/*! 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 67 Varying Coefficient of Friction.... [FREE SOLUTION] | 91影视

91影视

Log out

Learning Materials

Features

Discover

Chapter 6: Problem 67

Varying Coefficient of Friction. A box is sliding with a speed of 4.50 \(\mathrm{m} / \mathrm{s}\) on a horizontal surface when, at point \(P\) . it encounters a rough section. On the rough section, the coefficient of friction is not constant, but starts at 0.100 at \(P\) and increases linearly with distance past \(P\) , reaching a value of 0.600 at 12.5 \(\mathrm{m}\) past point \(P .(\text { a) Use the work-energy theorem to find how far this box }\) 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
(a) Solve a quadratic equation for distance; (b) Substitute into friction formula; (c) Use constant friction formula.

Step by step solution

01

Determine the work-energy relation

The work-energy theorem states \[W_{ ext{net}} = \Delta K\]where \( W_{\text{net}} \) is the net work done on the object and \( \Delta K \) is the change in kinetic energy. Since the box comes to rest, \( \Delta K = -\frac{1}{2} m v^2_0 \), where \( v_0 = 4.50 \mathrm{m/s} \) is the initial speed.
02

Define the force of friction

The frictional force \( f(x) \) is given by\[f(x) = \mu(x) \cdot N = \mu(x) \cdot mg\]where \( \mu(x) = 0.100 + \frac{0.600 - 0.100}{12.5}x \) is the linearly varying coefficient of friction and \( N = mg \) is the normal force.
03

Calculate the work done by friction

The work done by friction over some distance \( x \) is given by\[W = \int_{0}^{x} f(x') \, dx' = \int_{0}^{x} \mu(x') \cdot mg \, dx'\]Substituting for \( \mu(x') \), calculate \( W \) as \[W = mg\left[\int_{0}^{x} (0.100 + \frac{0.500}{12.5}x') \, dx' \right]\] which simplifies to \[W = mg \left[0.100x + \frac{0.500}{2\cdot 12.5}x^2 \right]\].
04

Solve for stopping distance using work-energy theorem

Set \( \Delta K = W \), or \[- \frac{1}{2} m v^2_0 = mg \left[0.100x + \frac{0.500}{2\cdot 12.5}x^2 \right]\] Divide by \( mg \) and solve for \( x \), \[x = 0.100x + \frac{0.500}{25}x^2 = - \frac{v^2_0}{2g}divide\]which is a quadratic equation.
05

Solve the quadratic equation

The equation is \[\frac{x^2}{25} + 0.100x + \frac{v^2_0}{2g} = 0\]Substitute \( v_0 = 4.50 \mathrm{m/s} \) and \( g = 9.81 \mathrm{m/s^2} \). Use the quadratic formula,\[x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}\]where \( a = \frac{1}{25} \), \( b = 0.100 \), and \( c = \frac{-v^2_0}{2g} \). Solve for \( x \).
06

Calculate coefficient of friction at stopping point

If you've found \( x \) in the previous step, substitute \( x \) into\[\mu(x) = 0.100 + \frac{0.500}{12.5}x\]to calculate the coefficient of friction at the stopping point.
07

Calculate distance if friction was constant

For a constant \( \mu = 0.100 \), the frictional work \[W = \mu mg x\] equal to\[x = - \frac{v^2_0}{2g \mu}\]Solve for \( x \) to find the distance if friction was constant.

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 Friction
Kinetic friction is a force that opposes the motion of an object sliding over a surface. It arises due to the interactions between the microscopic surface imperfections of the object and the surface it is moving on. In the given exercise, a box slides over a surface, encountering a rough patch where the coefficient of kinetic friction varies linearly from 0.100 to 0.600.

  • The force of kinetic friction, \( f(x) = \mu(x) \cdot N \), depends on both the coefficient of friction \( \mu(x) \) and the normal force \( N = mg \).
  • In this scenario, the coefficient varies as \( \mu(x) = 0.100 + \frac{0.600 - 0.100}{12.5}x \), representing a linear increase as the box moves across the rough patch.
  • The work done by the frictional force, which is the energy lost by the box, is calculated over the distance \( x \) the box slides.
This variable friction makes the problem interesting because the frictional force changes with distance, affecting how far the box will slide before stopping.
Kinetic Energy
Kinetic energy is the energy of motion. It is given by the formula \( K = \frac{1}{2}mv^2 \), where \( m \) is the mass and \( v \) is the velocity of the object. In the context of this exercise, the box's initial kinetic energy is \( \frac{1}{2} m (4.50)^2 \), since it starts its journey with a velocity of 4.50 m/s.

  • As the box slides over the surface, this kinetic energy is gradually reduced to zero due to the work done by kinetic friction.
  • The work-energy theorem connects the initial kinetic energy and the work done by friction: \( \Delta K = W_{\text{net}} \).
  • Given that the box comes to rest, the change in kinetic energy, \( \Delta K \), is negative, indicating that energy is removed from the system by friction.
Understanding how kinetic energy transforms, particularly under the influence of friction, is essential for analyzing how far an object will travel until it stops.
Quadratic Equations
Quadratic equations are mathematical expressions of the form \( ax^2 + bx + c = 0 \). They are crucial in physics problems like this one, where they often arise from the application of the work-energy theorem. In the exercise, after equating the change in kinetic energy to the work done by friction, a quadratic equation needs to be solved to find the stopping distance \( x \).

  • The equation derived from the work-energy relationship results in \( \frac{x^2}{25} + 0.100x + \text{constant} = 0 \).
  • This type of equation is solved using the quadratic formula: \( x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} \).
  • Inserting the values from the problem, such as the initial speed and gravitational acceleration, allows us to find the specific values of \( x \).
Mastering quadratic equations is key, as they frequently appear in physics when dealing with forces, motion, and energy relations.

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 5.00-kg package slides 1.50 \(\mathrm{m}\) down a long ramp that is inclined at \(12.0^{\circ}\) below the horizontal. The coefficient of kinetic friction between the package and the ramp is \(\mu_{\mathrm{k}}=0.310\) . Calculate (a) the work done on the package by friction; (b) the work done on the package by gravity, (c) the work done on the package by the normal force; (d) the total work done on the package. (e) If the package has a speed of 2.20 \(\mathrm{m} / \mathrm{s}\) at the top of the ramp, what is its speed after sliding 1.50 \(\mathrm{m}\) down the ramp?

You are a member of an Alpine Rescue Team. You must project a box of supphes up an incline of constant slope angle \(\alpha\) so that it reaches a stranded skier who is a vertical distance \(h\) above the bottom of the incline. The incline is slippery, but there is some friction present, with kinetic friction coefficient \(\mu_{\mathbf{k}}\) . Use the work-energy theorem to calculate the minimum speed you must give the box at the bottom of the incline so that it will reach the skier. Express your answer in terms of \(g, h, \mu_{k},\) and \(\alpha .\)

A moving electron has kinetic energy \(K_{1}\) . After a net amount of work \(W\) has been done on it, the electron is moving one-quarter as fast in the opposite direction. (a) Find \(W\) in terms of \(K_{1}\) . (b) Does your answer depend on the final direction of the electron's motion?

A \(0.800-\mathrm{kg}\) ball is tied to the end of a string 1.60 \(\mathrm{m}\) long and swung in a vertical circle. (a) During one complete circle, starting anywere, calculate the total work done on the ball by (i) the tension in the string and (ii) gravity. (b) Repeat part (a) for motion along the semicircle from the lowest to the highest point on the path.

The Grand Coulee Dam is 1270 \(\mathrm{m}\) long and 170 \(\mathrm{m}\) high. The electrical power output from generators at its base is approximately 2000 MW. How many cubic meters of water must flow from the top of the dam per second to produce this amount of power if 92\(\%\) of the work done on the water by gravity is converted to electrical energy? (Each cubic meter of water has a mass of 1000 \(\mathrm{kg}\) .)
See all solutions

Recommended explanations on Physics Textbooks

Solid State Physics

Read Explanation

Mechanics and Materials

Read Explanation

Electricity

Read Explanation

Wave Optics

Read Explanation

Electricity and Magnetism

Read Explanation

Magnetism and Electromagnetic Induction

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.