/*! 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 63 A car of mass \(987 \mathrm{~kg}... [FREE SOLUTION] | 91Ó°ÊÓ

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A car of mass \(987 \mathrm{~kg}\) is traveling on a horizontal segment of a freeway with a speed of \(64.5 \mathrm{mph}\). Suddenly, the driver has to hit the brakes hard to try to avoid an accident up ahead. The car does not have an ABS (antilock braking system), and the wheels lock, causing the car to slide some distance before it is brought to a stop by the friction force between the car's tires and the road surface. The coefficient of kinetic friction is \(0.301 .\) How much mechanical energy is lost to heat in this process?

Short Answer

Expert verified
Answer: The mechanical energy lost to heat in this process is 786909.16 Joules.

Step by step solution

01

Convert the initial velocity to m/s

Given that the initial velocity of the car is 64.5 mph, we need to convert this to meters per second (m/s). Formula to convert mph to m/s is: 1 mph = 0.44704 m/s So, 64.5 mph = 64.5 * 0.44704 m/s = 28.826 m/s
02

Compute initial kinetic energy of the car

We know that the mass of the car (m) is 987 kg and the initial velocity (v) is 28.826 m/s. The initial kinetic energy (KE) of the car can be calculated using the formula: KE = (1/2)mv^2 KE = (1/2) * 987 * (28.826)^2 = 393454.58 Joules
03

Compute the acceleration due to friction

We know that the coefficient of kinetic friction (μ) is 0.301. Since the car is on a horizontal surface, the only force acting on the car is the friction force. We can calculate the acceleration (a) due to friction using the formula: a = μg a = 0.301 * 9.81 = 2.9501 m/s^2
04

Compute the stopping distance

The car comes to a stop, hence the final velocity (vf) is 0 m/s. We can use the equation of motion \(v_f^2 = v_i^2 + 2as\) to find the stopping distance (s). Rearranging the equation for s, we get: s = (vf^2 - vi^2) / (2a) s = (0 - (28.826)^2) / (2 * -2.9501) = 140.98 meters
05

Compute work done by the friction force (W)

Now we will calculate the work done by the friction force (W) using the formula: W = -fs Since the friction force (f) is equal to m * a, W = -(m * a) * s W = -(987 * 2.9501) * 140.98 = -393454.58 Joules
06

Compute the mechanical energy lost to heat

The mechanical energy lost to heat is equal to the difference between the initial kinetic energy of the car and the work done by the friction force. Since the work done by the friction force is negative, the energy lost (ΔE) can be found by: ΔE = KE - W ΔE = 393454.58 - (-393454.58) = 786909.16 Joules The mechanical energy lost to heat in this process is 786909.16 Joules.

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Key Concepts

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

Kinetic Energy
Kinetic energy is the energy that an object possesses due to its motion. It is a fundamental concept in physics that helps us understand how energy is transferred and transformed. When dealing with kinetic energy, it is important to remember the formula:
  • KE = \( \frac{1}{2}mv^2 \)
where \( m \) is the mass of the object and \( v \) is its velocity.
In the exercise example, the car has a mass of 987 kg and an initial velocity of 28.826 m/s. By substituting these values into the formula, we calculate the initial kinetic energy to be 393,454.58 Joules.
This value represents the energy the car had while it was moving at that initial speed. When the car comes to a stop, its kinetic energy becomes zero. The difference between these two states of energy (initial and final) represents the amount of energy transformed into other forms, in this case, primarily into heat.
Friction Force
Friction force is a force opposing the motion of an object when it is in contact with another surface. It plays a critical role when discussing stopping mechanisms like brakes in a vehicle.
In our scenario, the friction force is what brings the car to a stop. When the driver applies brakes hard, and the wheels lock, friction between the tires and the road surface acts to reduce the car's speed until it completely stops.
The friction force depends on two factors:
  • The normal force, which in simple cases is equal to the weight of the object.
  • The coefficient of kinetic friction \( \mu \), which quantifies how much frictional resistance is encountered.
Work-Energy Principle
The work-energy principle is a key concept that links the concepts of work and energy together. It states that the work done by all forces acting on an object equals the change in the object’s kinetic energy. Mathematically, it can be expressed as:
  • \( W = \Delta KE \)
Here, \( W \) represents the work done, and \( \Delta KE \) is the change in kinetic energy.
In the exercise, the work done by the friction force is used to decrease the car's kinetic energy, ultimately bringing it to a stop. The friction force converts the car's initial kinetic energy into heat energy as the car stops. The total work done by this force can be calculated as \(-393,454.58\) Joules, which signifies the energy transformed from kinetic energy into heat, thus explaining why the car comes to rest.
Coefficient of Kinetic Friction
The coefficient of kinetic friction is a dimensionless value that quantifies the amount of frictional resistance present when one body is moving over another. It is denoted by \( \mu \) and varies based on the materials and surfaces in contact.
In this example, the coefficient of kinetic friction between the car's tires and the road surface is 0.301. A higher coefficient means greater frictional force and thus more energy converted into heat.
  • Friction force = \( \mu \times \) normal force
This coefficient plays a critical role in determining how quickly an object can stop, as it impacts the acceleration due to friction. In the exercise, we see that the work-energy principle hinges on this coefficient, which is directly used to calculate the necessary acceleration to bring the car to a stop, reflecting the effectiveness of the friction in converting kinetic energy into heat.

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Most popular questions from this chapter

You use your hand to stretch a spring to a displacement \(x\) from its equilibrium position and then slowly bring it back to that position. Which is true? a) The spring's \(\Delta U\) is positive. b) The spring's \(\Delta U\) is negative. c) The hand's \(\Delta U\) is positive. d) The hand's \(\Delta U\) is negative. e) None of the above statements is true.

A ball is thrown up in the air, reaching a height of \(5.00 \mathrm{~m}\). Using energy conservation considerations, determine its initial speed.

A spring with \(k=10.0 \mathrm{~N} / \mathrm{cm}\) is initially stretched \(1.00 \mathrm{~cm}\) from its equilibrium length. a) How much more energy is needed to further stretch the spring to \(5.00 \mathrm{~cm}\) beyond its equilibrium length? b) From this new position, how much energy is needed to compress the spring to \(5.00 \mathrm{~cm}\) shorter than its equilibrium position?

A body of mass \(m\) moves in one dimension under the influence of a force, \(F(x)\), which depends only on the body's position. a) Prove that Newton's Second Law and the law of conservation of energy for this body are exactly equivalent. b) Explain, then, why the law of conservation of energy is considered to be of greater significance than Newton's Second Law.

You have decided to move a refrigerator (mass \(=81.3 \mathrm{~kg}\), including all the contents) to the other side of the room. You slide it across the floor on a straight path of length \(6.35 \mathrm{~m}\), and the coefficient of kinetic friction between floor and fridge is \(0.437 .\) Happy about your accomplishment, you leave the apartment. Your roommate comes home, wonders why the fridge is on the other side of the room, picks it up (you have a strong roommate!), carries it back to where it was originally, and puts it down. How much net mechanical work have the two of you done together?

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