FIGURE EX6.19 shows the velocity graph of a 75 kg passenger in an elevator. What is the passenger's weight at t=1s? At 5 s? At 9 s?

Question

Velocity graph of a 75 kg passenger in an elevator over time, showing changes at 1s, 5s, and 9s.

Accepted Answer

Hi, everyone in this practice problem. We're being asked to calculate Ali's actual weight. When Ali is taking an elevator to get to the third floor, Ali's velocity versus time graph in the elevator is shown in the figure and if Ali's mass is 60 kg for being asked to calculate his weight first at D equals two seconds second at D equals to five seconds and third at D equals to seven seconds. So in the V Y versus D graph of Ali's velocity with V Y in meters per second and P in seconds, it is shown that from 0.0 to three seconds, Ali's velocity increases from 0 to 6 m per second. It remains constant at from 3 to 6 seconds at six m per second and it decreases from 6 to seconds from six m per second to zero m per second. The options given for Ali's actual weight are 1st 708 Newton, 2nd 588 Newton 3rd Newton B 1st 708 Newton 2nd 684 Newton and 3rd 348 Newton C 1st 645 Newton 2nd 455 Newton and 3rd 408 Newton lastly, D 1st 708 Newton 2nd Newton and 3rd 309 Newton. So from the velocity time graph, it can be seen that the velocity of the elevator increases at a constant rate for D equals to zero to T equals to three seconds. Namely, the velocity actually accelerates up and remains constant until six seconds. Namely the acceleration is going to be zero during that time. And finally, the velocity will decrease at a constant rate from T equal to six to T equal to eight seconds. Namely it will decelerated the slope of the first and the third part of the graph is constant. So that will mean that the acceleration is constant for both the first and the third part from the free body diagram, it can be seen that only two forces are going to be acting on ali on every single uh part of this question. Let's start with the first part. So with the first part of this question, the free body diagram is going to be represented by uh this graph right here. The sphere or the circle is going to be representing ali which we will assume to be a particle like. And then we will have F N pointing upwards and F G pointing downwards F G is essentially just uh W And in this case, from uh P equals to zero to T equals to three seconds or at P equals to two seconds, Ali will be accelerating up, which is seen from the figures uh shown in the problem statement. So a one or a uh the first acceleration will be pointing upwards. So according to Newton's second law of motion, we have Sigma F Y or the net force acting on the Y direction will equals to M multiplied by A Y. In this case, A Y is going to be a one and having the convention of anything going upwards to be positive, we will then have Sigma F Y to be equals to F N minus F G to be equals to M multiplied by a one just like. So, so then F G just M multiplied by G. So then F N equals to M multiplied by A one plus M multiplied by G. So F N will then equals to M multiplied by open parentheses. A one plus G close parentheses. So the normal force is actually the apparent weight experienced by the person or by ali as the elevator accelerates up. And hence the weight of the person for the first part of the graph can be calculated by this formula right here. So in this case, we know what M is, we know what G is, but we do not know what A one is. So we want to have to use the definition of acceleration. And the information given in the graph here to find the value for a one. So a one will equals to the change in velocity over time. So that will be a one will equals to V F minus V I divided by T F minus T I. In this case, V F is six m per second. We're just looking from 60.0 to 3 seconds. So V F is six V I is zero m per second. P F is three seconds and P I is zero seconds. So that will give us a one to be two m per second squared. Substituting that and everything that we know from the problem statement into F N into the F N equation that we got, we will get F N equals to M which is 60 kg multiplied that by open parenthesis. A one which is two m per second squared plus G which is 9.81 m per second squared. And that will give us F N or the apparent weight to be 708 Newton does. The weight of ali at P equals to two seconds is going to be 708 Newton. Awesome. I'm going to represent this with F N one so that it's easier for us to differentiate in the future. Now, we want to move on to the second part of this problem. So in the second part of the problem, we have determined previously that there is no acceleration because the velocity is constant. So in this case, A Y is going to equals to zero. So the free body diagram will then just be the sphere or the particle here which is going to be a, having uh F N pointing upwards and F G pointing downwards with no acceleration because A Y equals to zero. So in that case, using the same uh method which is the Newton second law will have Sigma F Y equals to M multiplied by A Y which will actually equals to zero. Because of that F N will equals to F G which will equals to W. So we can calculate F N by multiplying M and G just like normal because it is at equilibrium. So we will get M to be 60 kg and G to be 9.81 m per second squared, we will multiply that to get F N two to equal two, 588 Newton just like. So, so 588 Newton will be the weight of ali at the equals to five seconds. Awesome. So now we can actually move to the last part of this problem statement. So in the last part or in the third part, we have determined previously from the graph given that the elevator is going to be going uh is going to decelerated. So therefore, let's start with throwing the free body diagram in this case, we have again, the circle representing Oli F N pointing upwards and F G pointing downwards and the acceleration is going to be decelerating. So it will be pointing downwards and that will be a three just like so awesome. So now we will actually take similar approach which will be using Newton Second law Sigma F Y will equals to M multiplied by A Y which in this case will equals to M multiplied by a three Sigma F Y will still remains to be F N and F G only. And we will stick to the convention of F N being positive. So F N minus F G will equals to M multiplied by a three where F N simplifying everything will equals to M multiplied by a three plus F G which will be M multiplied by G. So therefore, F N will actually equals to M multiplied by open parentheses. A three plus G. We still do not know what A three is and doing what we did. For the first part, we have to look into our um graph given and calculate what A three is. So in this case, a three using the definition of acceleration will equals to F V F minus V I, all of that divided by T F minus T I. In this case, we are looking at uh 6 to 8 seconds because there's uh there will be constant acceleration throughout. And uh we are uh being asked to find the mass or the weight, the actual weight at T equals to seven seconds. So in this case, we will then have V F to be zero m per second V I to then be six m per second, the F to be eight seconds and the I to be six seconds. And that will give us a three value to be a negative three m per second squared just like so, and now we can substitute that into our F N equation. So F N will then equals to M multiplied by a three plus G in parentheses, which will equals to 60 kg multiplied by negative three m per second squared plus 9.81 m per second squared. And that will give us F N three value of 408 Newton just like so, so 408 will be the actual weight of Ali at T equals to seven seconds. So in this case, we have found uh Ali's actual weight at T equals two seconds, which will be 708 Newton at T equals to five seconds, which will be 588 Newton and at T equals to seven seconds, which will be 408 Newton. So that will be it for this video. All of those answers will correspond to option A in our answer choices, which will be the answer to this particular practice problem. So that will be it for this video. If you guys still have any sort of confusion, please make sure to check out our other lesson videos on similar topics and that will be it for this one. Thank you.

FIGURE EX6.19 shows the velocity graph of a 75 kg passenger in an elevator. What is the passenger's weight at t=1s? At 5 s? At 9 s?

Verified Solution

Key Concepts

Weight

Velocity Graph Interpretation

Net Force and Acceleration