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Stopping Distance. A car 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 \(\mu_{x}\) 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) \(d = \frac{v_0^2}{2 \mu_x g}\); (b) (i) Halved, (ii) Quadrupled, (iii) No change.

Step by step solution

01

Understanding the Problem

A car is moving on a level road with initial velocity \(v_0\). We need to calculate the minimum stopping distance using the work-energy theorem, with kinetic friction considered. We will express this distance in terms of \(v_0\), \(g\), and the friction coefficient \(\mu_x\).
02

Apply the Work-Energy Theorem

The work done by friction is equal to the change in kinetic energy of the car. Initially, kinetic energy (KE) is \(\frac{1}{2}mv_0^2\), and finally, it is 0 since the car stops. The work done by friction (W) is \(-f_k \cdot d\), where \(f_k\) is the friction force, and \(d\) is the stopping distance.
03

Express Friction Force

The friction force \(f_k\) is equal to \(\mu_x m g\), where \(\mu_x\) is the coefficient of kinetic friction and \(m\) is the mass of the car. Substitute this into the expression for work: \(-\mu_x m g \cdot d = \Delta KE = -\frac{1}{2}mv_0^2\).
04

Solve for Stopping Distance

Rearrange the previous equation to find \(d\):\[ \mu_x m g \cdot d = \frac{1}{2}mv_0^2 \]\[ d = \frac{v_0^2}{2 \mu_x g} \]This is the minimum stopping distance when brakes lock the tires causing them to slide.
05

Evaluate Changes in Stopping Distance

(i) If \(\mu_x\) is doubled, then \(d = \frac{v_0^2}{4 \mu_x g}\), so the stopping distance is halved.(ii) If \(v_0\) is doubled, then \(d = \frac{(2v_0)^2}{2 \mu_x g} = \frac{4v_0^2}{2 \mu_x g}\), so the stopping distance is quadrupled.(iii) If both \(\mu_x\) and \(v_0\) are doubled, then \(d = \frac{(2v_0)^2}{4 \mu_x g} = \frac{v_0^2}{2 \mu_x g}\), the stopping distance remains the same as the original.

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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 relative motion of two surfaces in contact. When a car brakes and its tires slide over the road, kinetic friction acts against the direction of motion. This force arises due to the interaction between the microscopic surface irregularities of the tires and the road.

The strength of kinetic friction is often described using the coefficient of kinetic friction, denoted as \(\mu_x\). This coefficient is a dimensionless number that represents the frictional force relative to the normal force, which is the force perpendicular to the contact surface. The frictional force can be calculated using the equation:
  • \( f_k = \mu_x \cdot F_N \)
  • Where \( F_N = m \cdot g \) is the normal force, with \( m \) being the mass of the object and \( g \) the acceleration due to gravity.
In the context of stopping a vehicle, the kinetic friction force is crucial in determining how quickly a car can come to a halt once the brakes are applied. The greater the coefficient of kinetic friction, the more effective is the frictional force in stopping the vehicle.
Stopping Distance
Stopping distance refers to the shortest distance a vehicle travels before coming to a complete stop after the brakes are applied. It's an important safety metric, especially for understanding the dynamics of vehicles when braking under different conditions.

Calculated through the work-energy theorem, stopping distance takes into account the initial kinetic energy of the moving vehicle, which is dissipated through work done by friction:
  • The initial kinetic energy is given by: \( KE_i = \frac{1}{2}mv_0^2 \)
  • The work-energy theorem tells us that the work done by kinetic friction (\( W = f_k \cdot d \)) equals the change in kinetic energy, resulting in \( W = -\frac{1}{2}mv_0^2 \).
For a car moving on a level surface with locked brakes, the stopping distance \( d \) is calculated as:\[ d = \frac{v_0^2}{2 \mu_x g} \]
This formula shows that stopping distance:
  • Increases with the square of the initial speed (\( v_0^2 \)).
  • Decreases with a higher coefficient of kinetic friction (\( \mu_x \)).
  • Is inversely proportional to gravitational acceleration \( g \).
Physics Problem Solving
Physics problem solving involves a logical and methodical approach to understanding and resolving physics questions or scenarios. It requires a clear comprehension of the principles at play, such as the work-energy theorem in the context of stopping distance.

Here’s how you can tackle a problem like calculating the stopping distance:
  • Understand the problem: Identify all given variables, like initial velocity \( v_0 \), gravitational acceleration \( g \), and coefficient of kinetic friction \( \mu_x \).
  • Apply relevant principles: Use the work-energy theorem here, which states that the work done on an object is equal to its change in kinetic energy.
  • Solve the equations: Manipulate the equations to solve for the unknown, which in this case is stopping distance \( d \).
  • Evaluate different scenarios: Consider how changes in factors like speed or friction coefficient affect the result.
This systematic approach ensures that you break down complex problems into smaller, manageable parts, leading to a solution based on fundamental physics principles.

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Most popular questions from this chapter

A \(4.80-\mathrm{kg}\) watermelon is dropped from rest from the roof of a 25.0 \(\mathrm{m}\) -tall building and feels no appreciable air resistance. (a) Calculate the work done by gravity on the watermelon during its displacement from the roof to the ground. (b) Just before it strikes the ground, what is the watermelon's (i) kinetic energy and (ii) speed? (c) Which of the answers in parts (a) and (b) would be different if there were appreciable air resistance?

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