Question
Integrated Concepts An elevator filled with passengers has a mass of 1700 kg. (a) The elevator accelerates upward from rest at a rate of $1.20 \textrm{ m/s}^2$ 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 $0.600 \textrm{ m/s}^2$ 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?

a) $1.87 \times 10^3 \textrm{ N}$

b) $1.67 \times 10^4 \textrm{ N}$

c) $1.56 \times 10^4 \textrm{ N}$

d) $19.4 \textrm{ m}$, $0 \textrm{ m/s}$

Solution Video

# OpenStax College Physics Solution, Chapter 4, Problem 49 (Problems & Exercises) (6:33)

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This is College Physics Answers with Shaun Dychko. This elevator and the people inside of it has a mass of 1700 kilograms, and there is a tension force due to the cable going upwards and the force of gravity going down. There are three different intervals of motion here during which there are different accelerations. So, in part A, we have an acceleration upwards of 1.2 meters per second squared and for time of 1.5 seconds. Our question is asking what is the tension force in the cable. Well the net force is all of the up forces minus all of the down forces. So that's tension force up minus force of gravity down, and that equals mass times acceleration. Then we can add force of gravity to both sides. I've also made a substitution of <i>mg</i> in place of <i>fg</i>. Then we have force of tension is <i>ma</i> plus <i>mg</i> and we can factor out the common factor <i>m</i> and it equals <i>m</i> times bracket <i>a</i> plus <i>g</i>. So that's 1700 kilograms times 1.2 meters per second squared acceleration upwards, plus acceleration due to gravity of 9.8 and that gives 1.87 times ten to the three Newtons is the tension force in the cable during this portion of its motion when it's accelerating upwards at 1.2 meters per second squared. Then the elevator goes at constant speed meaning acceleration is zero for 8.5 seconds. So the net force is still the same picture but now the acceleration is zero and so when we add force of gravity to both sides, we have force of gravity just by itself. So force of tension equals the force of gravity. We have substituted for <i>mg</i> there and so the force of tension is 1700 kilograms times the gravitational field strength 9.8 meters per kilogram, giving us 1.67 times ten to the four Newtons. Then in part C, the elevator decelerates which means its acceleration is directed downwards so it is negative 0.6 meters per second squared for three seconds. Then the force of tension, we're using the formula we figured out up here, it's mass times acceleration plus acceleration due to gravity. So that's 1700 kilograms, times negative 0.6 meters per second squared, plus 9.8 which gives 1.56 times ten to the four Newtons. Then in part D, we're asked to figure out what is the final vertical position of the elevator. So assuming that it starts at position zero, <i>y naught</i> equals zero, it'll then go to a position <i>y one</i> during a time interval of <i>delta t one</i>, which is 1.5 seconds and during this interval it has an acceleration <i>a one</i> of 1.2 meters per second squared. Then it goes to position <i>y two</i> for a time interval of 8.5 seconds with no acceleration, and then finally position <i>y three</i> which is what we want to find. During this interval of motion, we have acceleration three is negative 0.6 meters per second squared for a time <i>delta t three</i> of three seconds. So the final position <i>y three</i> is going to be the position before it, <i>y two</i>, plus the initial velocity when this interval started, which is the velocity at position <i>y two</i> and I've labeled that <i>v two</i>, times the time interval for going from two to three, which is <i>delta t three</i>. Then add to that one half times acceleration during interval three, times the time interval <i>delta t three</i> squared. Now we can't actually solve this because we don't know some of the things that are in this formula. We don't know <i>v two</i> yet and we don't know <i>y two</i>. So, we have to figure those out. Now <i>v two</i> is going to be equal to <i>v one</i> because there is no acceleration here and so the speed is constant. So whatever the velocity is at <i.y one</i> is going to be the velocity at <i> y two</i> as well. So that's going to be the velocity at <i>y zero</i> plus the acceleration during this interval here, plus the time of this interval <i>delta t one</i>. So that is 1.2 meters per second squared, times 1.5 seconds, which is 1.8 meters per second. So that gives us part of our formula for <i>y three</i>. We now know what <i>v two<i> is, it's 1.8 meters per second. We still need to figure out what <i>y two</i> is. Now, <i>y two</i> is going to be the position before it, <i>y one</i>, plus <i>v two</i> times <i>delta t two</i>, plus one half <i>a two</i> times <i>delta t two</i>. But there is no acceleration <i>a two</i>, it is zero. So this reduces to this formula <i>y one</i> plus the constant speed of <i>v two</i> times <i>delta t two</i>. We can't solve that either because we don't know what <i>y one</i> is. So we figure that out now. So <i>y one</i> is <i>y naught</i>, which is zero, we've taken that to be a reference level, plus <i>v naught</i> times <i>delta t one</i>, also this term is zero because there is no speed initially, plus one half times <i>a one</i> times <i>delta t one</i> squared. So that reduces to only this term, one half <i>a one</i> times <i>delta t one<i/> squared. So it's one half times 1.2 meters per second squared times 1.5 seconds squared and that gives 1.35 meters which we can then plug into <i>y two</i>. So we have 1.35 meters plus 1.8 meters per second, times the <i>delta t two</i>, 8.5 seconds, which is 16.65 meters and that in turn, we can finally plug in for <i>y two</i> in the formula for <i>y three</i>. So it's 16.65 meters plus 1.8 meters per second, times three seconds, this is the time interval <i>delta t three</i>, plus one half times negative 0.6 meters per second squared, times 3 seconds squared, giving us 19.4 meters is the final height of the elevator. The final speed <i>v three</i>, will be <i>v two</i> plus acceleration three, times <i>delta t three</i>, and<i>v two <i> we've already calculated as 1.8, and that's what we did here, and then we add to that 0.6 meters per second squared acceleration during interval three, times three seconds, and that give zero meters per second.