two-students-push-two-different-boxes-across-a-frictionless-floor-both-students-exert-a-horizontal-2

Question

Two students push two different boxes across a frictionless floor. Both students exert a horizontal force of 50 $\mathrm{N}$ . If the first box has mass $m_{1}=$
10 $\mathrm{kg}$ and the second box has mass $m_{2}=40 \mathrm{kg}$ , what is the value of
$a_{1} / a_{2} ?$
(A) 1$/ 4$
(B) 1$/ 2$
(C) 2
(D) 4

Salamat Ali · Accepted Answer

we have, ah, two equal forces. If one in if too home one is acted on the mass one, which is, uh, mass one with a produces expression one another force, which is are acted on the mast to and produces exploration two. So, uh, the ratio of exploration one to exploration to for the hearing problem, isn&#x27;t it? And pays in two over and one so and two we have here is a four four, and then one be ever won. So the option that is right is a D. So a 12 a two ratio is four ratio. So the answer is a d end off the problem. Thank you for

Kevin Hayakawa · Answer

okay. And this problem, we have two blocks that are side by Sign one another&#x27;s thie ground. We have block a on the left right here on the block. B is a little bit smaller, and it&#x27;s on the right right there. And they were being told that some exterior applied force is being ah, applied on block A. I&#x27;m sorry, actually going, Tio, draw two separate for your body diagrams for this problem. Just in order to better understand what is what is going on. I&#x27;m starting off with block A. So, of course, the first force that we&#x27;re going to include in their is the applied Force we want to think of it is being applied on the sensor of massive block A, which is in the sensor right there on the moving over Teo block B on the side over here because they&#x27;re in contact with one another. Block A is exerting a force on block B on. We&#x27;re going to call that F A B. Because block A is exerting a force on block B hover loosens. Their law dictates that every force has an equal and opposite reaction force. So in turn, the opposite direction of Block B is going to be exerting a force on Block A and is going to look a little something like that that completes our three body diagrams for this problem from&#x27;s. Now let&#x27;s go ahead and applying it in second Monte to the blocks. Um, and when you sum of the forces on block A. So if we set up our coordinate system so that X is increasing a CE, we go to the right. Then we can say that if we subtract off FDA from at that walk, give us a positive force in the end. Oh, and then we end up with the massive parquet anymore supplied by the acceleration. That&#x27;s the key here. Is that the blocks on contact one another. So one last requirement we have is that the two blocks air moving at the same accelerations. In other words, A is equal to a beam, so it&#x27;s just going to find that to be the same acceleration on then, sir, it&#x27;s some juice over here, Something new since second law looks like for block A, what does it look like for Block B? While the only force that&#x27;s being exerted is f of am being simply going to be equal to the mass of block being multiplied by that acceleration that we define, which is just lower case, eh? Was no, um, subscript So now it can combine. These two equations wants Do that over here. Basically, the two a&#x27;s are equal to one another. So let&#x27;s go ahead and highlight those stuff. We do some rearranging. Easiest way to do that would be to rearrange this expression on the right here and say that acceleration is equal. Tio half of a being over and be wait minus B a It&#x27;s then equal to I am a being multiplied by, uh, being for M B. What&#x27;s going on? Half of be a to both sides. That&#x27;s just from the right hand over here. Um and then we know that magnitude wise, the those equal opposite reaction forces are equal team one another. So in other words, F A B is equal to f b s. I can just go ahead and make that substitution over here on rearranged the mass is a little bit so that it makes more sense and a over and be on the inside here, um, so we can factor out these forces because they both have a doctor of f B A in front. So in other words, it&#x27;s for one. Plus I am a over B Andi. It&#x27;s actually basically with the end of the analytical solution. Looks like so go ahead and plug in numbers. There&#x27;s a two hundred fifty on the outside here on the inside. We have a one plus. Oh, sorry, sorry, sorry. We are self. The wrong thing we&#x27;re solving for FB. Aye, So there&#x27;s one last little transcription that we have to do. So I&#x27;m going to erase this. Come on, then we&#x27;re solving for FDA. So in order to move the masses to the other side, let&#x27;s just do a little switch roof factoring. So in other words, inside becomes f b f s I m being plus m a all over and be And again, this entire rights I looks like that&#x27;s now it&#x27;s the flip this equation in order to solve for B A communes they were just left with could be over and B plus and may Okay, so, running in a room, let&#x27;s go ahead and out of his life. So keep that in mind. We&#x27;re now going. Teo, plug in numbers, and this is our two hundred fifty. Newton&#x27;s goes on the inside. We have a five zero over five point zero plus twenty point zero. In other words, our final answer. We&#x27;LL have to significant figures, but numerically comes out to fifty, so it looks like it has won six thick, but, you know, actually has two. She put a big bucks from there, and now we&#x27;re don&#x27;ts probably only has one part.

Kathleen Tatem · Answer

Aloha. So for this problem, we are clones of Block that is caressing to forces at one and a two and moving in the ex direction told the Mass and by you off the giant it of F one. And it&#x27;s in the positive X direction. But we don&#x27;t know the value of F to That&#x27;s what we want to find and what we&#x27;re given for. Parts A through E is the value of the blocks acceleration. So you notice all this problem We use Newton&#x27;s second law, which is that the net force will equal to the mass times acceleration. The net force is the vector Some of these forces have long enough to So do you solve for F two, which is what we want to find. You can rearrange this equation. So the cap on May minus F one and this is really all happening in the ex direction so we can write this. Um, there&#x27;s m a X. That was a fun. It&#x27;s all in the extraction. So you did use the bank use of these forces. So then for part A, we&#x27;re told that exes 10 meters per second squared. So I have to be tika grams 10 meters per second squared minus 20 Nunes. And this of course kilograms times, meter per second stories and even says the 20 millions minus 20 Nunes or zero evens. So I have to zero in part A for part B, we have exes 20 meters per second squared. It&#x27;s actual be to kill grounds. Could be 40 Nunes minus 20 meetings. So that&#x27;s 20 new jeans, Um, as a vector. Also, the ex direction hurt, See x zero. So that means that we only have, um one term here Two is minus f lan which is minus 20. Mutants party The axes negative minus 10 meters per second squared. So have to be two kilograms times minus 10 years per second. Squared minus 29 years. So well, minus 20 minutes minus 20 evenings. So that&#x27;s minus 40. Newton&#x27;s work to you at the party X is minus 20 use for second squared. So just plugging in those numbers again, you should have minus 40. Nunes minus one is 20 times so minus 16 evens on the extraction

Alexander Allen · Answer

Okay, question. 40 to a 50 kilogramme box rests on the floor. The coefficients of static and kinetic friction between the bottom of the box and the floor, or nor 0.70 and no 0.50 respectively. What is the minimum force a person needs to exert on the box to start it sliding? Okay, this is I have the question. So we&#x27;re just going to make a few notes here. U S A. Is equal to 0.7. And, um, you okay? It&#x27;s going t equals in North Point fires. And so, at the limit, we&#x27;re going to have to take the idea that static friction is overcome intense. The force of friction when static friction has ever come is equal to the kerf. Isn&#x27;t static friction multiplied by the normal reaction force? Friction will resist motion on gwil effectively resist against any force that is less than the critical force required. A new But in this case, we&#x27;re at the limits. So we&#x27;re going to push at that point. And hence we just need need to multiply the coefficient of friction by the fact that we&#x27;re pushing on. Let me just remind you it&#x27;s a 50 kilogram books. So it&#x27;s where become that we&#x27;re going to have to take no 0.7 more to plot by 50 multiplied by 9.8, which is if we put it into a calculator equal to 343 Newtons to this part of the question, that&#x27;s the force which we need to use to start moving the box. So after the box start siding, the person continues to push it, exiting the same force. What is the acceleration of the box? Well, at this time, in this case, you are pushing with a force off 343 Newtons. And so your net force is your force that you&#x27;re pushing. Subtract off the force of friction on this is equal to M A. So the net force is equal to him. Hey! And therefore you take the force that you&#x27;re pushing with 343. These attract off the force of kinetic friction. No 0.5 multiplied by 50 most blood by 9.8. So this is the force that&#x27;s resisting. And finally this is going to equal m A. Which is your 50 multiplied by your A. It doesn&#x27;t take a genius to then say Let&#x27;s divide the whole left hand side by 50 toe have our acceleration and therefore our acceleration putting this into a calculator, remember? 343 subtract off half 50 multiplied by 9.8, dividing the whole thing by 50 we get an acceleration of 1.96 meters per second squared. Okay, that&#x27;s the acceleration of the box here on. That&#x27;s because once we&#x27;ve overcome static friction, kinetic friction takes over on its coefficient is usually less. Which means that the force that we&#x27;re now ever coming is serving to be greater than their frictional force. That is resisting our motion. Hence, the net force is not zero. We do have an acceleration, and that&#x27;s question 42.

Kristela Garcia · Answer

and this problem, you are given a force distance graph, and it tells you that one student applies a constant force of 30. Newton&#x27;s the entire time, and the other student starts off pushing at 40 Newton&#x27;s but then grows tired and eventually reduces its force overtime so we wouldn&#x27;t calculate or determine when the work done by each of the students is the same. So from a position are forced position graph. The area under a forest distance graph is equal to the work done. So if I look at my graph, I can visually see that this little segment fits in to this little segment. So if our to imagine this as having kind of flipped and scooted into that little space, the student who was pushing at 40 Newton&#x27;s has done this amount off work. They can actually calculate that value, but it actually will be necessary. So the student who was pushing at 40 has done this green amount of work where, as a student who was pushing 30 because he was pushing that 30 the entire time, he&#x27;s also done this same amount of work, so at position for the area&#x27;s under the graph are the same for both students. So the work done is the same for both students at position for

Charles Carter · Answer

Okay, So for this problem were given, um, a system of boxes. So we have box one in box two. We know that if we push on box, one of these could be different sizes. I don&#x27;t know. With a force of 12 Newtons, this system will accelerate at a rate of 1.51 point seven meters per second squared, so I can dio right away. I can see that this some of my masses m blanc plus into is just gonna be 12. About about 1.7 eso The some of our mass is gonna be around seven points 06 kilograms. Okay, so and we&#x27;re told that if we&#x27;re pushing on a box one, uh, then the leftover force it has for box to is going to be 8.5 Newtons. So, um, the box one kind of absorbs 3.5 Newtons and box two takes on 8.5 Newtons. And if I were to push it from the other direction, you know Box two takes on 8.5 and box one will take on 3.5 acceleration. Still same. So if I want to find my masses, I just I divide, like for box. Uh, make sure had it lined up, right? Yeah. Okay. So box one that we would do 3.5 divided, about 1.7. Okay, So 3.5. About about 1.7 to be 2.6 kilograms. You can amass two. We can either do the same process or just subtract that from 7.6 kilograms and we&#x27;ll get 5.0 kilograms. Thank you. No, that&#x27;s all I wanted. Okay, so that&#x27;s part a and part B. Thank you very much.

Problem

The purpose of Paragraph $4,$ as it relates to th…

Problem 13 Medium Difficulty

Answer

Related Courses

Cracking The AP Physics 1 Exam

Related Topics

Discussion

Top Educators

Recommended Videos

Watch More Solved Questions in Chapter

Video Transcript

Princeton Review

Cracking The AP Physics 1 Exam

What do you need help with?

Textbooks

Ask a question

Courses

AI Tutor

Cracking The AP Physics 1 Exam

Princeton ReviewCracking The AP Physics 1 Exam

Princeton Review

Cracking The AP Physics 1 Exam