College Physics
College Physics
11th Edition
ISBN: 9781305952300
Author: Raymond A. Serway, Chris Vuille
Publisher: Cengage Learning
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1. An elevator shown below filled with passengers has a mass of 1.7 10³ kg. The elevator does motions
(a) through (c) in succession.
T
y
L.
X
meg
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. For each part, are the
vectors for tension in the string and weight of the elevator of equal lengths or unequal lengths.
Owe back...
m₁
E
5 A
V
(a) The elevator accelerates upward from rest at a rate of 1.4
quantities are correct to 3 significant figures.
(i) Newton's Second Law in the y-direction can be written as:
Instruction: If a is the magnitude of the acceleration, pick "1" if the acceleration is upwards,
pick "-1" if the acceleration is downwards, and pick "0" if there is no acceleration.
✓x mea
ΣF=T-m₂g=
(ii) Calculate the tension in the cable supporting the elevator.
Enter to 3 significant figures
T=
N
(iii) How high has the elevator moved during this time?
Enter to 3 significant figures
Ay=
(iv) Calculate the velocity of the elevator after this time.
Enter to 3 significant figures
v(t = 1.5 s) =
m
m
S
for 1.5 s. Assume that all
(b) The elevator continues upward at constant velocity for 8.15 s. Assume that all quantities are
correct to 3 significant figures.
(i) Newton's Law in the y-direction can be written as:
Instruction: If a is the magnitude of the acceleration, pick "1" if the acceleration is upwards,
pick "-1" if the acceleration is downwards, and pick "0" if there is no acceleration.
x mea
ΣF₁=T-meg=
y
(ii) Calculate the tension in the cable supporting the elevator.
Enter to 3 significant figures
expand button
Transcribed Image Text:1. An elevator shown below filled with passengers has a mass of 1.7 10³ kg. The elevator does motions (a) through (c) in succession. T y L. X meg 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. For each part, are the vectors for tension in the string and weight of the elevator of equal lengths or unequal lengths. Owe back... m₁ E 5 A V (a) The elevator accelerates upward from rest at a rate of 1.4 quantities are correct to 3 significant figures. (i) Newton's Second Law in the y-direction can be written as: Instruction: If a is the magnitude of the acceleration, pick "1" if the acceleration is upwards, pick "-1" if the acceleration is downwards, and pick "0" if there is no acceleration. ✓x mea ΣF=T-m₂g= (ii) Calculate the tension in the cable supporting the elevator. Enter to 3 significant figures T= N (iii) How high has the elevator moved during this time? Enter to 3 significant figures Ay= (iv) Calculate the velocity of the elevator after this time. Enter to 3 significant figures v(t = 1.5 s) = m m S for 1.5 s. Assume that all (b) The elevator continues upward at constant velocity for 8.15 s. Assume that all quantities are correct to 3 significant figures. (i) Newton's Law in the y-direction can be written as: Instruction: If a is the magnitude of the acceleration, pick "1" if the acceleration is upwards, pick "-1" if the acceleration is downwards, and pick "0" if there is no acceleration. x mea ΣF₁=T-meg= y (ii) Calculate the tension in the cable supporting the elevator. Enter to 3 significant figures
(a) The elevator accelerates upward from rest at a rate of 1.4
ΣF₁=T-meg=
(ii) Calculate the tension in the cable supporting the elevator.
Enter to 3 significant figures
T=
✔N
quantities are correct to 3 significant figures.
(i) Newton's Second Law in the y-direction can be written as:
Instruction: If a is the magnitude of the acceleration, pick "1" if the acceleration is upwards,
pick "-1" if the acceleration is downwards, and pick "0" if there is no acceleration.
(iii) How high has the elevator moved during this time?
Enter to 3 significant figures
Ay=
(iv) Calculate the velocity of the elevator after this time.
Enter to 3 significant figures
v(t = 1.5 s) =
m
vxmea
m
S
m
for 1.5 s. Assume that all
x mea
ΣF₁=T-m₂g=
(ii) Calculate the tension in the cable supporting the elevator.
Enter to 3 significant figures
Owe back...
(b) The elevator continues upward at constant velocity for 8.15 s. Assume that all quantities are
correct to 3 significant figures.
(i) Newton's Law in the y-direction can be written as:
Instruction: If a is the magnitude of the acceleration, pick "1" if the acceleration is upwards,
pick "-1" if the acceleration is downwards, and pick "0" if there is no acceleration.
Enter to 3 significant figures
T=
N
(iii) How high has the elevator moved during this time?
Enter to 3 significant figures
Ay=
✔m
m
(c) The elevator decelerates at a rate of 0.4 for 3.4 s. Assume that all quantities are correct to 3
s²
significant figures.
(i) Newton's Law in the y-direction can be written as:
Instruction: If a is the magnitude of the acceleration, pick "1" if the acceleration is upwards,
pick "-1" if the acceleration is downwards, and pick "0" if there is no acceleration.
vxmea
ΣF₁=T-meg=
(ii) Calculate the tension in the cable supporting the elevator.
Enter to 3 significant figures
T=
N
(iii) How high has the elevator moved during this time?
Enter to 3 significant figures
Ay=
✔m
(d) What is the total distance the elevator has moved up?
Enter to 3 significant figures
Ay net
=
m
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?
expand button
Transcribed Image Text:(a) The elevator accelerates upward from rest at a rate of 1.4 ΣF₁=T-meg= (ii) Calculate the tension in the cable supporting the elevator. Enter to 3 significant figures T= ✔N quantities are correct to 3 significant figures. (i) Newton's Second Law in the y-direction can be written as: Instruction: If a is the magnitude of the acceleration, pick "1" if the acceleration is upwards, pick "-1" if the acceleration is downwards, and pick "0" if there is no acceleration. (iii) How high has the elevator moved during this time? Enter to 3 significant figures Ay= (iv) Calculate the velocity of the elevator after this time. Enter to 3 significant figures v(t = 1.5 s) = m vxmea m S m for 1.5 s. Assume that all x mea ΣF₁=T-m₂g= (ii) Calculate the tension in the cable supporting the elevator. Enter to 3 significant figures Owe back... (b) The elevator continues upward at constant velocity for 8.15 s. Assume that all quantities are correct to 3 significant figures. (i) Newton's Law in the y-direction can be written as: Instruction: If a is the magnitude of the acceleration, pick "1" if the acceleration is upwards, pick "-1" if the acceleration is downwards, and pick "0" if there is no acceleration. Enter to 3 significant figures T= N (iii) How high has the elevator moved during this time? Enter to 3 significant figures Ay= ✔m m (c) The elevator decelerates at a rate of 0.4 for 3.4 s. Assume that all quantities are correct to 3 s² significant figures. (i) Newton's Law in the y-direction can be written as: Instruction: If a is the magnitude of the acceleration, pick "1" if the acceleration is upwards, pick "-1" if the acceleration is downwards, and pick "0" if there is no acceleration. vxmea ΣF₁=T-meg= (ii) Calculate the tension in the cable supporting the elevator. Enter to 3 significant figures T= N (iii) How high has the elevator moved during this time? Enter to 3 significant figures Ay= ✔m (d) What is the total distance the elevator has moved up? Enter to 3 significant figures Ay net = m 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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