/*! 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 29 Stopping Distance. Acar is trave... [FREE SOLUTION] | 91Ó°ÊÓ

91Ó°ÊÓ

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?

Short Answer

Expert verified
a) Minimum stopping distance: \( d = \frac{v_0^2}{2\mu_k g} \). b) (i) Halved, (ii) Quadrupled, (iii) Doubled.

Step by step solution

01

Identify Forces and Work-Energy Principle

The car is initially moving at speed \( v_0 \), and we want to find the stopping distance \( d \). According to the work-energy principle, the work done by friction is equal to the car's initial kinetic energy.
02

State Initial Kinetic Energy

The initial kinetic energy of the car is given by: \( KE_i = \frac{1}{2}mv_0^2 \), where \( m \) is the mass of the car.
03

Work Done by Friction

The work done by the friction force is \( W = -f_k \cdot d \), where \( f_k = \mu_k mg \) is the frictional force, and \( d \) is the stopping distance.
04

Equate Work to Kinetic Energy

Using the work-energy theorem, \( -f_k \cdot d = -\frac{1}{2}mv_0^2 \). Simplifying gives \( \mu_k mgd = \frac{1}{2}mv_0^2 \).
05

Solve for Stopping Distance

Cancel \( m \) from both sides of the equation: \( \mu_k gd = \frac{v_0^2}{2} \). Now solve for \( d \): \( d = \frac{v_0^2}{2\mu_k g} \).
06

Analyze Effect of Doubling \( \mu_k \)

If \( \mu_k \) is doubled, the equation becomes \( d_{new} = \frac{v_0^2}{4\mu_k g} \). Thus, the stopping distance is halved.
07

Analyze Effect of Doubling \( v_0 \)

If \( v_0 \) is doubled, the equation becomes \( d_{new} = \frac{4v_0^2}{2\mu_k g} \). Thus, the stopping distance is quadrupled.
08

Analyze Effect of Doubling Both \( \mu_k \) and \( v_0 \)

If both \( \mu_k \) and \( v_0 \) are doubled, substitute into the equation: \( d_{new} = \frac{4v_0^2}{4\mu_k g} \), making the stopping distance doubled.

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.

Work-Energy Theorem
The Work-Energy Theorem is a fundamental concept in physics that links the net work done on an object to the change in its kinetic energy. Specifically, it states that the work done by all the forces acting on an object is equal to the change in its kinetic energy. This principle helps us understand how forces are transformed into motion and vice versa.
In the context of stopping distance, we can see this in how the frictional force does work on the car, reducing its kinetic energy until it stops. Initially, the car possesses kinetic energy due to its speed, and as it skids to a stop, the friction between the tires and the road performs negative work. This work reduces the car's kinetic energy from its initial value down to zero at the minimum stopping distance.
Coefficient of Kinetic Friction
The coefficient of kinetic friction (\( \mu_k \)) is a dimensionless value that compares the frictional force resisting motion to the normal force pressing two surfaces together. It varies depending on the materials in contact.
In our exercise, the coefficient of kinetic friction plays a crucial role in determining the stopping distance of the car. A higher coefficient means more frictional force is acting to decelerate the vehicle.
  • Doubling the coefficient (\( \mu_k \)) results in halving the stopping distance. This is because the increased friction works more efficiently to bring the car to a stop.
  • In contrast, a lower coefficient would increase the stopping distance, as the frictional force would be less effective. \( \mu_k \) depends on surfaces: for example, rubber on asphalt has a higher \( \mu_k \) than ice on metal.
Kinetic Energy
Kinetic energy is the energy an object possesses due to its motion. It is given by the formula \( KE = \frac{1}{2}mv^2 \), where \( m \) is the mass of the object and \( v \) is its velocity.
In the problem of a car coming to a stop, initial kinetic energy is crucial because it determines how much work must be done by friction to stop the vehicle.
  • If the initial speed doubles, kinetic energy increases by a factor of four. This is critical because stopping distance will depend significantly on the velocity squared. A higher speed results in a much larger stopping distance.
  • Thus, small increases in speed can greatly increase the stopping distance, highlighting the importance of speed limits for safety.
Frictional Force
Frictional force is the resistive force that acts opposite to the relative motion or the tendency of such motion of two contacting surfaces. The force of friction \( f_k \) is calculated by the equation \( f_k = \mu_k mg \), where \( \mu_k \) is the coefficient of kinetic friction and \( mg \) is the normal force.
For a car skidding to a stop, the frictional force is the key player that opposes the car's motion and converts kinetic energy into heat, slowing the car down.
  • In this exercise, it's crucial to understand that friction is what allows the work-energy theorem to bring the car to a halt. Without enough frictional force, as determined by the coefficient of kinetic friction, stopping the car quickly would be much harder, or impossible in some surfaces like icy roads.
  • The frictional force ensures the work done against motion matches the reduction in kinetic energy needed to stop the car.

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

Automotive Power I. A truck engine transmits 28.0 \(\mathrm{kW}(37.5 \mathrm{hp})\) to the driving wheels when the truck is traveling at a constant velocity of magnitude 60.0 \(\mathrm{km} / \mathrm{h}\) (37.3 \(\mathrm{mi} / \mathrm{h} )\) on a level road. (a) What is the resisting force acting on the truck? (b) Assume that 65\(\%\) of the resisting force is due to rolling friction and the remainder is due to air resistance. If the force of rolling friction is independent of speed, and the force of air resistance is proportional to the square of the speed, what power will drive the truck at 30.0 \(\mathrm{km} / \mathrm{h} ?\) At 120.0 \(\mathrm{km} / \mathrm{h} ?\) Give your answers in kilowatts and in horsepower.

A block of ice with mass 2.00 \(\mathrm{kg}\) slides 0.750 \(\mathrm{m}\) 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? You can ignore friction.

CALC A net force along the \(x\) -axis that has \(x\) -component \(F_{x}=-12.0 \mathrm{N}+\left(0.300 \mathrm{N} / \mathrm{m}^{2}\right) x^{2}\) is applied to a 5.00 \(\mathrm{-kg}\) object that is initially at the origin and moving in the \(-x\) -direction with a speed of 6.00 \(\mathrm{m} / \mathrm{s} .\) What is the speed of the object when it reaches the point \(x=5.00 \mathrm{m} ?\)

The spring of a spring gun has force constant \(k=400 \mathrm{N} / \mathrm{m}\) and negligible mass. The spring is compressed \(6.00 \mathrm{cm},\) and a ball with mass 0.0300 \(\mathrm{kg}\) is placed in the horizontal barrel against the compressed spring. The spring is then released, and the ball is propelled out the barrel of the gun. The barrel is 6.00 \(\mathrm{cm}\) long, so the ball leaves the barrel at the same point that it loses contact with the spring. The gun is held so the barrel is horizontal. (a) Calculate the speed with which the ball leaves the barrel if you can ignore friction. (b) Calculate the speed of the ball as it leaves the barrel if a constant resisting force of 6.00 \(\mathrm{N}\) acts on the ball as it moves along the barrel. (c) For the situation in part (b), at what position along the barrel does the ball have the greatest speed, and what is that speed? (In this case, the maximum speed does not occur at the end of the barrel.)

When its \(75-\mathrm{kW}(100 \mathrm{-hp})\) engine is generating full power, a small single-engine airplane with mass 700 \(\mathrm{kg}\) gains altitude at a rate of 2.5 \(\mathrm{m} / \mathrm{s}(150 \mathrm{m} / \mathrm{min}\) , or 500 \(\mathrm{ft} / \mathrm{min}\) ). What fraction of the engine power is being used to make the airplane climb? (The remainder is used to overcome the effects of air resistance and of inefficiencies in the propeller and engine.)

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.