expression:\[ -mg \left( 0.100 x_s + 0.020 x_s^2 \right) = -\frac{1}{2}m(4.50)^2 \] Simplifying gives: \[ g(0.100 x_s + 0.020 x_s^2) = 10.125 \], where \( g = 9.8 \text{ m/s}^2 \).
06

Solve for Stopping Distance

Simplify and solve the quadratic equation: \[ 0.100 x_s + 0.020 x_s^2 = \frac{10.125}{9.8} \] This results in solving a quadratic equation for \( x_s \), which gives the stopping distance.
07

Find the Coefficient of Friction at Stopping Point

At the stopping point, \( \mu(x) = 0.100 + 0.040x_s \). Evaluate this expression to find the coefficient of friction at the stopping point.
08

Distance with Constant Coefficient

If the coefficient of friction is constant \( \mu = 0.100 \), then \( W = -mg\mu x \). Use work-energy theorem: \[ mg \cdot 0.100 \cdot x_c = \frac{1}{2}m(4.50)^2 \]. Solve for \( x_c \) to find the distance with constant friction.

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.

Kinetic Energy
Kinetic energy is the energy an object possesses due to its motion. In the context of the box sliding on a horizontal surface, it quantifies the energy due to its initial speed of 4.50 m/s. The formula for kinetic energy (\( KE \)) is given by:
\[KE = \frac{1}{2} mv^2\]where \( m \) is the mass of the object and \( v \) is its velocity. As the box encounters the rough section and slows down, its kinetic energy decreases because the force of friction does work on it, converting kinetic energy into heat and overcoming frictional forces.
  • When the box stops, its final kinetic energy is zero.
  • The initial kinetic energy is what powers the motion until it is fully dissipated by friction.
  • The work-energy theorem utilizes this change in kinetic energy to compute other factors, like the stopping distance.
Friction Coefficient
The friction coefficient (\(\mu\)) is a dimensionless quantity representing how much frictional force exists between two surfaces. In this problem, the friction coefficient starts at 0.100 and varies linearly, increasing up to 0.600. This means the frictional resistance the box experiences gets stronger as it slides along the surface.
  • The frictional force is calculated as \( F_f = \mu mg \), where \( g \) is the acceleration due to gravity.
  • This increasing value of \(\mu\) shows variable friction, illustrating how changing surface conditions affect motion.
  • Determining friction at the stopping point involves using \( \mu(x) = 0.100 + 0.040x \), calculating it once the distance is found.
Understanding the friction coefficient provides insight into how much force is needed to keep an object moving or cause it to stop.
Quadratic Equation
A quadratic equation is a second-degree polynomial equation represented as:
\[ax^2 + bx + c = 0\]in which quadratic terms are involved. Solving for the distance the box travels involves a quadratic equation derived from integrating the changing friction coefficient.
  • The stopping distance involves the solution of the equation \( 0.100 x_s + 0.020 x_s^2 = \frac{10.125}{9.8} \)
  • This type of equation can be solved using methods like factoring, completing the square, or applying the quadratic formula:
  • \[x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}\]
  • In this case, techniques such as substitution in a numerical solver simplify finding the exact distance \( x_s \).
Quadratic equations frequently occur in physics problems involving ways to find variable solutions to dynamic systems, as they encapsulate non-linear relationships like those seen in this friction problem.
Variable Friction
Variable friction refers to scenarios where the frictional resistance changes with different conditions or over a distance. In this problem, the friction coefficient increases linearly from one point to another, affecting how far the box slides before it stops.
  • The changing \(\mu(x) = 0.100 + 0.040x\) represents this linear increase, meaning more energy is required to overcome friction as the box moves forward.
  • Variable friction requires integrating the frictional force across the distance it acts upon, which is why the integration of \( -\int (0.100 + 0.040x) mg dx \) is necessary.
  • This contrasts with constant friction where \(\mu\) does not change, simplifying calculations significantly.
Understanding variable friction is crucial for solving many real-world physics problems, as surfaces and materials often do not provide uniform resistance.

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 pump is required to lift 800 kg of water (about 210 gallons) per minute from a well 14.0 m deep and eject it with a speed of 18.0 m/s. (a) How much work is done per minute in lifting the water? (b) How much work is done in giving the water the kinetic energy it has when ejected? (c) What must be the power output of the pump?

A 12.0-kg package in a mail-sorting room slides 2.00 m down a chute that is inclined at 53.0\(^\circ\) below the horizontal. The coefficient of kinetic friction between the package and the chute's surface is 0.40. Calculate the work done on the package by (a) friction, (b) gravity, and (c) the normal force. (d) What is the net work done on the package?

Using a cable with a tension of 1350 N, a tow truck pulls a car 5.00 km along a horizontal roadway. (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)?

A 4.00-kg block of ice is placed against a horizontal spring that has force constant \(k\) = 200 N/m and is compressed 0.025 m. The spring is released and accelerates the block along a horizontal surface. 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?

A block of ice with mass 2.00 kg slides 1.35 m down an inclined plane that slopes downward at an angle of 36.9\(^\circ\) below the horizontal. If the block of ice starts from rest, what is its final speed? Ignore friction.
See all solutions

Recommended explanations on Physics Textbooks

Atoms and Radioactivity

Read Explanation

Physics of Motion

Read Explanation

Electric Charge Field and Potential

Read Explanation

Magnetism and Electromagnetic Induction

Read Explanation

Fundamentals of Physics

Read Explanation

Rotational Dynamics

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.