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Physics / College Physics 1 / Chapter 4 / Problem 49

College Physics | 1st Edition | ISBN: 9781938168000 | Authors: Paul Peter Urone, Roger Hinrichs

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Integrated Concepts An elevator filled with passengers has

Chapter 4, Problem 49

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QUESTION:

Problem 49PE

Integrated Concepts An elevator filled with passengers has a mass of 1700 kg. (a) The elevator accelerates upward from rest at a rate of 1.20 m/s2 for 1.50 s. Calculate the tension in the cable supporting the elevator. (b) The elevator continues upward at constant velocity for 8.50 s. What is the tension in the cable during this time? (c) The elevator decelerates at a rate of 0.600 m/s2 for 3.00 s. What is the tension in the cable during deceleration? (d) How high has the elevator moved above its original starting point, and what is its final velocity?

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QUESTION:

Problem 49PE

Integrated Concepts An elevator filled with passengers has a mass of 1700 kg. (a) The elevator accelerates upward from rest at a rate of 1.20 m/s2 for 1.50 s. Calculate the tension in the cable supporting the elevator. (b) The elevator continues upward at constant velocity for 8.50 s. What is the tension in the cable during this time? (c) The elevator decelerates at a rate of 0.600 m/s2 for 3.00 s. What is the tension in the cable during deceleration? (d) How high has the elevator moved above its original starting point, and what is its final velocity?

ANSWER:

STEP 1 OF 8

a)

STEP 2 OF 8

The net force due to tension and weight

${F}_{net}=T-w$

$ma=T-mg$

Mass of the passengers, m=1700 kg

Acceleration of the elevator(upward)= $1.20\ m/{s}^{2}$

Acceleration due to gravity.g= $9.8\ m/{s}^{2}$

We have to find the tension (T)

$T=ma+mg$

$=m(a+g)$

$=(1700\ kg)(1.20\ m/{s}^{2})+9.80\ m/{s}^{2}$

$=1.87\times{}{10}^{4}N$

So, tension in the cable is $1.87\times{}{10}^{4}N$

STEP 3 OF 8

b)When the velocity is constant , the acceleration , $a=0\ m/{s}^{2},\$

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