Textbook Question
How much work does an elevator motor do to lift a 1000 kg elevator a height of 100 m? How much power must the motor supply to do this in 50 s at constant speed?
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Ch 09: Work and Kinetic Energy
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 9, Problem 36b
How much work must you do to push a 10 kg block of steel across a steel table at a steady speed of 1.0m/s for 3.0 s? What is your power output while doing so?
Verified step by step guidance1
Step 1: Identify the forces acting on the block. Since the block is moving at a steady speed, the net force is zero. The force you apply must balance the kinetic friction force. The formula for the force of kinetic friction is: , where is the coefficient of kinetic friction, is the mass of the block, and is the acceleration due to gravity.
Step 2: Determine the coefficient of kinetic friction for steel on steel. From standard tables, the coefficient of kinetic friction for steel on steel is approximately 0.6. Use this value in the formula for the frictional force.
Step 3: Calculate the work done. Work is given by the formula: , where is the distance the block is pushed. To find , use the relationship , where is the speed (1.0 m/s) and is the time (3.0 s).
Step 4: Calculate the power output. Power is the rate at which work is done and is given by the formula: , where is the work done and is the time (3.0 s).
Step 5: Substitute the known values into the formulas to compute the work done and the power output. Ensure all units are consistent (e.g., mass in kilograms, distance in meters, time in seconds).
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Key Concepts
Here are the essential concepts you must grasp in order to answer the question correctly.
Work
Work is defined as the product of force and displacement in the direction of the force. In physics, it is calculated using the formula W = F × d, where W is work, F is the force applied, and d is the distance moved. When pushing an object at a constant speed, the work done is equal to the force of friction multiplied by the distance traveled.
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Friction
Friction is the resistive force that opposes the motion of an object when it is in contact with another surface. It is influenced by the nature of the surfaces in contact and the normal force acting between them. For a block sliding on a table, the frictional force can be calculated using the coefficient of friction and the weight of the block.
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Power
Power is the rate at which work is done or energy is transferred over time. It is calculated using the formula P = W/t, where P is power, W is work, and t is the time taken. In this scenario, once the work done to push the block is determined, the power output can be calculated by dividing the work by the time taken to push the block.
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Related Practice
Textbook Question
Justin, with a mass of 30 kg, is going down an 8.0-m-high water slide. He starts at rest, and his speed at the bottom is 11 m/s. How much thermal energy is created by friction during his descent?
Textbook Question
An 8.0 kg crate is pulled 5.0 m up a 30° incline by a rope angled 18 ° above the incline. The tension in the rope is 120 N, and the crate's coefficient of kinetic friction on the incline is 0.25. What is the increase in thermal energy of the crate and incline?
Textbook Question
A baggage handler throws a 15 kg suitcase along the floor of an airplane luggage compartment with a speed of 1.2 m/s. The suitcase slides 2.0 m before stopping. Use work and energy to find the suitcase's coefficient of kinetic friction on the floor.
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Textbook Question
How much work does an elevator motor do to lift a 1000 kg elevator a height of 100 m?
Textbook Question
A 70 kg human sprinter can accelerate from rest to 10 m/s in 3.0 s. During the same time interval, a 30 kg greyhound can go from rest to 20 m/s. What is the average power output of each? Average power over a time interval ∆t is ∆E/∆t.
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