/*! 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 \(\mathrm{A} 1000 \mathrm{kg}\) ... [FREE SOLUTION] | 91Ó°ÊÓ

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\(\mathrm{A} 1000 \mathrm{kg}\) car traveling at a speed of \(40 \mathrm{m} / \mathrm{s}\) skids to a halt on wet concrete where \(\mu_{\mathrm{k}}=0.60 .\) How long are the skid marks?

Short Answer

Expert verified
The length of the skid marks is approximately 136.05 meters.

Step by step solution

01

Identify Given Variables

The car's mass \(m = 1000 kg\), its speed \(v = 40 m/s\), and the coefficient of kinetic friction \(\mu_k = 0.60\) are given.
02

Calculate Initial Kinetic Energy

The initial kinetic energy of the car is calculated using the equation for kinetic energy, \(KE_{initial} = 1/2mv^2\). Substituting given values into the equation yields \(KE_{initial} = 1/2 * 1000kg * (40m/s)^2 = 800000J\).
03

Calculate Friction

Next, calculate the friction force. According to Newton's law of motion, the friction force is equal to the product of the car's mass, the coefficient of kinetic friction, and the acceleration due to gravity \(g = 9.8 m/s^2\). Hence, the friction force \(F_k = \mu_kmg\), from which we find that \(F_k = 0.60 * 1000kg * 9.8 m/s^2 = 5880N\).
04

Calculate Work Done by Friction

The work \(W\) done by the friction force is equal to the change in kinetic energy. Since the car comes to a halt, its final kinetic energy is zero. Therefore, \(W = KE_{final} - KE_{initial} = 0 - 800000J = -800000J\). The work done by the friction force is negative because it acts opposite to the direction of motion.
05

Calculate Distance

The work done is also defined as the product of the force and the distance over which it acts, hence \(W = F_d\). From this, we can solve for the distance \(d = W/F\), therefore the distance (which is the skid marks left by the car) can be calculated as \(d = -800000J / 5880N = 136.05m\).

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Key Concepts

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

Understanding Kinetic Energy
Kinetic energy is a type of energy an object possesses due to its motion. Mathematically, it's represented by the equation for kinetic energy, \( KE = \frac{1}{2}mv^2 \), where \( m \) is the mass of the object in kilograms and \( v \) is the velocity of the object in meters per second.

In our example, a car with a mass of \( 1000 kg \) traveling at \( 40 m/s \) has a significant amount of kinetic energy. By substituting the given values into the kinetic energy equation, the car's energy before the skid can be calculated, showing the energy it must lose to come to a stop. Understanding this energy helps us comprehend why the car leaves skid marks and why they are as long as they are – it’s all about the energy conversion from kinetic to heat and sound as the car skids to a halt.
Coefficient of Kinetic Friction Explained
The coefficient of kinetic friction, usually represented by \( \mu_k \), is a dimensionless quantity that measures the ratio of the friction force between two objects in motion relative to each other and the normal force pressing them together. A higher \( \mu_k \) value indicates more resistance to sliding.

In our exercise, the wet concrete provides a \( \mu_k = 0.60 \) when interacting with the car's tires. This value is crucial for calculating the force of friction that ultimately stops the car by counteracting its kinetic energy. It's interesting to note that different surfaces will have different coefficients affecting the distance of the skid marks – a concept that highlights the importance of understanding material properties and their interactions in physics.
Newton's Laws of Motion and Their Role
Newton's laws of motion are three fundamental principles that describe the relationship between the motion of an object and the forces acting upon it. In particular, Newton's second law is crucial for this problem, stating that \( F = ma \) where \( F \) is force, \( m \) is mass and \( a \) is acceleration.

During the skidding process, there is a frictional force that opposes the motion of the car, essentially an application of Newton's third law - for every action, there is an equal and opposite reaction. This frictional force is calculated by multiplying the mass of the car, the coefficient of kinetic friction, and the acceleration due to gravity. It's the force that does the work to reduce the kinetic energy of the car to zero, ultimately stopping the car and determining the length of the skid marks, directly relating to the principles inscribed by Sir Isaac Newton.

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Most popular questions from this chapter

Blocks with masses of \(1.0 \mathrm{kg}, 2.0 \mathrm{kg},\) and \(3.0 \mathrm{kg}\) are lined up in a row on a frictionless table. All three are pushed forward by a \(12 \mathrm{N}\) force applied to the \(1.0 \mathrm{kg}\) block. How much force does the \(2.0 \mathrm{kg}\) block exert on (a) the \(3.0 \mathrm{kg}\) block and (b) the \(1.0 \mathrm{kg}\) block?

Josh starts his sled at the top of a \(3.0-\mathrm{m}\) -high hill that has a constant slope of \(25^{\circ} .\) After reaching the bottom, he slides across a horizontal patch of snow. Ignore friction on the hill, but assume that the coefficient of kinetic friction between his sled and the horizontal patch of snow is \(0.050 .\) How far from the base of the hill does he end up?

In the winter sport of curling, two teams alternate sliding \(20 \mathrm{kg}\) stones on an icy surface in an attempt to end up with the stone closest to the center of a target painted on the ice. During one turn, a player releases a stone that travels \(27.9 \mathrm{m}\) before coming to rest. The friction force acting on the stone is \(2.0 \mathrm{N}\). What was the speed of the stone when the player released it?

A \(1000 \mathrm{kg}\) car pushes a \(2000 \mathrm{kg}\) truck that has a dead battery. When the driver steps on the accelerator, the drive wheels of the car push backward against the ground with a force of \(4500 \mathrm{N}.\) a. What is the magnitude of the force of the car on the truck? b. What is the magnitude of the force of the truck on the car?

In the sport of parasailing, a person is attached to a rope being pulled by a boat while hanging from a parachute-like sail. A rider is towed at a constant speed by a rope that is at an angle of \(15^{\circ}\) from horizontal. The tension in the rope is \(2300 \mathrm{N}\). The force of the sail on the rider is \(30^{\circ}\) from horizontal. What is the weight of the rider?

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