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An automobile with passengers has weight \(16400 \mathrm{~N}\) and is moving at \(113 \mathrm{~km} / \mathrm{h}\) when the driver brakes, sliding to a stop. The frictional force on the wheels from the road has a magnitude of \(8230 \mathrm{~N}\). Find the stopping distance.

Short Answer

Expert verified
The stopping distance is approximately 100 meters.

Step by step solution

01

Convert Speed to Meters per Second

The initial speed is given in kilometers per hour, need to convert it to meters per second. Use the conversion: \(1 \text{ km/h} = \frac{1}{3.6} \text{ m/s}\). For an initial speed of \(113\text{ km/h}\), we calculate: \(113 \text{ km/h} \times \frac{1}{3.6} = 31.39 \text{ m/s}\).
02

Calculate the Deceleration

Use Newton's second law to find the deceleration caused by the frictional force. The force \( F = ma \) can be rewritten as \( a = \frac{F}{m} \). First, find the mass \( m \) from the weight \( W = 16400 \text{ N} \). Using \( W = mg \), where \( g = 9.8 \text{ m/s}^2 \), we get \( m = \frac{16400}{9.8} = 1673.47 \text{ kg}\). Then,\( a = \frac{8230}{1673.47} = 4.92 \text{ m/s}^2\). Since friction is acting against the motion, the deceleration is \(-4.92 \text{ m/s}^2\).
03

Find the Stopping Distance

We will use the equation of motion \( v^2 = u^2 + 2as \), where \( v = 0 \text{ m/s} \) (final speed), \( u = 31.39 \text{ m/s} \) (initial speed), \( a = -4.92 \text{ m/s}^2\), and \( s \) is the stopping distance. Substitute in the values: \( 0 = (31.39)^2 + 2(-4.92)s \). This simplifies to \( (31.39)^2 = 2 \times 4.92 \times s \), giving \( s = \frac{(31.39)^2}{2 \times 4.92} \approx 100.0 \text{ meters}\).

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Key Concepts

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

Newton's Second Law of Motion
Newton's Second Law plays a crucial role in understanding the mechanism of stopping an automobile in motion. It states that the force acting on an object is equal to the mass of the object multiplied by its acceleration: \( F = ma \). This means, when a force is applied, like the frictional force between the road and the tires of a car, it leads to an acceleration or deceleration depending on the direction of the force. In our example, the frictional force causes the car to decelerate, not accelerate, as it moves in the opposite direction to its motion.

To calculate the deceleration, we first need the mass of the automobile, which can be determined from its weight using the relation \( W = mg \) where \( g \) is the gravitational acceleration \( 9.8 \ \text{m/s}^2 \). This allows us to establish the deceleration as \( a = \frac{F}{m} \). The negative sign indicates a deceleration, or a slowing down of the car, caused by the frictional resistance.
Understanding Kinematics
Kinematics is the branch of physics dealing with the motion of objects, specifically defining the concepts of velocity, acceleration, and distance without considering the masses or forces that cause the motion. In the context of our problem, it allows us to relate the initial velocity of the automobile, its deceleration, and the stopping distance.

To resolve the initial speed into a usable form, it's essential to convert it from kilometers per hour to meters per second. With this conversion done, the kinematic equation \( v^2 = u^2 + 2as \) helps find the stopping distance \( s \), where \( u \) is the initial velocity, \( v \) the final velocity (0 in the case of stopping), and \( a \) the acceleration which is negative here because it's a deceleration.

Using these relationships, kinematics simplifies predicting how far the car will travel before it halts completely.
Role of Frictional Force
Frictional force is vital in bringing the automobile to a stop. In this situation, the force acts opposite to the direction of motion, resulting in its deceleration. The frictional force on the car wheels is given as 8230 N, which provides a measure of how strong the resistance is against the forward motion of the car.

Without frictional force, a moving car would continue in motion indefinitely as per Newton’s First Law of Motion, often referred to as the law of inertia. The magnitude of the frictional force influences the rate of deceleration, impacting how quickly and over what distance the car can stop.

This force is calculated as a part of the second law (\( F = ma \)), considering both the surface type and vehicle weight. Hence, understanding this concept not only informs about stopping distances but also highlights the significance of adequate tire-road friction, especially for safety in real-world situations.
Calculating Deceleration
Deceleration is a measure of how quickly an automobile can come to a stop. Calculating deceleration involves using the frictional force to determine how quickly the speed decreases. The formula \( a = \frac{F}{m} \) derived from Newton's second law allows us to calculate this directly as \( 4.92 \ \text{m/s}^2 \), with the automobile's mass and the frictional force both considered.

Deceleration needs to be understood as negative acceleration; it acts to reduce the speed, rather than increase it. The calculation provides critical insights into vehicle braking effectiveness and road safety parameters.

In our problem, the negative value, \(-4.92 \ \text{m/s}^2 \), showcases the fact that the speed of the vehicle is reducing due to the opposing frictional force, helping us predict its stopping distance when brought to rest under certain conditions.

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Most popular questions from this chapter

A \(0.63 \mathrm{~kg}\) ball thrown directly upward with an initial speed of \(14 \mathrm{~m} / \mathrm{s}\) reaches a maximum height of \(8.1 \mathrm{~m} .\) What is the change in the mechanical energy of the ball-Earth system during the ascent of the ball to that maximum height?

Two blocks, of masses \(M=2.0 \mathrm{~kg}\) and \(2 M,\) are connected to a spring of spring constant \(k=200 \mathrm{~N} / \mathrm{m}\) that has one end fixed, as shown in Fig. \(8-69 .\) The horizontal surface and the pulley are frictionless, and the pulley has negligible mass. The blocks are released from rest with the spring relaxed. (a) What is the combined kinetic energy of the two blocks when the hanging block has fallen \(0.090 \mathrm{~m} ?\) (b) What is the kinetic energy of the hanging block when it has fallen that \(0.090 \mathrm{~m} ?\) (c) What maximum distance does the hanging block fall before momentarily stopping?

A uniform cord of length \(25 \mathrm{~cm}\) and mass \(15 \mathrm{~g}\) is initially stuck to a ceiling. Later, it hangs vertically from the ceiling with only one end still stuck. What is the change in the gravitational potential energy of the cord with this change in orientation? (Hint: Consider a differential slice of the cord and then use integral calculus.)

A \(0.42 \mathrm{~kg}\) shuffleboard disk is initially at rest when a player uses a cue to increase its speed to \(4.2 \mathrm{~m} / \mathrm{s}\) at constant acceleration. The acceleration takes place over a \(2.0 \mathrm{~m}\) distance, at the end of which the cue loses contact with the disk. Then the disk slides an additional \(12 \mathrm{~m}\) before stopping. Assume that the shuffleboard court is level and that the force of friction on the disk is constant. What is the increase in the thermal energy of the disk-court system (a) for that additional \(12 \mathrm{~m}\) and (b) for the entire \(14 \mathrm{~m}\) distance? (c) How much work is done on the disk by the cue?

A block of mass \(m=2.0 \mathrm{~kg}\) is dropped from height \(h=40 \mathrm{~cm}\) onto a spring of spring constant \(k=1960 \mathrm{~N} / \mathrm{m}\) (Fig. 8 -39). Find the maximum distance the spring is compressed.

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