/*! 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 30 (a) If the coefficient of kineti... [FREE SOLUTION] | 91Ó°ÊÓ

91Ó°ÊÓ

(a) If the coefficient of kinetic friction between tires and dry pavement is \(0.80,\) what is the shortest distance in which you can stop an automobile by locking the brakes when traveling at \(29.1 \mathrm{~m} / \mathrm{s}\) (about \(65 \mathrm{mi} / \mathrm{h}) ?\) (b) On wet pavement, the coefficient of kinetic friction may be only \(0.25 .\) How fast should you drive on wet pavement in order to be able to stop in the same distance as in part (a)? (Note: Locking the brakes is not the safest way to stop.)

Short Answer

Expert verified
(a) The shortest stopping distance on dry pavement is 54.1 m; (b) On wet pavement, you should drive at 16.3 m/s to stop in the same distance.

Step by step solution

01

Identify the Known Values

We are given the coefficient of kinetic friction for dry pavement \( \mu_k = 0.80 \) and for wet pavement \( \mu_k = 0.25 \). Additionally, the initial speed of the car is \( v = 29.1 \, \mathrm{m/s} \). We need to find the shortest stopping distance for part (a) and the speed for part (b) that results in the same stopping distance.
02

Calculate Stopping Distance on Dry Pavement

When the brakes are locked, the only force stopping the car is the frictional force, given by \( F_{friction} = \mu_k \cdot m \cdot g \). Using the work-energy principle, the work done by friction is equal to the change in kinetic energy: \( \mu_k \cdot m \cdot g \cdot d = \frac{1}{2} m v^2 \). Simplifying, we find \( d = \frac{v^2}{2 \mu_k g} \), where \( g \approx 9.81 \, \text{m/s}^2 \). Substituting the values for dry pavement: \( d = \frac{(29.1)^2}{2 \times 0.80 \times 9.81} \approx 54.1 \, \text{m} \).
03

Determine Speed on Wet Pavement for Same Stopping Distance

To stop in the same distance on wet pavement, the stopping distance formula \( d = \frac{v^2}{2 \mu_k g} \) should remain equal to 54.1 m. Re-arrange the formula to solve for initial speed on wet pavement: \( v = \sqrt{2 \mu_k g d} \). Substitute \( \mu_k = 0.25 \) for wet pavement and solve: \( v = \sqrt{2 \times 0.25 \times 9.81 \times 54.1} \approx 16.3 \, \text{m/s} \).

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 91Ó°ÊÓ!

Key Concepts

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

Stopping Distance
Stopping distance is an essential concept to understand when dealing with vehicle safety and dynamics. It refers to the total distance a car travels before it comes to a complete stop after the brakes are applied. Several factors influence stopping distance, including the friction between the tires and the road, vehicle speed, and braking technique. Additionally, environmental conditions such as road surface, weather, and vehicle load play significant roles.

When discussing stopping distance, we consider two main components:
  • Reaction Distance: The distance your car travels during your reaction time (the time it takes you to recognize a hazard and begin braking).
  • Braking Distance: The distance from the point at which you start to apply the brakes until the vehicle stops.
In our exercise, the focus is on braking distance, particularly how it is affected by different coefficients of friction under dry and wet conditions. Understanding these components not only improves safety awareness but also aids in making informed decisions while driving.
Coefficient of Friction
The coefficient of friction is a crucial factor in determining how quickly a vehicle can stop. It's a measure of the "grip" between two surfaces, in this case, the car tires and the road. This measure is a dimensionless number typically ranging between 0 and 1.

There are two types of friction:
  • Static Friction: Which is the frictional force preventing motion when the surfaces are at rest relative to each other.
  • Kinetic Friction: Which occurs when the surfaces are sliding past each other. This is what we focus on when considering stopping a moving vehicle.
In our example, the change in the coefficient of kinetic friction from 0.80 on dry pavement to 0.25 on wet pavement indicates lesser grip in wet conditions. It's important to remember that a lower coefficient means less friction, leading to longer stopping distances. Therefore, understanding this concept can guide safer driving practices, especially under adverse weather conditions.
Work-Energy Principle
The work-energy principle is a fundamental concept in physics that connects the work done by forces applied to an object and the change in its kinetic energy. In simpler terms, when work is done on an object, it either gains or loses energy.

For a decelerating vehicle, like in our exercise, the work done by the frictional force is what causes the vehicle to stop. The kinetic energy formula is expressed as:
  • \[ \text{Kinetic Energy} = \frac{1}{2} m v^2 \]
In this context, the work done by friction to stop the car can be equated to the change in kinetic energy:
  • \[ \mu_k \cdot m \cdot g \cdot d = \frac{1}{2} m v^2 \]
Solving for the stopping distance \(d\), we derive:
  • \[ d = \frac{v^2}{2 \mu_k g} \]
This crucial equation relates stopping distance to initial speed, friction coefficient, and gravitational force—demonstrating how energy principles apply in real-world physics problems.
Physics Problem Solving
Physics problem solving involves breaking down complex scenarios into manageable parts, using scientific principles to find solutions. The exercise provided is a classic case, and here is a methodical approach to tackling such problems:

Step-by-Step Layout:
  • Identify Known Values: Start by listing all given quantities and constants, such as coefficients of friction and initial speeds, which provide a foundation for problem-solving.
  • Use Appropriately Linked Equations: Equations like those derived from the work-energy principle help connect different knowns and unknowns, facilitating comprehensive solutions.
  • Re-arrange Formulas: Be comfortable manipulating equations to isolate the variable of interest, whether it's a stopping distance or initial speed.
  • Substitute Values Cautiously: After structuring the relevant formula, substitute the known values to find the unknowns accurately.
Solving physics problems is akin to piecing a puzzle together, where understanding each small piece is crucial to find the whole picture. With practice, insights gained from exercises like this improve not only physics knowledge but also logical reasoning and analytical skills.

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 pickup truck is carrying a toolbox, but the rear gate of the truck is missing, so the box will slide out if it is set moving. The coefficients of kinetic and static friction between the box and the bed of the truck are 0.355 and \(0.650,\) respectively. Starting from rest, what is the shortest time in which this truck could accelerate uniformly to \(30.0 \mathrm{~m} / \mathrm{s}(\approx 60 \mathrm{mi} / \mathrm{h}) \quad\) without causing the box to slide? Include a free-body diagram of the toolbox as part of your solution. (Hint: First use Newton's second law to find the maximum acceleration that static friction can give the box, and then solve for the time required to reach \(30.0 \mathrm{~m} / \mathrm{s}\).)

People who do chin-ups raise their chin just over a bar (the chinning bar), supporting themselves only by their arms. Typically, the body below the arms is raised by about \(30 \mathrm{~cm}\) in a time of \(1.0 \mathrm{~s},\) starting from rest. Assume that the entire body of a \(680 \mathrm{~N}\) person who is chinning is raised this distance and that half the \(1.0 \mathrm{~s}\) is spent accelerating upward and the other half accelerating downward, uniformly in both cases. Make a free-body diagram of the person's body, and then apply it to find the force his arms must exert on him during the accelerating part of the chin-up.

A person pushes on a stationary \(125 \mathrm{~N}\) box with \(75 \mathrm{~N}\) at \(30^{\circ}\) below the horizontal, as shown in Figure 5.61 . The coefficient of static friction between the box and the horizontal floor is 0.80 . (a) Make a free-body diagram of the box. (b) What is the normal force on the box? (c) What is the friction force on the box? (d) What is the largest the friction force could be? (e) The person now replaces his push with a \(75 \mathrm{~N}\) pull at \(30^{\circ}\) above the horizontal. Find the normal force on the box in this case.

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?

A stockroom worker pushes a box with mass \(11.2 \mathrm{~kg}\) on a horizontal surface with a constant speed of \(3.50 \mathrm{~m} / \mathrm{s}\). The coefficients of kinetic and static friction between the box and the surface are 0.200 and \(0.450,\) respectively. (a) What horizontal force must the worker apply to maintain the motion of the box? (b) If the worker stops pushing, what will be the acceleration of the box?

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.