Explanation Given data: Mass of the person ( m ) is 95 kg. The person stands on a bathroom scale in an elevator. The scale reads 830 N for the first 3.0 s after the elevator starts to move, then 930 N for the next 3.0 s. That is, apparent weight of the person for first 3.0 s ( w app-1 ) after the elevator starts to move is 830 N. The apparent weigh of the person for next 3.0 s ( w app-2 ) is 930 N. Formula used: Write the expression for the weight of the person (normal weight due to gravity of the earth) as follows. w = m g (1) Here, m is the mass of the person, and g is the acceleration due to gravity of the earth, which is 9.8 m s 2 . Write the expression for final velocity by using the kinematic equation as follows. ( v y ) f = ( v y ) i + a y Δ t (2) Here, ( v y ) i is the initial speed of the person, ( v y ) f is the final speed of the person, a y is the acceleration of the elevator. and Δ t is the acceleration duration. Explanation: Calculation of the weight of the person ( w ) : Substitute 95 kg for m , and 9.8 m s 2 for g in equation (1), w = ( 95 kg ) ( 9.8 m s 2 ) = 931 kg ⋅ m s 2 = 931 N { ∵ 1 kg ⋅ m s 2 = 1 N } From the given data, the apparent weight of the person for first 3.0 s of the elevator is 830 N, which is less than the actual weight of the person (931 N). As the apparent weight of the person is less than the actual weight, the motion of the elevator is in the downward direction. Write the expression for apparent weight of the person when the elevator moves in the downward direction as follows. w app = w − m a y (3) Here, w is the actual weight of the person. For the first 3.0 s of the elevator motion from its starting: Modify the expression in equation (3) for the apparent weight of the person for first 3.0 s of the elevator motion from its starting as follows. w app-1 = w − m ( a y ) 1 Rearrange the expression to find the acceleration of the elevator for the first 3.0 s of its motion as follows. ( a y ) 1 = w − w app-1 m Substitute 931 N for w , 830 N for w app-1 , and 95 kg for m , ( a y ) 1 = 931 N − 830 N 95 kg = 101 N 95 kg = 101 kg ⋅ m s 2 95 kg = 1.0631 m s 2 Therefore, the acceleration of the elevator for the first 3.0 s of its motion is 1.0631 m s 2 . Calculation of velocity of the elevator at the end of 3.0 s [ ( v y ) f-1 ] : As the elevator moves in the downward direction, the initial speed of the elevator is 0 m s . ( v y ) i-1 = 0 m s Modify the expression in equation (2) for the final velocity of the elevator at the end of the 3.0 s as follows. ( v y ) f-1 = ( v y ) i-1 + ( a y ) 1 ( Δ t ) 1 Substitute 0 m s for ( v y ) i-1 , 1

College Physics: A Strategic Approach (4th Edition)

4th Edition

ISBN: 9780134609034

Author: Randall D. Knight (Professor Emeritus), Brian Jones, Stuart Field

Publisher: PEARSON

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Question

Chapter 5, Problem 60GP

To determine

To find: The velocity of the elevator at the time of 6.0 s after it starts to move.

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Chapter 5, Problem 59GP

Chapter 5, Problem 61GP

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Chapter 5 Solutions

College Physics: A Strategic Approach (4th Edition)

Ch. 5 - An object is subject to two forces that do not...Ch. 5 - Are the objects described here in static...Ch. 5 - Prob. 3CQ Ch. 5 - Prob. 4CQ Ch. 5 - An astronaut takes his bathroom scale to the moon...Ch. 5 - A light block of mass m and a heavy block of mass...Ch. 5 - a. Can the normal force on an object be directed...Ch. 5 - A ball is thrown straight up. Taking the drag...Ch. 5 - You are going sledding with your friends, sliding...Ch. 5 - Suppose you are holding a box in front of you and...

Ch. 5 - You are walking up an icy slope. Suddenly your...Ch. 5 - Three objects move through the air as shown in...Ch. 5 - Prob. 13CQ Ch. 5 - Raindrops can fall at different speeds; some fall...Ch. 5 - An airplane moves through the air at a constant...Ch. 5 - Is it possible for an object to travel in air...Ch. 5 - For Questions 17 through 20, determine the tension...Ch. 5 - For Questions 17 through 20, determine the tension...Ch. 5 - For Questions 17 through 20, determine the tension...Ch. 5 - For Questions 17 through 20, determine the tension...Ch. 5 - In Figure Q5.21, block 2 is moving to the right....Ch. 5 - The wood block in Figure Q5.22 is at rest on a...Ch. 5 - Prob. 23MCQ Ch. 5 - Prob. 24MCQ Ch. 5 - Prob. 25MCQ Ch. 5 - While standing in a low tunnel, you raise your...Ch. 5 - A 5.0 kg dog sits on the floor of an elevator that...Ch. 5 - A 3.0 kg puck slides due east on a horizontal...Ch. 5 - Eric has a mass of 60 kg. He is standing on a...Ch. 5 - Prob. 30MCQ Ch. 5 - A football player at practice pushes a 60 kg...Ch. 5 - Two football players are pushing a 60 kg blocking...Ch. 5 - Land Rover ads used to claim that their vehicles...Ch. 5 - A truck is traveling at 30 m/s on a slippery road....Ch. 5 - The three ropes in Figure P5.1 are tied to a...Ch. 5 - Prob. 2P Ch. 5 - Prob. 3P Ch. 5 - A construction crew would like to support a 1000...Ch. 5 - When you bend your knee, the quadriceps muscle is...Ch. 5 - An early submersible craft for deep-sea...Ch. 5 - Prob. 7P Ch. 5 - A 65 kg student is walking on a slackline, a...Ch. 5 - Prob. 9P Ch. 5 - The forces in Figure P5.10 are acting on a 2.0 kg...Ch. 5 - The forces in Figure P5.11 are acting on a 2.0 kg...Ch. 5 - A horizontal rope is tied to a 50 kg box on...Ch. 5 - Prob. 13P Ch. 5 - In a head-on collision, a car stops in 0.10 s from...Ch. 5 - An astronauts weight on earth is 800 N. What is...Ch. 5 - A woman has a mass of 55.0 kg. a. What is her...Ch. 5 - Prob. 17P Ch. 5 - a. How much force does an 80 kg astronaut exert on...Ch. 5 - It takes the elevator in a skyscraper 4.0 s to...Ch. 5 - Riders on the Power Tower are launched skyward...Ch. 5 - Zach, whose mass is 80 kg, is in an elevator...Ch. 5 - Prob. 22P Ch. 5 - Figure P5.23 shows the velocity graph of a 75 kg...Ch. 5 - Prob. 24P Ch. 5 - A 23 kg child goes down a straight slide inclined...Ch. 5 - Prob. 26P Ch. 5 - Two workers are sliding a 300 kg crate across the...Ch. 5 - A 4000 kg truck is parked on a 7.0 slope. How big...Ch. 5 - A 1000 kg car traveling at a speed of 40 m/s skids...Ch. 5 - It is friction that provides the force for a car...Ch. 5 - The rolling resistance for steel on steel is quite...Ch. 5 - Prob. 34P Ch. 5 - Prob. 35P Ch. 5 - Prob. 36P Ch. 5 - Prob. 38P Ch. 5 - Prob. 40P Ch. 5 - A 1000 kg car pushes a 2000 kg truck that has a...Ch. 5 - A 2200 kg truck has put its front bumper against...Ch. 5 - Blocks with masses of 1.0 kg, 2.0 kg, and 3.0 kg...Ch. 5 - Prob. 44P Ch. 5 - Prob. 45P Ch. 5 - Prob. 46P Ch. 5 - Each of 100 identical blocks sitting on a...Ch. 5 - A 500 kg piano is being lowered into position by a...Ch. 5 - Dana has a sports medal suspended by a long ribbon...Ch. 5 - Prob. 51GP Ch. 5 - Your forehead can withstand a force of about 6.0...Ch. 5 - A 50 kg box hangs from a rope. What is the tension...Ch. 5 - Prob. 54GP Ch. 5 - A 50 kg box hangs from a rope. What is the tension...Ch. 5 - A fisherman has caught a very large, 5.0 kg fish...Ch. 5 - Riders on the Tower of Doom, an amusement park...Ch. 5 - Prob. 58GP Ch. 5 - Seat belts and air bags save lives by reducing the...Ch. 5 - Prob. 60GP Ch. 5 - A 20,000 kg rocket has a rocket motor that...Ch. 5 - Youve always wondered about the acceleration of...Ch. 5 - Prob. 63GP Ch. 5 - An impala is an African antelope capable of a...Ch. 5 - Josh starts his sled at the top of a 3.0-m-high...Ch. 5 - Prob. 67GP Ch. 5 - Prob. 68GP Ch. 5 - Prob. 69GP Ch. 5 - Prob. 70GP Ch. 5 - Prob. 71GP Ch. 5 - Prob. 72GP Ch. 5 - Prob. 73GP Ch. 5 - Its possible for a determined group of people to...Ch. 5 - Prob. 75GP Ch. 5 - Prob. 77GP Ch. 5 - Prob. 79GP Ch. 5 - Prob. 80GP Ch. 5 - Prob. 82GP Ch. 5 - Prob. 83GP Ch. 5 - Prob. 84GP Ch. 5 - MCAT-Style Passage Problems Sliding on the Ice In...Ch. 5 - MCAT-Style Passage Problems Sliding on the Ice In...Ch. 5 - MCAT-Style Passage Problems Sliding on the Ice In...Ch. 5 - MCAT-Style Passage Problems Sliding on the Ice In...

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