/*! 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 48 Consult Multiple-Concept Example... [FREE SOLUTION] | 91Ó°ÊÓ

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Consult Multiple-Concept Example 10 in preparation for this problem. Traveling at a specd of \(16.1 \mathrm{m} / \mathrm{s},\) the driver of an automobile suddenly locks the wheels by slamming on the brakes. The coefficient of kinetic friction between the tires and the road is \(0.720 .\) What is the speed of the automobile after 1.30 \(\mathrm{s}\) have clapsed? Ignore the effects of air resistance.

Short Answer

Expert verified
The final speed of the car after 1.30 s is approximately 7.91 m/s.

Step by step solution

01

Understand the Given Information

We have an initial speed of the car as \( v_i = 16.1 \text{ m/s} \) and the coefficient of kinetic friction \( \mu_k = 0.720 \). The time elapsed is \( t = 1.30 \text{ s} \). We need to find the final speed \( v_f \) of the car.
02

Apply the Formula for Frictional Force

The frictional force \( f_k \) can be determined using \( f_k = \mu_k \cdot m \cdot g \), where \( m \) is the mass of the car and \( g \approx 9.81 \text{ m/s}^2 \). This simplifies to the acceleration due to friction, \( a = \mu_k \cdot g \), since \( f_k = m \cdot a \).
03

Calculate the Acceleration

Using the formula for frictional force, the acceleration \( a \) is \( a = \mu_k \cdot g = 0.720 \cdot 9.81 \). Solving gives \( a = 7.0632 \text{ m/s}^2 \) acting against the motion (negative direction).
04

Use the Kinematic Equation

Use the equation \( v_f = v_i + a \cdot t \) to find the final speed. Substitute \( v_i = 16.1 \), \( a = -7.0632 \), and \( t = 1.30 \) seconds.
05

Calculate the Final Speed

Substitute the values into the equation: \( v_f = 16.1 - 7.0632 \cdot 1.30 \). Calculate to find \( v_f \approx 7.91376 \text{ m/s} \).

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Key Concepts

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

Kinematic Equations
Kinematic equations are incredibly useful for solving problems related to motion, especially when acceleration is involved. These equations help us find unknown variables such as final velocity, time, displacement, or acceleration when we have sufficient initial information. There are four primary kinematic equations, but one of the most commonly used is:\[v_f = v_i + a \cdot t\]- **Variables Explanation:** - \(v_f\): final velocity - \(v_i\): initial velocity - \(a\): acceleration - \(t\): time elapsedBy using these equations, we can determine how an object moves under constant acceleration. In the given exercise, we applied this formula to find the car’s speed after the brakes were applied, showcasing its practical utility in solving real-world problems.
Coefficient of Friction
The coefficient of friction is a critical factor in determining the force required to move one object over another. It is a dimensionless quantity that represents the frictional force between two objects in contact. This coefficient is represented by the symbol \( \mu \) and varies based on the two surfaces interacting.- **Types of Friction:** - **Static Friction:** Prevents motion; usually higher than kinetic. - **Kinetic Friction:** Opposes motion of a sliding object; used in out exercise.In our case, the coefficient of kinetic friction \( \mu_k \) is used, which is 0.720. This value is critical as it helps us calculate the force of friction which, in turn, influences the car's deceleration. It's fascinating to see how various surfaces, like rubber on asphalt, have different friction coefficients leading to varied braking efficiencies.
Acceleration Due to Friction
Acceleration due to friction occurs when an object slows down as it slides across a surface. This form of acceleration is different from usual acceleration, as it works against the motion of the object. To calculate this, we multiply the coefficient of kinetic friction \( \mu_k \) by the acceleration due to gravity \( g \).- **Formula to Calculate:** \[ a = \mu_k \cdot g \]- **Understanding the Terms:** - \( \mu_k \): coefficient of kinetic friction - \( g \): acceleration due to gravity \(\approx 9.81 \text{ m/s}^2\)In the exercise, this calculation gave us an acceleration of \( 7.0632 \text{ m/s}^2 \). This negative value indicates that it acts in the opposite direction of the car's initial movement, effectively slowing it down. Understanding this concept helps us appreciate how different factors play roles in deceleration and why the material of the tires and the road can be so crucial for safety.

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Most popular questions from this chapter

The central ideas in this problem are reviewed in Multiple-Concept Example 9. One block rests upon a horizontal surface. A second identical block rests upon the first one. The coefficient of static friction between the blocks is the same as the coefficient of static friction between the lower block and the horizontal surface. A horizontal force is applied to the upper block, and the magnitude of the force is slowly increased. When the force reaches 47.0 \(\mathrm{N}\), the upper block just begins to slide. The force is then removed from the upper block, and the blocks are returned to their original configuration. What is the magnitude of the horizontal force that should be applied to the lower block so that it just begins to slide out from under the upper block?

Multiple-Concept Example 17 reviews the basic concepts involved in this problem. Air rushing over the wings of high-performance race cars generates unwanted horizontal air resistance but also causes vertical downforce, which helps the cars hug the track more securely. The coefficient of static friction between the track and the tires of a \(690-\mathrm{kg}\) race car is \(0.87 .\) What is the magnitude of the maximum acceleration at which the car can speed up without its tires slipping when a \(4060-\mathrm{N}\) downforce and an \(1190-\mathrm{N}\) horizontal-air-resistance force act on it?

ssm mmh A 1580-kg car is traveling with a speed of 15.0 m/s. What is the magnitude of the horizontal net force that is required to bring the car to a halt in a distance of 50.0 m?

ssm Three forces act on a moving object. One force has a magnitude of 80.0 N and is directed due north. Another has a magnitude of 60.0 N and is directed due west. What must be the magnitude and direction of the third force, such that the object continues to move with a constant velocity?

Two particles are located on the \(x\) axis. Particle 1 has a mass \(m\) and is at the origin. Particle 2 has a mass 2\(m\) and is at \(x=+L\) . A third particle is placed between particles 1 and 2. Where on the x axis should the third particle be located so that the magnitude of the gravitational force on both particle 1 and particle 2 doubles? Express your answer in terms of \(L\) .

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