a-conveyor-belt-800-mathrmm-long-moves-at-025-mathrmm-mathrms-if-a-package-is-placed-at-one-end-fi-3

Name: a conveyor belt 800 mathrmm long moves at 025 mathrmm mathrms if a package is placed at one end fi 3
Uploaded: 2021-06-07
Duration: 2 min 2 s
Channel: Numerade
Description: A conveyor belt $8.00 \mathrm{m}$ long moves at $0.25 \mathrm{m} / \mathrm{s}$. If a package is placed at one end, find its displacement from the other end as a function of time.

Question

A conveyor belt $8.00 \mathrm{m}$ long moves at $0.25 \mathrm{m} / \mathrm{s}$. If a package is placed at one end, find its displacement from the other end as a function of time.

Kian Manafi · Accepted Answer

So we&#x27;re told that this conveyor belt here, I&#x27;ll just draw a quick sketch of, it&#x27;s going to be, our conveyor belt is eight m long And we&#x27;re told that if we set something on this conveyor belt, it moves at two point or sorry, .25 meters per second. So every second we&#x27;re moving .25 m. And we want to figure out a equation for the displacement from the other side of our conveyor belt. So that means that initially are displacement is going to be eight m since we&#x27;re eight m away from the other side of the conveyor belt when we place an object on there. So we can say that our initial to play our initial displacement, The zero is equal to eight m. And so our displacement equation is just going to be our initial displacement and then plus um our velocity. So in this case it&#x27;s negative .25 m per second Times T. And the reason it&#x27;s negative .25 m is because we&#x27;re getting closer and closer to the other side of our compare belt, so our displacement is actually decreasing as time goes on Since we start with initial displacement of eight m from the other side of our conveyor belt. And then we&#x27;re moving towards our conveyor belt at .25m/s. So our displacement gets um minus, or we get closer by 0.25 m every second. So that&#x27;s why our displacement is going to be equal to our initial displacement plus negative 0.25 times T. So then we can just say our displacement equation is equal to 8 -25 times t. And again, the reason that we have this minus 0.25 times T. Is because we&#x27;re going towards the other side of the conveyor belt At a speed of .25 m/s.

Vishal Gupta · Answer

hi Initially speed off the conveyor belt is given us for meter per second and finally, speed, as finally that can were built is coming to rest. So it&#x27;s finally speed should be zeal. The coefficient off static friction between this belt and the package put over it is given a 0.2. So for so friction experienced by that package over this conveyor belt, we begin by new into MG and said, and as its direction is opposite to the direction of motion. So that is given us negative and using Newton&#x27;s second law motion that can also be given us am in tow. A. There is the retardation produced in the motion off that package which hasn&#x27;t kept over duck and were built. Or you can see the retardation in the motion off the belt conveyor belt. Canceling this M retardation off the belt will come out to B minus mu into G means minus 0.2 in 29.8 meter. First, secondly, square, which comes out to be minus 1.96 m per second squared. So we have to find here the minimum time in which this belt should come to rest so that the package does not start moving over this belt. So for which we will use first equation of motion which says, finally, speed VF is given us initially, speed v I was a into t a za finally speeding zero. We I that was four meter per second for eight. This is minus 1.96 meter for second square into time, which is missing. So this time will be given by 1.96 and two T is equal to four. So time comes out to be four by 1.96 seconds. Or we can say this is 2.4 seconds and here it becomes the answer. What a given problem means in this much time the bell should come to rest. And in that case, the package put over it will not start sliding over the bed. Thank you.

Sheh Lit Chang · Answer

in this question. Okay. We have a package on the conveyor belt. Yeah. So it&#x27;s moving at five m/s. And then Okay. And then coming about suddenly stops. We want to find out how far uh that the package remove before coming to rest. Okay? And we are given that the coefficient of kinetic question is Their .15. What? So to solve this problem, we need the acceleration of the package also to find acceleration. We need a free body diagram after package mm. So everybody diagram, we look like this. Okay? We have to wait. Yeah, but normal for us. And now we have that friction, which is you kn thanks to the right. So uh the acceleration yes to the right Using your 12th hour. Mhm. Okay. Um submission of fossils along the exit goes to have a okay? And we know that W. Is equal to end. And so uh okay. And okay. And G. Is equal to M. A. The the mass cancer out. Okay, so the acceleration is 0.15 times 9.81. We get 1.472 m/s squared. Then we use the kingdom attics equation using the square goes through, you know, square plus to a to access to data and sister distance traveled by the package. Okay. And uh the square yes zero, You know. Yes 5m/s. And then he decided to be negative 1.472. Yeah, it&#x27;s for a 2nd square. Okay, so um so by rearranging, we have our delta S. Is we go through? I don&#x27;t know The square and has been on square divide by two a minus five square divide by negative two times -1.4. That&#x27;s normal. Thank goodness. I yeah, because I pulled the negative sign out Um 1.572. Mhm. And calculate this you get as to be 8.49 m. Okay, so this is the answer for this question and that&#x27;s all

Kristela Garcia · Answer

So we have this walk that&#x27;s being dropped in a conveyor boat, so it&#x27;s going to have an initial velocity of 1.5 meters per second and a Susan. As soon as it hits the conveyor belt, it&#x27;s going to have a friction that it&#x27;s to the left of this is coefficient of This is a Connecticut friction. So we have my normal force by force of gravity, and we want to know how long it takes to stop. So, using Newton&#x27;s second law, I can say my friction is equal to M. E. And my friction is my coefficient of kinetic friction times, my normal force and my normal force in this case is my mess times gravity. Oh, the block. My mess is canceled and so my acceleration is my coefficient of kinetic friction times. Gravity is my acceleration. So using a kid a magic equation, I can then use that to sell for how long it will take. Until it comes to a stop. These negatives will cancel. Zero is not there. So I have 1.5 meters per second, divided by 0.7 times 9.8 gives us a time of zero point two one seconds. And if we know the time it takes to a slide to a stuff, then we can use that part B, which asks System by the distance. Same idea. I can use a different cinematic equation to sell for distance. It moves again. Final Velocity is zero. We know my acceleration from part, eh? Iss negative, you gene. So I get 1.5 squared, divided by two Ted&#x27;s 0.7 times 9.8 gives us a distance traveled of their 0.16 leaders.

Nicholas Majtenyi · Answer

question 79 states. Hey, boxes placed on a conveyor belt that moves with the constant speed of 1.25 meters per second. The coefficient. Kinetic friction between the box and the plate. 0.780 Part of the question asks, How much time does it take for the boxes? Stop sliding relative to the belt and B. How far is the box move in this time? Okay, so the new it&#x27;s the scenario is shown here were able box on this conveyor belt that&#x27;s moving. Stay to the right direction is indicated. Here. I finally got my axes. Just so it&#x27;s someone a little more clear about why next, when the boxes placed it will, he&#x27;s slipping, essentially until stops relative to the belt. So his connect friction is what stops that box removing so it moves along with the belt. The question asks, How long does this take in terms of semi seconds? And how far will this box travel before these objects stops slipping? Right, So you have the scenario here? First thing we should do, um, we should draw 1/4 body diagram acting on the object just to see how what&#x27;s happening to it. Obviously, if it&#x27;s on the belt underneath it, the normal force would be pointing up course of gravity, really pulling it down. It states that&#x27;s is movie with a constant speed. So there&#x27;s no net force acting on the object, However, as is moving to very there&#x27;ll be opposing force of friction. I&#x27;ll call little F K force of kinetic friction and will be again pulling to the right hand side. So we look at our forces in the wind direction. It&#x27;s simply the normal force, minus a force of gravity. That, of course, is tasked equal zero. So we&#x27;re left with an R Y component. Are normal forces just there bouncing with the force of gravity in our exes? A similar story. The force is pulling the object along. Why is the force of kinetic friction equaling zero with the force of it being pulled? Is there a force of connect friction? If remember, it&#x27;s just you, kay times and which is just as he found in her white component in UK Times MG. So we&#x27;re left with this. If we&#x27;re saying it&#x27;s moving to the right hand side in this scenario that we have this mu kmg equally ref Nichols and may Because it is moving, our mass is cancelled, which is convenient because we don&#x27;t have I mass given to us soon. See, the object is accelerating You Okay, g, This mu k is coefficient connect actions brought it to us and in turn, the object is accelerating 7.65 meters per second or I should say decelerating, really causing it to stop. So okay, we can see how quickly this box of stopping But we want to know how long that takes. So we solve for the exploration. We have time, have a velocity. So attribute indication that we need to use our kid a Matic equations, cinematic equations. Katie With all the variables provided to us, we know that it will be stopping. So we have a final velocity equals an initial velocity plus 80. However, again we&#x27;re saying while we solved for the magnitude here we are describing a slowing down of the object. So our a placing it in terms of ire initially in terms of magnitude, we apply that is recognize the situation that the object is DX already. This is our scenario again. The box will be stopping its motion so its final velocity will be zero So determined that are we not? No, sir. You know you&#x27;re not. That&#x27;s provided our time. Could be simply put rearranged. Maybe not minus sorry. Divided by acceleration exhibition resolve our initial velocity is given to us. So the man of time provided to three significant for years is just 0.163 seconds. Great part B of this question asked how far they travel again we have all these other parameters So we do We should recognize that this again we need to deal with Candy Matic equations There&#x27;s multiple you could choose now that we have the time solved However, I&#x27;m just gonna choose the one we have initial and final velocities Final velocity squared equals initial velocity squared I can hear minus two X because our exploration offside Desiree exploration of magnitude should be negative again. Our final velocity will be zero term goes to zero. You can rearrange its equation to solve For the distance the box will traveling, which is just to be not squared divided by two times a Hey, we calculated Be not is given so simply the distance debacle travel will be 0.102 meters or 102 centimeters again. 236 pigs

David Zhang · Answer

So for this question, um, start by looking at the problem in determining that from this point, when using the second law, some of the forces schools mass and acceleration the wind direction because it&#x27;s going down. So this is on the inclined inclined plane and this point the exploration zero. So you can also find the normal force, which is equal to wait. Times coasting data are costing 30 and we can determine that it is equal to 8.5 times. Um, so when you find the fictional force, you can determine that it&#x27;s just the coefficient of the connect friction, and it&#x27;s moving. Does your point too multiplied by that normal force? 3.5 times m. You get 1.7 times for the frictional force. The work done by the friction from A to B did use this equation or times sine 30 on this friction. Let&#x27;s call it that far one. You multiply that by the distance so you can simplify this down. Thio mg sign 30 minus 1.7 en just the friction force we found or the work done by the frictional force done we found earlier. You multiply that by six, which is d And then we should get 19.23 times ever. So once it hit the surface level, um, there is now a new normal. So okay, went through here. We can take this new normal, the second normal force. That&#x27;s just end because a flat surface that&#x27;s 9. 21 times. Um, yeah. In this case, when you find the frictional force Second Fiction Force, use the same co official friction. And now you multiply that new normal force that you get you 1.962 times them now to find the work done by the friction on this flat surface. Take work. Come back. Fiction too negative, multiplied by the distance X That&#x27;s seven. So when we plug in the numbers 1.692 times and times seven, that should get your negative 13.7 34 times m So I&#x27;m looking at this. You have a connect energy? A which is theory one half and B squared equation. Last time velocity squared. Plug in the values we have. That&#x27;s the velocities one. So it&#x27;s one times and one squared and that is your 10.5 times. Then this is connected via a kinetic energy at sea at the end. Same equation? Yeah, you don&#x27;t know the mass or the velocity it, so we can just leave them as variables way, assuming that no energy has lost that be so using the conservation of energy. No, we would have this equation. It&#x27;s just the work. Yeah. Can I count your A? And this is more looking for is connected idiocy. And we can find that by plugging we already know. So we found before connect energy a five times m plus 19.23 m plus minus negative. 13.7 34 a m equals 0.5 m m a c squared. So since we have most everything we need, we get the animal, cancel up and now we just have to solve for BC. And if you solve that, we just add up or subtract these numbers, then divide by 0.5 on each side you built isolate V C G. B C squared equals 11.992 When you square with bet, you get the glass. A C equals 3.46 m per second. That&#x27;s the final velocity now to find the distance. Yeah, Mhm. You want to find the work done by the friction on the belt. But first, we&#x27;re gonna find the frictional force to be the coefficient of friction times normal force, which is equal to 0.2 mass times gravity go to 1.962 m. Now to find the work done, it&#x27;s multiply the negative friction times the distance going to the belt. You get negative 1.962 times m times x dot Yeah. Oh, now, using conservation of energy in the kinetic energy of See that we found earlier plus work e m The kinetic energy of the belt negative one half e was plug in the values that we know once We don&#x27;t know we keep as variables called Once again we divide through by M. Adams. Cancel out, we don&#x27;t need the masses and we just I mean, that&#x27;s left. Don&#x27;t you get that you should be able to solve for this able expelled and you should get 2.3 m

Problem

During each cycle, the velocity $v$ (in $\mathrm{…

Problem 5 Easy Difficulty

A conveyor belt $8.00 \mathrm{m}$ long moves at $0.25 \mathrm{m} / \mathrm{s}$. If a package is placed at one end, find its displacement from the other end as a function of time.

Answer

$s=8.00-0.25 t$

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$s=8.00-0.25 t$

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Allyn J. Washington, Richard S. EvansBasic Technical Mathematics with Calculus

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