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Chapter 6: Problem 21

A \(1500 \mathrm{kg}\) car skids to a halt on a wet road where \(\mu_{\mathrm{x}}=0.50\) How fast was the car traveling if it leaves \(65-\) m-long skid marks?

Short Answer

Expert verified
The car was initially traveling at roughly \(32 \mathrm{m/s}\).

Step by step solution

01

Calculate the force of friction

Calculate the force of friction (\(F_{\mathrm{f}}\)) using the equation \(F_{\mathrm{f}}=\mu_{\mathrm{x}} \cdot F_{\mathrm{g}}\), where \(F_{\mathrm{g}}\) is the gravitational force. As the car is on a horizontal road, the gravitational force is \(F_{\mathrm{g}}=m \cdot g\), where \(m = 1500 \mathrm{kg}\) is the mass of the car and \(g = 9.8 \mathrm{m/s^2}\) is the acceleration due to gravity. Therefore, \(F_{\mathrm{g}} = 1500 \mathrm{kg} \cdot 9.8 \mathrm{m/s^2} = 14700 \mathrm{N}\). Substituting \(\mu_{\mathrm{x}}=0.50\) and \(F_{\mathrm{g}} = 14700 \mathrm{N}\) into the equation for \(F_{\mathrm{f}}\), we find \(F_{\mathrm{f}} = 0.50 \cdot 14700 \mathrm{N} = 7350 \mathrm{N}\).
02

Calculate the work done by friction

Calculate the work (\(W\)) done by the force of friction stopping the car using the equation \(W = F_{\mathrm{f}} \cdot d\), where \(d = 65 \mathrm{m}\) is the distance of the skid marks. Therefore, \(W = 7350 \mathrm{N} \cdot 65 \mathrm{m} = 477750 \mathrm{J}\) However, as the friction force is a stopping force, the work done by it is negative, so \(W = -477750 \mathrm{J}\).
03

Determine the initial speed of the car

The work done by the force of friction is equal to the change in the car's kinetic energy. As the car comes to a halt, its final kinetic energy is 0. Hence, the initial kinetic energy of the car equals the work done by friction. Using the formula for kinetic energy \(K = 0.5 \cdot m \cdot v^2\), where \(v\) is the speed of the car, we set \(-W = K\). Therefore, \(v = \sqrt{-2 \cdot W / m}\). Substituting \(W = -477750 \mathrm{J}\) and \(m = 1500 \mathrm{kg}\), we find \(v \approx 32 \mathrm{m/s}\).

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Key Concepts

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

Kinetic Friction Coefficient
The kinetic friction coefficient, usually denoted as \( \mu_k \), is a dimensionless number that represents the amount of friction between two objects in motion relative to each other. Its value is determined experimentally, and it varies depending on the materials involved and the condition of the surfaces. In our example, the coefficient of \( \mu_x = 0.50 \) is relatively high, reflecting the significant resistance a car tire encounters when skidding on a wet road. This coefficient plays a crucial role in calculating the force of friction, which is crucial to determining how the car will come to a stop.

Work-Energy Principle
The work-energy principle is a fundamental concept in physics asserting that the work done on an object is equal to the change in its kinetic energy. When you apply a force to an object over a distance, you are doing work on it. In the case of our skidding car, the work done by the friction force is negative because it acts in the opposite direction to the car's motion. So, the negative work done by friction decreases the car's kinetic energy to zero, effectively stopping the car.

Kinetic Energy Calculation
Kinetic energy, represented by the equation \( K = 0.5 \cdot m \cdot v^2 \), is the energy an object possesses due to its motion, where \(m\) is the object's mass and \(v\) is its velocity. In our problem, we were tasked with determining the car's speed before the skid started. By utilizing the work-energy principle, we took the work done by friction (which equaled the loss of kinetic energy) and calculated the car's initial speed. Remember, this calculation assumes that all the work done by friction converted the car's kinetic energy to heat and sound, bringing it to a stop, and that there are no other forces contributing to the car's deceleration.

Gravitational Force
Gravitational force is the attraction between two objects with mass. On the surface of the earth, this force causes objects to have weight. It can be calculated using the formula \( F_g = m \cdot g \), where \(m\) is the mass and \(g\) is the acceleration due to gravity (\(9.8 \:\mathrm{m/s^2}\) near Earth's surface). In our physics problem, the gravitational force acts on the mass of the car and is a critical component in determining the force of friction since the frictional force is directly proportional to the object's weight (mass times the acceleration due to gravity).

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Most popular questions from this chapter

It's a snowy day and you're pulling a friend along a level road on a sled. You've both been taking physics, so she asks what you think the coefficient of friction between the sled and the snow is. You've been walking at a steady \(1.5 \mathrm{m} / \mathrm{s}\), and the rope pulls up on the sled at a \(30^{\circ}\) angle. You estimate that the mass of the sled, with your friend on it, is \(60 \mathrm{kg}\) and that you're pulling with a force of \(75 \mathrm{N}\). What answer will you give?

A horizontal rope is tied to a \(50 \mathrm{kg}\) box on frictionless ice. What is the tension in the rope if: a. The box is at rest? b. The box moves at a steady \(5.0 \mathrm{m} / \mathrm{s} ?\) c. The box has \(v_{x}=5.0 \mathrm{m} / \mathrm{s}\) and \(a_{x}=5.0 \mathrm{m} / \mathrm{s}^{2} ?\)

Sam, whose mass is \(75 \mathrm{kg}\), takes off down a 50 -m-high, \(10^{\circ}\) slope on his jet-powered skis. The skis have a thrust of \(200 \mathrm{N}\). Sam's speed at the bottom is \(40 \mathrm{m} / \mathrm{s}\). What is the coefficient of kinetic friction of his skis on snow?

You've entered a "slow ski race" where the winner is the skier who takes the longest time to go down a \(15^{\circ}\) slope without ever stopping. You need to choose the best wax to apply to your skis. Red wax has a coefficient of kinetic friction \(0.25,\) yellow is 0.20 green is \(0.15,\) and blue is \(0.10 .\) Having just finished taking physics, you realize that a wax too slippery will cause you to accelerate down the slope and lose the race. But a wax that's too sticky will cause you to stop and be disqualified. You know that a strong headwind will apply a \(50 \mathrm{N}\) horizontal force against you as you ski, and you know that your mass is 75 kg. Which wax do you choose?

Sam, whose mass is 75 kg, takes off across level snow on his jet-powered skis. The skis have a thrust of \(200 \mathrm{N}\) and a coefficient of kinetic friction on snow of \(0.10 .\) Unfortunately, the skis run out of fuel after only 10 s. a. What is Sam's top speed? b. How far has Sam traveled when he finally coasts to a stop?
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