Textbook Question
A 20,000 kg rocket has a rocket motor that generates 3.0 x 105 N of thrust. Assume no air resistance. What is the rocket's initial upward acceleration?
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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 18c
It takes the elevator in a skyscraper 4.0 s to reach its cruising speed of 10 m/s. A 60 kg passenger gets aboard on the ground floor. What is the passenger's weight after the elevator reaches its cruising speed?
Verified step by step guidance1
Step 1: Understand the problem. The passenger's weight is the force exerted by gravity on the passenger. Once the elevator reaches its cruising speed, it is moving at a constant velocity, meaning there is no acceleration. The passenger's weight will be determined solely by the gravitational force acting on them.
Step 2: Recall the formula for weight. Weight is given by the equation: , where is the mass of the passenger and is the acceleration due to gravity (approximately 9.8 m/s² on Earth).
Step 3: Substitute the given values into the formula. The mass of the passenger is 60 kg, and the acceleration due to gravity is 9.8 m/s². The equation becomes: .
Step 4: Note that the elevator's cruising speed does not affect the passenger's weight. Since the elevator is moving at a constant velocity, there is no additional force acting on the passenger due to acceleration. The weight remains the same as calculated using the gravitational force.
Step 5: Perform the multiplication to find the passenger's weight. The result will be in newtons (N), as weight is a force. This step involves calculating .
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Key Concepts
Here are the essential concepts you must grasp in order to answer the question correctly.
Weight
Weight is the force exerted by gravity on an object, calculated as the product of mass and the acceleration due to gravity (approximately 9.81 m/s² on Earth). For a 60 kg passenger, the weight can be calculated using the formula W = m * g, where W is weight, m is mass, and g is the acceleration due to gravity.
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Equilibrium in Motion
When an object moves at a constant speed in a straight line, it is said to be in a state of dynamic equilibrium. In the case of the elevator reaching its cruising speed, the forces acting on the passenger (weight and normal force from the elevator floor) are balanced, meaning the passenger does not experience any net force or acceleration.
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Normal Force
The normal force is the support force exerted by a surface perpendicular to the object resting on it. In the context of the elevator, once it reaches cruising speed, the normal force acting on the passenger equals their weight, ensuring that the passenger feels a sensation of weightlessness, but their actual weight remains unchanged at 588 N (60 kg * 9.81 m/s²).
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Related Practice
Textbook Question
A woman has a mass of 55 kg. What is her weight while standing on earth?
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Textbook Question
The position of a 2.0 kg mass is given by x = (2t3 - 3t2) where t is in seconds. What is the net horizontal force on the mass at t = 1s?
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Textbook Question
It takes the elevator in a skyscraper 4.0 s to reach its cruising speed of 10 m/s. A 60 kg passenger gets aboard on the ground floor. What is the passenger's weight before the elevator starts moving?
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Textbook Question
FIGURE EX6.19 shows the velocity graph of a 75 kg passenger in an elevator. What is the passenger's weight at t=1s? At 5 s? At 9 s?
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Textbook Question
Zach, whose mass is 80 kg, is in an elevator descending at 10 m/s. The elevator takes 3.0 s to brake to a stop at the first floor. What is Zach's weight before the elevator starts braking?
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