Textbook Question
Below what speed does a 3.0-mm-diameter ball bearing in 20°C air experience linear drag?
Ch 06: Dynamics I: Motion Along a Line
Knight Calc5th EditionPhysics for Scientists and EngineersISBN: 9780137344796
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Table of contents
- Ch 01: Concepts of Motion35
- Ch 02: Kinematics in One Dimension75
- Ch 03: Vectors and Coordinate Systems58
- Ch 04: Kinematics in Two Dimensions60
- Ch 05: Force and Motion23
- Ch 06: Dynamics I: Motion Along a Line53
- Ch 07: Newton's Third Law27
- Ch 08: Dynamics II: Motion in a Plane52
- Ch 09: Work and Kinetic Energy56
- Ch 10: Interactions and Potential Energy50
- Ch 11: Impulse and Momentum56
- Ch 12: Rotation of a Rigid Body71
- Ch 13: Newton's Theory of Gravity52
- Ch 14: Fluids and Elasticity44
- Ch 15: Oscillations55
- Ch 16: Traveling Waves37
- Ch 17: Superposition39
- Ch 18: A Macroscopic Description of Matter53
- Ch 19: Work, Heat, and the First Law of Thermodynamics60
- Ch 20: The Micro/Macro Connection53
- Ch 21: Heat Engines and Refrigerators34
- Ch 22: Electric Charges and Forces40
- Ch 23: The Electric Field39
- Ch 24: Gauss' Law32
- Ch 25: The Electric Potential63
- Ch 26: Potential and Field57
- Ch 27: Current and Resistance42
- Ch 28: Fundamentals of Circuits39
- Ch 29: The Magnetic Field47
- Ch 30: Electromagnetic Induction60
- Ch 31: Electromagnetic Fields and Waves32
- Ch 32: AC Circuits63
- Ch 33: Wave Optics32
- Ch 34: Ray Optics57
- Ch 35: Optical Instruments30
- Ch 36: Special Relativity38
- Ch 37: The Foundations of Modern Physics25
- Ch 38: Quantization49
- Ch 39: Wave Functions and Uncertainty31
- Ch 40: One-Dimensional Quantum Mechanics35
- Ch 41: Atomic Physics18
- Ch 42: Nuclear Physics27
Knight Calc5th EditionPhysics for Scientists and EngineersISBN: 9780137344796Not the one you use?Change textbook
Chapter 6, Problem 31
A 1500 kg car skids to a halt on a wet road where = 0.50. How fast was the car traveling if it leaves 65-m-long skid marks?
Verified step by step guidance1
Step 1: Identify the forces acting on the car. The car is skidding, so the force of kinetic friction is the only horizontal force acting to stop it. The force of kinetic friction can be calculated using the formula: , where is the coefficient of kinetic friction, is the mass of the car, and is the acceleration due to gravity.
Step 2: Use Newton's second law to find the acceleration of the car. The net force acting on the car is the force of kinetic friction, which causes the car to decelerate. Using , substitute for to find the acceleration: . The negative sign indicates deceleration.
Step 3: Use the kinematic equation to relate the car's initial velocity, final velocity, acceleration, and distance. The equation is: , where is the final velocity (0 m/s, since the car comes to a halt), is the initial velocity, is the acceleration, and is the distance (65 m). Rearrange the equation to solve for : .
Step 4: Substitute the values for and into the equation. Use and m. Remember that is approximately 9.8 m/s².
Step 5: Perform the square root operation to find the initial velocity . This will give you the car's speed before it started skidding.
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Key Concepts
Here are the essential concepts you must grasp in order to answer the question correctly.
Friction and Coefficient of Kinetic Friction
Friction is the force that opposes the relative motion of two surfaces in contact. The coefficient of kinetic friction (μk) quantifies this force, representing the ratio of the frictional force to the normal force. In this scenario, μk = 0.50 indicates that the frictional force acting on the car is half of its weight, which plays a crucial role in determining how quickly the car can stop.
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Kinematics and the Equation of Motion
Kinematics is the branch of physics that describes the motion of objects without considering the forces that cause the motion. The equation of motion, particularly the one relating initial velocity, final velocity, acceleration, and distance, is essential here. It allows us to calculate the initial speed of the car based on the distance it skidded and the deceleration caused by friction.
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Deceleration and Newton's Second Law
Deceleration refers to a decrease in velocity, which in this case is caused by the frictional force acting on the car. According to Newton's Second Law, the net force acting on an object equals its mass times its acceleration (F = ma). The deceleration can be calculated from the frictional force, allowing us to determine how quickly the car slowed down during the skid.
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Related Practice
Textbook Question
A rubber-wheeled 50 kg cart rolls down a 15° concrete incline. What is the magnitude of the cart's acceleration if rolling friction is (a) neglected and (b) included?
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Textbook Question
A 4000 kg truck is parked on a 15° slope. How big is the friction force on the truck? The coefficient of static friction between the tires and the road is 0.90.
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Textbook Question
A 50,000 kg locomotive is traveling at 10 m/s when its engine and brakes both fail. How far will the locomotive roll before it comes to a stop? Assume the track is level.
Textbook Question
Above what speed does a 3.0-mm-diameter ball bearing in 20°C water experience quadratic drag?
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Textbook Question
A medium-sized jet has a 3.8-m-diameter fuselage and a loaded mass of 85,000 kg. The drag on an airplane is primarily due to the cylindrical fuselage, and aerodynamic shaping gives it a drag coefficient of 0.37. How much thrust must the jet's engines provide to cruise at 230 m/s at an altitude where the air density is 1.0 kg/m3
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