5

Block (1 kg) is located the surface of table hand pushes horizontally the right on block with norma force 0f N: The coefficient of kinetic friction between the bloc...

Question

Block (1 kg) is located the surface of table hand pushes horizontally the right on block with norma force 0f N: The coefficient of kinetic friction between the block and the surface equals sheet of paper; draw the free body diagram for block using tne two-subscript notz Ition from class_ After comp eting the free body diagram entcr below cach force and its components_ Remember that the x-component the component and the component is the component:FORCES on BLOCK 1 Weight force on block by Earth W

Block (1 kg) is located the surface of table hand pushes horizontally the right on block with norma force 0f N: The coefficient of kinetic friction between the block and the surface equals sheet of paper; draw the free body diagram for block using tne two-subscript notz Ition from class_ After comp eting the free body diagram entcr below cach force and its components_ Remember that the x-component the component and the component is the component: FORCES on BLOCK 1 Weight force on block by Earth WIE Normal force on block by Surface N1s Normal force NIh block by Hand Frictional force on block by Surface f1S SubmnitArs"cr Tries 0/3 What is the acceleration of block 1? m/s? Submit Anser Tries 0/3



Answers

A block of mass $5.00 \mathrm{~kg}$ sits on top of a second block of mass $15.0 \mathrm{~kg}$, which in turn sits on a horizontal table. The coefficients of friction between the two blocks are $\mu_{s}=0.300$ and $\mu_{k}=0.100$. The coefficients of friction between the lower block and the rough table are $\mu_{s}=$ $0.500$ and $\mu_{k}=0.400$. You apply a constant horizontal force to the lower block, just large enough to make this block start sliding out from between the upper block and the table. (a) Draw a free-body diagram of each block, naming the forces acting on each. (b) Determine the magnitude of each force on each block at the instant when you have started pushing but motion has not yet started. (c) Determine the acceleration you measure for each block.

Okay, so we've got a block draw it here, I'm going to call it M. One and it's going to sit on another block M. Two which sits on a table we've got M. One is five kg. M two is 15 kilograms. Um And then you between the blocks static, is this mu between the blocks kinetic? Is this um Actually that first one should have been a three? Um You with the table static. Is this mu table kinetic? Is this a free body diagram of each block? Naming the forces? Um There's going to be, oh I forgot the force F. Here. Okay. I. M. One. There's going to be the weight gravitational force, there's going to be the normal force and then there's going to be a frictional force that's going to caused that block to move in this direction. Okay. Now in block number two there's going to be uh gravitational force. There's also going to be uh huh. F. G. One from above. There is also going to be a normal force, there's also going to be a frictional force from above and then there's gonna be a frictional force with the table. Okay now I did notice that the answers are in the back of the book, so I see a normal force on one, friction on one and wait on one. Very good 12 I see friction to frictions. Oh I forgot to put the main force, the pushing force, yep. Okay, I've got all of them labeled now. Um Next part of the question is just before it moves at the instant when you've started pushing but we haven't quite started to move yet. So in B F G one is just going to be M one times G F G two is going to be M two times G. So F N one is going to be F. G. One because they oppose each other. F N two is going to be F. G. Two plus F. G. One. She too. Okay, so let's make sure that I get those, putting this in a calculator, M one is five. M. Two is 15. You B. S. Is three. New B. K. Is 0.1. That should have been 0.3. Yeah. Okay. New T. S. Is 0.5. You T K. Is 0.4. Um G is 9.81 Okay, now let's get some of these. M. One G. So F do you want is M. One G. That's 49.1. F. G two is M two G. 1 47. Fn two equals F. G. One. Yes. F G two 1 96. So let's make sure that I got these right. 49.11 47. 1 96. 1 96. 49.11 47. Okay, very good. I've got the vertical ones correct Now, horizontal ones. F. F B looks easy. Is just going to be force of friction. Yeah, in the B direction. Hm. Well that's going to equal M one times a one. Let me look at the question again. If no motion has started, if M one is not accelerating at all, then FFB would have to be zero because it equals M. One times acceleration F F t would have to oppose F. So F F t minus F would have to be zero. So F would be F F. T. Which would be um U sub s mm times times the normal force on two, but the normal force on two is fN two. F. N two. Okay. Um And this would be F with the table S or mute table S. So F is going to equal mu Abel S Times F. And two. Just 98.1. So this is part B. And in part B. I do see the 98.1. That's good. But it's showing me that there is a friction force above which is 14.7. So it's showing a friction force here here. Why isn't it drawing here of 14.7. So if nothing's moving if the lower block has not even moved. Well let's just see mu sub S. Musa bs times the normal force times F. And one. So I guess that still makes sense. So musa bs times F. N. One. Where am I? So F F B equals you? B. S. Mhm. Times F. N. One. Okay but didn't I get FN one already? Oh F N one equals F. G. One. So I didn't declare that yet, there's the 14.7. Okay, so that makes sense that even when it's not moving, there's still some friction there. Okay. See with the acceleration for each block, what's the acceleration for each block? And why? And be does is there an answer of 113 Newton's in the back of the book? Let's look at B again. Yeah, I don't know why in part B there's an answer of 113 Newton's that seems, I think it's not answering anything. So let's go to uh see so now everything is moving. So um if everything is moving then uh F the some of the forces on yeah. Block one, which is F F. B. Has to equal mass times acceleration. So that FFB would have to equal M. One A. One. So A is gonna be F F. B over round one. See what that kiss me A one equals F F B over M one. Okay, but when everything is moving then this would have to change two kinetic friction. So F F B would now be um you B k f N one, New b k F and one which is the same as F G one to That would be correct for the motion of B because now it's kinetic. Okay, so acceleration of block one is 0.981 meters per second squared. New kinetic um over times F N one for M one. Okay. 0.981 meters per second squared. Okay, so now I need to get the acceleration of M two. So that would be f minus F F b minus F f T f minus F F B. Yeah, T is going to equal M two times A. This is a one. This is a two, so a two is going to be f minus f f B minus f F t over em too. Okay, and everything is kinetic, so just to make sure that I get this right, I'm gonna basically redo the calculations um F I do have minus F F B I have right above it, which is gonna be new. BK I'm F G one Okay, times F F T. But F F T was mu t k times FN two. Mhm. Okay. No, I want to divide by M two. What are the chances that it would be the exact same thing? Yeah. Yeah. Okay and I was thinking about it and this does make sense but the acceleration one is going to be relative to acceleration too, so a one is really going to be this plus a to all of this, erase, oops, trying to erase probably an easier way to do this. Um Plus a two plus eight to because it's going to be relative to +82 And so that's going to give us the 0.981 here. But it's gonna give us going back to the calculator here. Plus a to whoops two 1.96 here. Mhm. All right, so thank you for watching.

Or this question. We have to use Newton's second law. So let me. I loved my reference frame as these one. This is my Y axis, and this is my sex is now. Let us apply Newton's second law to the case. Eight for the block number one. So we are engaged A and then for block number one x axis. We have the following the net force acting in block number one in the X direction. Is it close to the mask off that block times its acceleration? The net force on the X axis That's acting block number one is composed by only one force, which is this contact force exerted by block number two on block number one. Then the contact forced is because of the mass off the block number one times its acceleration, which is that No, no, we can apply neutral second law on the Y axis. By doing that, we get the following the Net forces acting on the vertical direction. Is it close to the mass off block number? World Times acceleration on the vertical direction Since it's not moving and we will not move from the vertical direction, its acceleration is close to zero, then normal number one minus wait. Number one is there close to zero. Meaning that normal number one is the cost of the weight. Number one. Now we do the same for block number two. For a block number two on the X axis, we get the net. Force acting is equal to the mass off block number. Two times he's acceleration, which is equal to the acceleration off the block number one because they're moving together, Then the net force is composed No, by true forces, the Contact Force and the Frictional force. So will you get contact force that points in the positive direction minus fictional force that points the negative direction is because of the mask off block number two times its acceleration before moving further. That I correct something that I have forgot this week. This force isn't positive because there is the contact force that Block Number two exerts on block number one and it's points to the left. So in our reference frame, it is a negative force. So miners FC is equals to m one times acceleration in the X direction now continuing for a block for block number two why direction we get the following the net force acting on that block in that direction. Is it close to the mass off the block times its acceleration on that direction, it's It will not move in that direction. It's not moving, so acceleration is it goes to zero. There are only two forces acting on the vertical direction, which are the normal force, pointing to the positive direction on the weight force point in the negative direction. Then the normal number choo is it goes to the weight number two. No, By doing that, we got four equations, one shoe tree, four off course. We have to discover what is the Contact force. And then what is the acceleration? So we must work with these two equations that includes both of these factors. Let me call this equation one. And these older equation, too. If we add equation number one and the question of virtue, we get minus F C plus F C minus F F Zico's true M one plus and true times a X before the contact forces are. Simplify it and we end up with minus F f is equals to M one plus M two times a x them the acceleration That's direction is given by minus frictional force divided by M one plus and truth. This equation will be very user food during this question. So I put it up there on the board. Now that we know the acceleration, we can go back either to equation number one or equation number two and calculate what is the contact force. It's much simpler if we use equation number one. Then you'll be using this equation. So apart into equation number one, the Contact Force is given by and one we from minus sign Times aux. But the X he's given by minus f f divided by m one plus m too. So we got em one times f f divided by m one plus and truth. So these are the equations for the Contact Force and for the acceleration on both cases. So now we just have to use these equations and calculate for Earth items A and B Let me clear the board and finish the question. The contact forced in situation A is given by SI is equal to M one times 5.8, divided by m one plus and true So this is six. Then the contact force is. It goes to 5.8 divided by True because it can simplify. Here on Deasy's 2.9 Newtons, these is the force with which one block pushes against the order for the second situation. Contact Force is given by and one which now is he goes to six times 5.8 reaches the frictional force divided by and one place and true, which is a question nine. We can again divide by treat so that we have three and true, then 11.6 divided by three. And finally this results in approximately 3.9 Newtons. Now, for the second item, we went to complete the acceleration. So the first situation acceleration he's given by minus 5.8, provided by three plus tree, which is six. These results in approximately minus 0.97 meters per second squared on situation be. The acceleration is given by minus 5.8, divided by nine and this is approximately minus 0.624 meters per second squared. And then he finished the question. Acceleration is negative because the blocks are being is going down until they stop

In this question we have given that up for. Yeah. Oh that are two blocks that is placed at one another. And we need to find that we need to find the different uh scenarios from the given. Okay so now we have even two blocks right? So basically one each Uh one is bigger so that we draw it. This is the first block, This is the second floor right? This is block V. And A P. And Okay. And we are pulling with the full support of our okay in the first scenario draw. And we have given that The top of the block and you place the longer block to start, it's moving after about you don't .25 seconds. It is moving in a constant stream and the block on the top it is not to draw all level free body diagram for the two blocks during at the time when they are accelerating space high all the forces basically. So the free body diagram for this to do nothing. But in the first part For the blockade there will be MG four and over here one more force because we eat also at the uh so that means the M G M G M A G plus um to NBG when we actually because they are also facing the same thing and no one will be these two guys, you know what I mean? Just wants another religious ones like this one. Now if there is a fictional forces that well if Pablo is moving in this direction, so there will be a flicks and fourth due to A on B and in this direction right? And for the block week there will be and M V G downward the blockade. And obviously there will be 1/4 over here also due to the block with the MBG and normal as well. So for her MBG and normal as well. And now there will be a force since the christian forces good in this direction. Selection for this will be in this direction. Right. And this included in the diagram and then they are moving with the constant velocity in the second scenario draw the level politics graham during the time when they are moving in the question the space when they are constantly with that means That means that lesson is zero. Let's listen to zero. That means there is no full network. So that wins. And and I was there when we fix an age well over here they will be F over here and over here in this recommend clicks and quotes in this mechanism and in this scenario if this is a and this is me. So F over here that will fix and force that will be equal and this will be a total and one M plus M. V. G. And over here there will be a real normal because they are in the question, you know you experience and okay. And obviously there will be some contact contact for over here that will that will fit in in this direction and normal instrument is to believe landlord because there is no explanation. And the next part well support the bottom block have mass 0.4. Right? And the coefficient of friction between the block and the table is 0.3. In the top block has 0.1. Mask The contradiction between the globe is now what force you need to start to keep the block moving at a constant speed of 20 points again. And so basically okay for that we can use that phone up that Total mass of the block is nothing but the bigger blog we have given 0.4 and 0.1 I guess 0.4 and 0.1. So net mass will be nothing but 0.5. So that for several years that will be 0.5. So normal will be nothing. But I'm into the 0.5-10. It was 500. The frictional force will do nothing. But we have given sexual condition that they blocked on the table as you .3 six and 45 in the country. That means at one point have you done that? And at the same time you see the frictional force. So what is the middle force to require to move at the speed of the second question is which? And for the second block, uh not all four for the second along with the 0.1 into 10. So one Newton and of course the 600 in them is given. And the difficult to That is equals two. Yeah, you don't want to so you want to so, uh, the the that we do not want to introduce you to a point to new towns. So now we need to find a course of what we do, the immune forces. So this is the friction force. So they should be doing all right. And this one was unforgivable. And that is nothing but Mars Excellent. And basically minimum force. That thing. What are we talking about this month? So that maybe 1.5 million. Okay, Thank you.

In the party of this problem we have to describe in words the motion of the block as it moves from it. To be The blog experiences are not constant force of five m from a. to b. So it will have a constant exploration during this journey between the two points. But as the block interest in the next section of the path, it experiences a frictional force which they accelerates it and the bloke comes to rest it. See okay, in part B of this problem, we have to drive free body diagram where the block is at point P. So this time we need to draw three body diagram when the when the block is that point P. At point B. There are full for suspecting on this block, It's what is acting downward which is equals two MG. We have normal force on it which is upward. five neutral forces acting towards white direction and the frictional forces acting towards the left direction, Which is equals 2 μ KMG. Now, in the party of this problem, we have to find a direction of the exploration of the block at sea. So this time we need to write the direction of the exploration at sea. That's right according to calculate the expiration. So here we can white it minus milk MG plus five new john Is equals 2 3 Kg into. Hey, in this case this first term, which is equals two minus milk. Into here, we have said three kg Until nine pointed 1 metaphors can square Plus 5 to 10 is equals 23 kg into a. Yeah, here it seems that this time will be greater than this one in magnitude. So the exploration will be negative. I don't know. So the direction of this exploration will be towards the left side. Let's move to the party of this problem. Yeah, From B to C, the net force appeared on the block is constant. So there will be constant exploration in the block in the party of this problem, we have to find network and on the block as it moves from A to B, B to C and uh and B to C. So we need to calculate rigged in when the blog moves from A to B, which is W A B and Brendan, corresponding the point B to C. So let's write like waiting for work done, which is a W Ab. So W abs equals two F dot G. And this is equal to f typical this to exercise in the same direction And this is equal to five new turn into one, divide by two m. So here we will get $2.5 mm. Similarly, we can wait for W B two C as a W B to C is equal to have much dot de. So this is equal to F. D minus uh um You care M G. D. In this case this force is opposite to the displacement. So we have to put a negative sign here. So W B C is equal to five new chan into one x 2 m minus milk. Into three kg Into 9.8. 1 metal parts can square. So here we have at Into one, divide by two m. So here we will get WBC's equals two 2.5 minus 14.7 milk. And to reach out. So this is a WBC. Okay now let's move to the part F of this problem. In this case we have to calculate the corruption to friction which is denoted by muche. Let's kill with the acceleration from B to C. For this case, calculate the speed at B. So from it to be we can white as as equals to have divided by M. And this is equals 25 new 10, divide by tricky. So we will get here 167 major person can square. This is the exploration of the block from A to B. All the speed at B is written as we B squared minus. We are squared is equals to two A to B and D. So here we can. Why did this will be square minus zero m. Person can square first. Can Whole square is equal to two into 1.67. 7 metal parts can square into half meter. So half of the media From here we can wipe the value for this will be as we B is equal to 1.67 major personality. Our skill cleared the exploration of the block as it moves from B to C. For this case we can ride it with his square minus will be square is equals two true B. C. This is the exploration when the block is moving from B to C. And here we have so we can write you that this B C. Is equal to. We see square minus will be square divide by two D. So this originals zero m per scan. All squared minus into 1.67 m/s. Hold square, Divide by two into 1, divide by two m. So here we will get the value for this. Abc. As ABC. Z equals 2 -2.79 metaphors can square. We can white. That F minus Milky MG is equals to em A B C. Let's set the values into this question. So here we have five new 10 minus mu care which is unknown into the mess, which is equal to three kg in 29 point at one m per second squared, is equals two. Three kg into Every switches equals 2 -2.7 nine m can square Formula. We will get the value for this music as μ KZ equals two 0.45. This is the required answer and of the problem. Thank you.


Similar Solved Questions

5 answers
"7ls &18 CLEL functicm / : B" E i Ajpoef? assoecosprisdrtd abltevisted @sc} TMnc #rif %33 MBre 012i ETuL? 7[zab tlw; I-% 2 0 {w} $ fc) + # AE aakjua dlafniicn % nole ix Jone: deasetosprisssics fkv} Ib j:R" _R be giv2n, Frvnte tbet ic o:rintyz &; #Al ~k i_ Dol 3c931 Isc &
"7ls &18 CLEL functicm / : B" E i Ajpoef? assoecosprisdrtd abltevisted @sc} TMnc #rif %33 MBre 012i ETuL? 7[zab tlw; I-% 2 0 {w} $ fc) + # AE aakjua dlafniicn % nole ix Jone: deasetosprisssics fkv} Ib j:R" _R be giv2n, Frvnte tbet ic o:rintyz &; #Al ~k i_ Dol 3c931 Isc &...
5 answers
New element which consists of two isotopes_ The first isotope 2. You discover The second isotope 248X 241X (mass 242.45 u) comprises 40.00% of the total: What - would be the average atomic mass (mass 247.11 u) accounts for the rest for your new element?
new element which consists of two isotopes_ The first isotope 2. You discover The second isotope 248X 241X (mass 242.45 u) comprises 40.00% of the total: What - would be the average atomic mass (mass 247.11 u) accounts for the rest for your new element?...
5 answers
Copper to copper nitrateCopper nitrate to copper hydroxide
Copper to copper nitrate Copper nitrate to copper hydroxide...
4 answers
Four point charges are placed at tle fOur cOmers of square that is }.0 GI cach side. Find the electric potential at the center of the square if the four charges are cach ? LC a) k7 MV 6) ;+MV c) 6.8 MV d) /+MV 2) 0
Four point charges are placed at tle fOur cOmers of square that is }.0 GI cach side. Find the electric potential at the center of the square if the four charges are cach ? LC a) k7 MV 6) ;+MV c) 6.8 MV d) /+MV 2) 0...
5 answers
Point chatge hxed onthcyaxis: 1 mcgative point charge = 28 NC a Y1 third point chargc q +019mand a positive point charge G2atY2 - +0.32m '9,9uCis fxed at the Origin Thenet electrostatic force exerted on the charge qby the other twochargeshas. magnitude of 30 Nand points direction Delermine the magnitude of 92NumberUnits
point chatge hxed onthcyaxis: 1 mcgative point charge = 28 NC a Y1 third point chargc q +019mand a positive point charge G2atY2 - +0.32m '9,9uCis fxed at the Origin Thenet electrostatic force exerted on the charge qby the other twochargeshas. magnitude of 30 Nand points direction Delermine the ...
5 answers
Evaluate. Use a capital C for any constant:6x2€ 'e2r8 dx e2x3 3
Evaluate. Use a capital C for any constant: 6x2€ 'e2r8 dx e2x3 3...
5 answers
ProblemPREVIEW ONLY ANSWERS NOT RECORDED(10 points) Find Ine sum 0f the first 88 terms of the arithmetic sequence;3,FI4-19,-27-35.Answer;
Problem PREVIEW ONLY ANSWERS NOT RECORDED (10 points) Find Ine sum 0f the first 88 terms of the arithmetic sequence; 3,FI4-19,-27-35. Answer;...
5 answers
In Exercises 15-18,find a basis for the space spanned by the given vectors , V1 V5 0-3 242-3 ~8 ' -6 71t:,d.| ' 9V5.-3 22 ~36
In Exercises 15-18,find a basis for the space spanned by the given vectors , V1 V5 0 -3 24 2 -3 ~8 ' -6 71t:,d.| ' 9 V5. -3 2 2 ~3 6...
5 answers
[V] The figure below shows snapshots of a traveling wave at t-0 gad F] &5U >Howimany of the following functions could represcnt the traveling wave? Select all that apply DEA sin[kox + Ct)] B D = Acos[k(x + Ct)] . D =A cos[kix + Ct) Do] D D =A sin[k(x Ct]l
[V] The figure below shows snapshots of a traveling wave at t-0 gad F] & 5U > Howimany of the following functions could represcnt the traveling wave? Select all that apply DEA sin[kox + Ct)] B D = Acos[k(x + Ct)] . D =A cos[kix + Ct) Do] D D =A sin[k(x Ct]l...
5 answers
2)A particle has a rest mass of 6.95×10^−27 kg and a momentum of6.41×10−18 kg·m/s.Determine the total relativistic energy E of the particle.𝐸=_____JJFind the ratio of the particle's relativistic kinetic energy 𝐾to its rest energy Erest.𝐾/𝐸rest=_______
2) A particle has a rest mass of 6.95×10^−27 kg and a momentum of 6.41×10−18 kg·m/s. Determine the total relativistic energy E of the particle. 𝐸=_____JJ Find the ratio of the particle's relativistic kinetic energy 𝐾 to its rest energy Erest. 𝐾/ð...
5 answers
Sample? notfthe chipsA cards cendd lot contains Printed are areed defective defective 88 orcards the the are lot, and 3 what is what is are placed in the the selected Hi| test that that afte any two the L being cards sample the populated are first defective with 3 are the semiconductor defective but
sample? notfthe chipsA cards cendd lot contains Printed are areed defective defective 88 orcards the the are lot, and 3 what is what is are placed in the the selected Hi| test that that afte any two the L being cards sample the populated are first defective with 3 are the semiconductor defective but...

-- 0.065369--