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An elevator shown below filled with passengers has a mass of 1850 kg. The elevator does motions (a) through (c) in succession.  The elevator accelerates upward from rest at a rate of 1  ms2 for 1.9 s. (ii) Calculate the tension in the cable supporting the elevator. (iv) Calculate the velocity of the elevator after this time. The elevator continues upward at constant velocity for 8.8 s.   (iii) How high has the elevator moved during this time?

An elevator shown below filled with passengers has a mass of 1850 kg. The elevator does motions (a) through (c) in succession.  The elevator accelerates upward from rest at a rate of 1  ms2 for 1.9 s. (ii) Calculate the tension in the cable supporting the elevator. (iv) Calculate the velocity of the elevator after this time. The elevator continues upward at constant velocity for 8.8 s.   (iii) How high has the elevator moved during this time?

College Physics
College Physics
11th Edition
ISBN:9781305952300
Author:Raymond A. Serway, Chris Vuille
Publisher:Raymond A. Serway, Chris Vuille
Chapter1: Units, Trigonometry. And Vectors
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Problem 1CQ: Estimate the order of magnitude of the length, in meters, of each of the following; (a) a mouse, (b)...
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An elevator shown below filled with passengers has a mass of 1850 kg. The elevator does motions (a) through (c) in succession.

 The elevator accelerates upward from rest at a rate of 1  ms2 for 1.9 s.

(ii) Calculate the tension in the cable supporting the elevator.

(iv) Calculate the velocity of the elevator after this time.

The elevator continues upward at constant velocity for 8.8 s.

  (iii) How high has the elevator moved during this time?

(c) The elevator decelerates at a rate of 0.7  ms2 for 2.85 s.

 (ii) Calculate the tension in the cable supporting the elevator.

(iii) How high has the elevator moved during this time?

(d) What is the total distance the elevator has moved up?

 

1. An elevator shown below filled with passengers has a mass of 1850 kg. The elevator does motions (a) through (c) in succession. mg For each of the parts below draw a free body diagram of the elevator in your notebook for each of the parts (a) to (C). Draw the acceleration and velocity vectors in the boxes. L. a V (a) The elevator accelerates upward from rest at a rate of 1 for 1-9 s. (i) Newton's Second Law in the y-direction can be written as EF, =-1 Vmeg+ 1 T=1 vVm,a (ii) Calculate the tension in the cable supporting the elevator. T= 18130 (iii) How high has the elevator moved during this time? Ay = 1.8 (iv) Calculate the velocity of the elevator after this time. v(t = 1.9 s) =.95 m (b) The elevator continues upward at constant velocity for 8-8 s. () Newton's Law in the y-direction can be written as
Transcribed Image Text:1. An elevator shown below filled with passengers has a mass of 1850 kg. The elevator does motions (a) through (c) in succession. mg For each of the parts below draw a free body diagram of the elevator in your notebook for each of the parts (a) to (C). Draw the acceleration and velocity vectors in the boxes. L. a V (a) The elevator accelerates upward from rest at a rate of 1 for 1-9 s. (i) Newton's Second Law in the y-direction can be written as EF, =-1 Vmeg+ 1 T=1 vVm,a (ii) Calculate the tension in the cable supporting the elevator. T= 18130 (iii) How high has the elevator moved during this time? Ay = 1.8 (iv) Calculate the velocity of the elevator after this time. v(t = 1.9 s) =.95 m (b) The elevator continues upward at constant velocity for 8-8 s. () Newton's Law in the y-direction can be written as
(a) The elevator accelerates upward from rest at a rate of 1 for 1-9 s. (1) Newton's Second Law in the y-direction can be written as EF,=1 (ii) Calculate the tension in the cable supporting the elevator. meg+1 VT=1 mea T= 18130 (ii) How high has the elevator moved during this time? Ay = 1.8 (iv) Calculate the velocity of the elevator after this time. v(t = 1.9 s) = 95 (b) The elevator continues upward at constant velocity for 8.8 5. (1) Newton's Law in the y-direction can be written as EF, =1 (ii) Calculate the tension in the cable supporting the elevator. meg+[1 T= -1 X mea T= 18130 (ii) How high has the elevator moved during this time? Ay = 38.72 (C) The elevator decelerates at a rate of 0.7 for 2-85 S. (i) Newton's Law in the y-direction can be written as EF,=1 (ii) Calculate the tension in the cable supporting the elevator. meg+[1 T=-1 ma T= 18130 (iii) How high has the elevator moved during this time? Ay = 1.99 (d) What is the total distance the elevator has moved up? Ay net = 14.595 Reflect: In which case, is the tension in the chord pulling the elevator the greatest? Why? Does the tension always equal the weight of the elevator?
Transcribed Image Text:(a) The elevator accelerates upward from rest at a rate of 1 for 1-9 s. (1) Newton's Second Law in the y-direction can be written as EF,=1 (ii) Calculate the tension in the cable supporting the elevator. meg+1 VT=1 mea T= 18130 (ii) How high has the elevator moved during this time? Ay = 1.8 (iv) Calculate the velocity of the elevator after this time. v(t = 1.9 s) = 95 (b) The elevator continues upward at constant velocity for 8.8 5. (1) Newton's Law in the y-direction can be written as EF, =1 (ii) Calculate the tension in the cable supporting the elevator. meg+[1 T= -1 X mea T= 18130 (ii) How high has the elevator moved during this time? Ay = 38.72 (C) The elevator decelerates at a rate of 0.7 for 2-85 S. (i) Newton's Law in the y-direction can be written as EF,=1 (ii) Calculate the tension in the cable supporting the elevator. meg+[1 T=-1 ma T= 18130 (iii) How high has the elevator moved during this time? Ay = 1.99 (d) What is the total distance the elevator has moved up? Ay net = 14.595 Reflect: In which case, is the tension in the chord pulling the elevator the greatest? Why? Does the tension always equal the weight of the elevator?
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