questions-1-2-and-3-are-short-free-response-questions-that-require-about-13-minutes-to-answer-and--2

Question

Questions 1, 2, and 3 are short free-response questions that require about 13 minutes to answer and are worth 8 points. Questions 4 and 5 are long free-response questions that require about 25 minutes each to answer and are worth 13 points each. Show your work for each part in the space provided after that part.

A small box of mass $m$ is placed on an inclined plane with an angle of incline of $	heta$ . There is a coefficient of kinetic friction $\mu_{\mathrm{k} 	ext { between the inclined plane and }}$ the small box. The small box is attached to a much heavier box of mass 3$m$ by a pulley system shown below.
(a) Draw free body diagrams of both masses, including all of the forces acting on each.
(b) Assuming a frictionless, massless pulley, determine the acceleration of the blocks once they are released from rest in terms of $\mu_{k} g,$ and $	heta$
(c) If $\mu_{k}=0.3$ and $	heta=45^{\circ},$ what distance is traveled by the blocks 3 $\mathrm{s}$
after being released from rest?

Vishal Gupta · Accepted Answer

Hi in the given problem theories an inclined plane, The rough inclined plane which is making an angle of tita waved. The horizontal a block of mass M. Has been kept over it which is tied with the help of a thread. The threat is passing over a frictionless. Massless fully and another block of masks. three M. is tied with the help of the threat. Now we have to plot, we have to draw the free body diagrams of both the blocks. So first of all the weight of this block which is kept over the inclined plane will be acting vertically downward like this. You have this is M. G. Its components, there will be two components of this weight if this angle is he, that this angle will also be theater. So the component of the weight perpendicular to the inclined plane will be MG. Cause theta and its component along the inclined plane will be MG sine Tito. Normal normal reaction applied by the inclined plane to this block will be N is equal to MG cost heat up this three and the block of Mastery um will be having a tendency to Slide to move downward under the influence of its weight. three MG. Due to which this block kept on the inclined plane, inclined plane will be having a tendency to move up so our force of friction will be acting on it in a downward direction. There will be a tension in that thread which will remain same in both of its segments, attention teeth. So here these are the free body diagrams of both the bloods like this, which is the answer for the first part of the problem. So free body diagrams off both the blocks are shown in the adjoining figure. And here we will have to show the acceleration also as both the blocks are connected with each other, so the acceleration a will be same in both of these blocks. Now, in the second part of the problem, we have to find that acceleration for which was to fall. We will use free body diagram of the block of mass three M. As this block is under the influence of two forces three M. G and T. And as it is moving down, so three MG should be more. So the net force acting on it will be three and G minus T. And using Newton&#x27;s second law of motion, this net force should be created by the product of mass with the acceleration. So here it becomes first equation. Now we will use free body diagram of the block kept on the inclined plane which is having a mass of him. And as this block is also under the influence of two forces B upward along the inclined plane and MG. Sine theta And force of friction. So this is under the influence of three forces MG santa and air force of friction downward. A long time climb plane. But as the block is moving up so attention T. Should be more. So the net forces acting on it will be T minus the sum of mg. Scientific to with the force of friction, and that should be equal to M. Into a Again expert Newton&#x27;s 2nd law of motion. No, as for so friction is given us the coefficient of friction, mu k times the normal reaction and normal reaction here is MG cost theta. So finally this equation comes out to be T minus mg sine tita minus mu UK MG. Because tita is equal to me. That is the question number two. And if we right here, equation one again, that was three M. G minus T is equal to three M. A. And now if we add these two equations, the tension will be cancelled, leaving behind three mg minus mg. Sine tita minus mu k mg because tita is equal to me. So the acceleration here will come out to be canceling this empty from all the terms. The expression for acceleration is 3G -G sign tita. And here, also, as we have added decorations, it will come out to be for me. So this is four M. E. So here this is 3G -G scientist to minus mu K G because tita divided by four. And here it is the answer for the second part of the problem expression for the acceleration of the system. Now, finally, in the third part of the problem, Coefficient of friction has been given a 0.3 Angle Theta has been given us 45° in time. T because 23 seconds we have to find the distance traveled by the object of mastery. M given that it has started from rest. So initially speed, you will be taken as zero so to find it. First of all, you will find acceleration of the system. Using the answer we found in the second part of the problem. So that is three G means G can be taken here as a common out. So this is three minus sine. Theta means signed 45 degree minus mu game, which is 0.3 G. Has been taken as a comin out Cost histamines cause 45°.. Here this is G means 9.8 m per second squared, divided by four. So finally this acceleration comes out to be 5.2 m per second squared. Hence using the second equation of motion as is equal to U. T Plus have 80 square that it stands traveled by. The object will be zero plus half into 5.2 m per 70 square into three square. So finally this distance comes out to be 23.4 Meta, which is the answer for the 3rd part of the problem. Thank you.

Kristela Garcia · Answer

in this problem, we&#x27;re asked to draw a free body diagram and calculate the acceleration for two masses on an incline plane. So we&#x27;re gonna start with our diagram and these messes air connected over a frictionless pulley. And this has a massive three M, and this has a mess of em. And so for part A, we just want to draw the free body diagram. So we have normal force for particular tour surface mg straight down the ramp, straight down, we have a tension going up the ramp and a friction down the room before mess em. And then for mastery M, we have the attention going up and if three mg going down. And so for part B, when we&#x27;re asked us offering its acceleration, this is just gonna be in terms of variables, and we&#x27;re actually gonna treat this is the system. So we&#x27;re gonna take all the all of the forces that would affect its motion. So if you look at M G, it&#x27;s gonna have two components. It&#x27;s gonna have a component that&#x27;s adjacent and a component that is opposite. And these components MG signed data friction. These tensions and three mg are all going to affect the acceleration. So we&#x27;re gonna take this direction down for three. We&#x27;re gonna take that direction is positive. Sobering to say three mg minus T plus T minus F minus mg signed data. And this is Newton&#x27;s second law. The sum of the forces? No, we&#x27;re just treating it as a system. So I&#x27;m gonna write some of the forces on the system equals mass times, exhilaration, and our mass. Here is our total mass. So that&#x27;s three em plus m. And then the acceleration is going to be for the system. And here you can notice those those tensions cancel out, and friction can be replaced with the coefficient of friction times my normal force. Now, in this case, my normal force is going to be equal to m g co sign. So friction can be replaced with the coefficient of kinetic friction times, mg coast and data. So replacing all of my variables. With that, we have three m g minus mu m geico sank data minus mg sign fatum all divided by three m plus m, which is four em equals a And my accelerations here can actually counsel. I&#x27;m sorry, my masses here can actually counsel. And I&#x27;m love with three G minus mu. Geico signed Tatum minus She signed data all divided by four. And then part see asks Well, if we have a coefficient of kinetic friction given as 0.3 and an angle of 45 degrees, how far are we going to travel in three seconds? So for that we can assault for exoneration, given our values. So using 10 for gravity, I end up with three times 10 minus point three times 10 times. He co signed a 45 minus 10 times. He signed a 45 divided by four gives me an acceleration of 5.56 meters per second squared and then, using a cinematic equation, I can assume that the initial velocity was zero since it started from rest and then our distance traveled will be 5.56 times when half times our time squared of about 25 meters

Ryan Hood · Answer

in this problem. We have block one on an inclined plane of angle Alfa and tethered to block too, which is hanging over the edge. If we draw the free body diagram for block one, we have attention from rope to clean up the ramp. This way, our normal force is normal to the rams surface. So it&#x27;s this way the force of friction as opposing the motion. So if Block One is going up the ramp, then the force of her shins pushing this way and then we also have our force of gravity downward since the angle of our ramp is Alfa Window, the singles Alfa and that means by trigonometry this must be equal to mg sign Alfa and this I must be equal to the energy co sign. Time&#x27;s up, Costa. Now to solve part, eh? We need to notice that even when Block one is moving a constant glossy, its acceleration a zero and so its net force is zero. So we have the sum of the forces is equal to zero. And since it&#x27;s moving in this direction, we know that some of the forces in that direction is equal to zero. So we needed to his decompose all our forces into the ramp direction which I&#x27;ll just call the extraction. And then this could be the wind direction. And so we need the ex components of all our forces were going to someone up and sell me Go zero. That&#x27;s the condition for a constant velocity blocks lighting in this direction. So we know that the tension points in the positive direction. But the force of Persian points in the name of direction also the contribution from the normal forceful point in the next in the negative direction. Well, actually, script Sorry, just got confused. The normal force is in the white direction on Lee. It&#x27;s the ex component of the gravity force which will be considering. And so when we set this up, we will have This implies negative force of friction, employing the negative out of force of friction because I already know it must be negative. So when we calculate this, we have to remember not to make it negative again. And we&#x27;re going to subtract him one times gravity times sign of Alfa, which is the component here from from the gravity. And then we&#x27;re gonna add in the force of tension, which is purely the force here, which is into times G. That tension gets transmitted all along the road so tea is equal to mpg. That&#x27;s positive because it&#x27;s pointing the positive extraction This is equal zero because of some of the force is equal to zero. Now the force, due to friction when the block is moving, is equal to the coefficient of kinetic friction times the normal force. In this case, the normal force is equal to this component. So it&#x27;s mg ten percent off like that. And then whatever you add this in here and then sulfur him too. You get the phone or is old, which is into is equal to me. Okay, Em one Casanova lesson one I&#x27;m signing off. I also cancel the factor, Gene because each term shared and that&#x27;s the final answer for our day. For Part B, it is very similar, except in this case, the block is moving downward. And so the tension is going to be exactly the same. The contribution from the gravity is going to be exactly same. The only difference is that the contribution from the Friction force is going to be off sit inside. And so whatever we write this down, we can pretty much faro This result here except this term has to change sign because this is a term that came from the court position of Stack First, our kinetic friction running connect. And so we can say that I am too is equal to negative UK in one times Casanova plus M one sign of Alfa. Nor is this charmed and change sign. The only thing that changed was this and so that&#x27;s the final answer for party and Park Si The block is no longer moving. So after a place the coefficient of kinetic friction with the coefficient of static friction and we also need to find a range of values. And so, in this case, the range comes from block one sliding up versus a block one sliding down. Those two cases will produce the range of values we still need. The sum of the forces equal zero because the block is still not accelerating. And so because of that, we can borrow these results here, except we need to make a few replacements and so will see that the range is from em. Two is equal to negative me less, but pretty much borrowing this result, except on replacing the coefficient of kinetic friction with the coefficient of static friction. Everything else remains the same because we still have these contributions from those forces. So that&#x27;s the lower bound for into if into is less than that. Then the coefficient of static friction cannot, uh, stop. The block on the block has to move at that point. And then the higher value is this value here. Except we made make that change you. So him too. Pickles. Mu s. So I was in one coastline, Alfa less than one. I&#x27;m signing off, and that&#x27;s your range, and that&#x27;s you.

Dading Chen · Answer

Okay, let&#x27;s take a look at the $35 for each block. So they freed by government for the hanging block is something like this. Which means that the hang of hope or experience a graph to you from the Earth and the ground, these evil MSG images a massive hanging Bob, and you also experience attention. It was going upward in the direction of an air force on the hanging block. Should be going downward is because the hanging block is trying to move down. Okay. And the everybody that was full of block on inclined plane should be something that is, and the angle between the inclined plane and the grounds. 45 degree. Any block yourself experience of gravity from the Earth, which is, um a g m I am eyes the mass of a block from inclined plane, and it experienced attention that was, uh, moving upward. But along the direction of the inclined plane and only, uh, political direction, you each friends a normal force there was moving upward and which is perpendicular to the inclined plane. And lastly, you also experienced the friction. There was allow the inclined plane, but in opposite direction when compared with the direction of attention and in the air Force off the block on an incline plan should be moving upward, but a long incline play. Okay, so you&#x27;ve reconsidered. Okay, Upward direction on an incline plan is the deposit X axis and the direction that is perpendicular to the incline planes in upward direction is the Y axis. So therefore can decompose the gravity which you will have the s component of the gravity and the y component with gravity. OK, As you can tell, no more force is equal to the gravity times cause I&#x27;m 45 degree which is equal to oh, I m i g because I 45 degree. And we also know the friction in his case sixties kinetic friction. So it Ziegel Toby coefficient of kinetic friction passed a normal force, which is you go to meal times M I. G cause I 45 degree and an air force only vertical direction that was perpendicular to the incline plants is equal. Zero is because it blocked in the moving up or down. He only moved along the inclined plane. Okay, so therefore we can say that the net force equation for the block on in climb were just net. I is equal to the tension minus the friction. An M minus name. Ask component over the gravity. OK, which is going which is also going to the opposite direction when you compare with that direction of attention, which is, um, Jane sign for inviting me, which is you go to is the question we put the t minus mu n minus McGee. Sign for my degree. If we expend it will have t minus mule m i g because I 45 degree minus MRG sigh or if I&#x27;d agree. And we also know that the net force off the block on Inclined Plan is equal to the mass off the block on an inclined plane has the acceleration A and is or you go to t minus. Um, I gee, because I 45 degree, the I minus MRG sign 45 degrees. So it seems that we don&#x27;t know how much tension is. Remember, the block on incline plan was connected. Ah, with uh, hanging block to real strength. Okay. Which means that if we take a look at everybody that were funny hanging block, we can have the therefore see questionable hanging ball. Is there definite H music? Green pencil? I have no h. You see? You go to T minus t. OK, because it&#x27;s going Donald were okay, This is equal to Mohd finest he and then we&#x27;ll have attention. T is equal to MSG minus f in their age. Which is he going to MSG minus, um age eight. Okay, because then their force on hanging glass is equal to the mass of the hanging above has the acceleration and attention finally hanging Block is you go to the tension on a block off inclined play. Okay, is because they&#x27;re connected with each other and there in the sand. Sister. Therefore Okay. Along these inclusion here, back into here and we&#x27;ll have and my a equal to m h g minus. Um, my j Yeah. Minus mule. Am I G cause I 45 degree and then minus am I signed 45 degrees. Okay, I am convinced of the m e J two left side. No have. And my a plus I m a j is able to msg minus you. Am I Jane? Course I 45 degree minus. Am I sigh 45 degree. Okay, So now I can determine acceleration off this system, which is you can take a out. We&#x27;ll have eight times in my plus image is equal to MSG minus, um, I g cause I 45 degree minus. Am I signed for by degree, you know, have a which is the acceleration. So you go to MSG minus, Um, I oh, j cause I 45 degree minus. And why so everybody me divided by and my plus a mate And from the question we know the Mbai, which is, uh, mass of the block on an inclined plane is 2.0, kilograms. And the mess overhanging block image is able to three point. I&#x27;ll kill where a coefficient of a kinetic friction is given a 0.1 I and we know the acceleration of gravity, you see, go to 9.8 meter a second square. We know, cause I&#x27;m 45 degree and signed 45 degrees right now. So cause I or if I agree, is you go to So for if I&#x27;d agree, which is square to over to. Okay, So if you&#x27;re plugging these values back into the equation here we&#x27;ll have. Acceleration is equal to, um three point. Oh, you were kinds nine point a meter per second square minus Joe 0.1 I times 2.0 killer worms times Now I&#x27;m going eight meter per second square and end Times Square to over to minus to a bono kilograms times 9.8 meters per second square 10 times square to over to and it&#x27;s holding here, divided by 2.1 kilograms plus 3.1 kilograms and this will give us acceleration is equal to 2.6 meter But seconds were okay Simple question Be things when we know the acceleration of on a question A Now we can determine attention on us on a string OK, was his tea is equal to remember Fun question a worry Uh, this the equation pretension which is right here. Attention is evil toe msg minus Emma Jay Gray. That&#x27;s right down. So its able to msg minus amazing A Which can we go toe image G minus a. We know the mass overhanging block image is 3.1 kilograms 10 times G minus eight, which is the XRs on ground D minus the acceleration off the block, which is not going a university square minus 2.6 University Square. And this will give us attention is about 21. Only six new times. If we run into the tens digit, we&#x27;ll have 22 newshounds. I&#x27;m sorry. No intense digits. Integer here. So is about 22 new times. Okay, so this is the answer for this question. Thank you.

Surjit Tewari · Answer

question. What stoop? I&#x27;ve two blocks are with Aram like this. This is a poorly and one block. It&#x27;s hanging that are that broke is on that end this component off the weight forces and one G side alpha. This is normal. This company it&#x27;s and one day Forssell. This is M G. So now considered the system off two blocks, we can light the net force on that Two blocks is G minus. And when g sign up for minus new times and one G course and on this night, force is equal to zero because the system is moving with constant velocity. So X elation in the quarter zero So we can se i m is a little and one times sign up for my last new times. Well, so far. Similarly, for Part B, when the most time when it&#x27;s sliding downwards, then we can die And don&#x27;t is a photo. Um, one times sign up for minus new times, because so alpha, this is the value of them. To Vanda la Campbell, it&#x27;s sliding downwards, for the hard part of the question, were to find that any of values off them, too, since when the logs are addressed, then that in chalk will lose off. M two is is between and one sign up for minus new times cause self I and the lowest for Lewis and one times sign up for less New York Times course. And and one belongs. So this sense, this is the danger. And one I am so sorry. This is that NGOs and

Guilherme Barros · Answer

In this exercise, we have the system that shown here in the figure of the left that consists of two blocks of mass and one equals 2 kg and and to equal to 6 kg that are connected by, ah, string that passes through a pulley that has a mass of 10 kg and a radius of zero points. 25 m. The second block and two is on an inclined plane that makes an angle off 30 degrees with the horizontal. And both blocks are subject to friction. And because the friction coefficient new is equal to 0.36 and I go first of all is to draw the free body diagram for both blocks and the police. So for the first block in one, we have him the bucket one subject to the force of gravity that&#x27;s equal to him one g a normal force that I&#x27;m just gonna told man in. Then we have the tension floors pointing to the right, and we also have the frictional force which is just gonna call laugh, pointing to ah to the left notice that F is equal to the normal force times mu and the normal forces just in one g. So having one g mute, this is a frictional force for the second block. We have the gravitational force that can be decomposed into two components, one that is perpendicular to the surface and has a value, as we know from inclined plane studies that this is mg times the cool sign of theater. And this here&#x27;s sorry, this here&#x27;s data as the force Ah, parallel to the surface, Which is it which would end one g sign of data. We also have attention force actually, let me just rather normal force first. Then we have the tension floors t two. That&#x27;s attention that the string exerts over the block and we also have the original force f. And I&#x27;m just gonna call it f two and I&#x27;m called the first frictional force from the first look at floor. So f two is equal to the normal, which is in I Geico sine theta times mu. And finally, for the Cooley we have the gravitational force and a normal force g there. Oh, no one just gonna call in. And we also have the attention to you on that acts, uh, to the left on the Pooley and attention T two that acts to the rat every it acts downwards and to the right. Okay, so these are the free body diagrams for the three object and in question A. We have to determine the exploration off the to blocks. So let&#x27;s move on Depression A and we&#x27;re gonna use the these three body diagrams. So, first of all, I&#x27;m gonna write an equation for the acceleration of the first block. Notice that ah, horizontally that the block external accelerates horizontally and the horizontal force is equal to t minus F one. That&#x27;s the net horizontal force or have t minus F one is equal to m one a. Then actually it should be t one. This is the Yeah, you should have just put a one here that this is the attention of the string upon the first block. Then for the second block, we have that along prophetic. I&#x27;m sorry, Parlow. To the the surface that is parallel to the exploration we have that the total forces in one g times the call sign. Okay, the science data minus the official force F two minus teach you and this is able to him two times the acceleration a notice that the exploration of both blocks is the same. And for the Pooley, instead of writing a force equation for the center mass, I&#x27;m gonna write a torque equation. Notice that, um, the Force t two causes the the pulley to rotate clockwise. So the torque we were positive, so have to t two times are the force t one causes the fully to rotate counterclockwise. So it&#x27;s negative. Mind at the torque in negative minus t one are. And this is equal to the moment of inertia, of the holy times, the linear acceleration. So we have these three leaves. Okay, I&#x27;m gonna write them a little differently because I&#x27;m gonna, uh, substitute everything that I can as explicitly as I can to have tea one minus f one. And they said before F one is just and one gene you and this is able to m one a. Then we have and one G science data minus f to which is and one Geico. Science data sounds mu minus t two, and this is equal to him to a then we have de to Manistee one are and this is equal to the moment of inertia, of the ability which is n r squared over two. That&#x27;s the moment of inertia off a cylinder times the the angular acceleration Alfa, which is the leader Acceleration A divided by the radius of the pool. Notice that the radius here castles out. So we&#x27;re left only with two minus. T one is equal to m A over to. So this equation to this week in here I&#x27;m going to give the name one to this. I&#x27;m gonna give meaning to to this one. Here, the name three. Then what I&#x27;m gonna do is to some the three equations one plus two plus three. So, miss, um, through them, and notice that the teeth cancel out because we have 21 minus C, 2% u minus 21 So we only have to care about the other terms. So I have minus and one g nu plus M one g, scientific data minor applied. This should be in a two year I team two because this is the mass of the second block, not the first one. I just carried out a mistake, but I should be in too. So this here should be him too minus and to times g times a cool sign of favorite times mu iss and this is equal to and one pleasant to less came over to a so a times in one pleasant to plus an over to is equal to G times am two times side of data minus mu time School side of data because ser minus and one mu so a is equal to g times and to science data minus mu Cool science data Klein is in one you divided by and one wasn&#x27;t too. Plus I am over to and now we can substitute the numbers explicitly. So ji is not 8 m per second square times and to which is 6 kg. So it is a sign of 30. Mine is your 0.36 times the co sign. Ah, 30. Mine is in one which is 2 kg. Times is your 0.36 divided by 6 kg was 2 kg plus 10 kg of I device to. So if we carry out the copulation, A is equal to 0.31 m per second square. So this is the exploration of that. Your blocks, Any question be we have to calculate the tensions in the two sides of the of the string. So t wanted to to s. So basically, what I&#x27;m gonna do is to go back to the equations that I wrote before. So I&#x27;m going to use first the first ingredient in order to find t one. So we have we just just copy The clay didn&#x27;t up here. So is he. What? Minus him Once she knew is equal Chem one A. So T Wat is able to end one A because G mute. So this is 2 kg times a which is, uh, 1 31 m per second square plus g which is not what, 8 m per second squared times you&#x27;re going 36. So it&#x27;s here. One is able to 7.68 newness and then t to in order to fight you two just gonna use Equation three, which says that t two minutes to one in which we may over to so two to minus t. One is a royal A over to so teach two is equal to t one plus and a or two no t one, as we found, is people to 7.68 Newtons plus an, which is uh, 10 kg divided by two times is your point 31 m per second squared. So t two is able to 9.23 Newtons. This is the value off the attention of the second part of the stream and they include enough over question.

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A cannon is used to launch a speaker emitting 3 k…

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