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Question

In the figure above, two blocks of mass 3$m$ and 2$m$ and 2$m$ are attached together. The plane is frictionless and the pulley is frictionless and massless. The inclined portion of the plane creates an angle $	heta$ with the horizontal floor. What is the acceleration of the block 2$m$ if both blocks are released from rest (gravity $=g ) ?$
(A) 2$m g$
(B) $\left(\frac{2}{5}ight) g \sin 	heta$
(C) $\left(\frac{2}{3}ight) g \sin 	heta$
(D) $\left(\frac{3}{5}ight) g \sin 	heta$

Vishal Gupta · Accepted Answer

Hi in the given problem here. This is a smooth election lists horizontal surface which is joined with the inclined surface which is inclined at an angle He top with the horizontal. There is a block having a mass to mm and another block kept here. Over the horizontal surface. It&#x27;s mars is three a.m. There is a frictionless massless bully and a string is connecting these two blogs passing over the pulley like this. No component of weight of this block which is kept on the inclined plane. The wheat is two MG acting vertically downward and one of the component, if this angle is heat at this angle will also be theater. This component will be too MG cost leader and have this component that is to MG sine teacup which will be sliding this block in downward direction. Suppose with an acceleration A. So attention T. Will be created in this spring equal in magnitude in a positive direction about this bully like this. So taking everybody die ground of the smaller block, smaller block which has been kept over the inclined surface, taking it as the block is moving downward. So out of the two forces acting on it. UnG Scientist A and T. These two MG scientists should be more so two MG sine. Theta minus tension is equal to using net force is equal to m into a using Newton&#x27;s second law of motion. So this is the net force acting on the blog will be kept equal to mass, the product of mass with acceleration and the masses to him. Exhibition support. This is a which will be seen on this block also which is kept on the horizontal surface. This is the question number one. Now, taking the free body diagram of the amir of lock means block having a mass of three um which has been kept kept over the horizontal surface and in the direction of motion there is only one force acting on it which is equal to the tension in the string. So we can say it is only for speedy again. Using Newton&#x27;s second law of motion, they should be able to three M. Into A. So now putting this value of T. From question to question one, we get two MG sine tita minus three M. A. Is equal to M. A. Or if we shift this three may towards the right hand side, it will become less positive. So two MG sine theta will come out to be equal to five M. A. And cancelling these mars. An expression for the acceleration in the system comes out to be G. Scientists are divided by five which is an answer for this given problem and we can say option B is correct. Thank you.

Ze-Han Lee · Answer

All right, so in this problem, we have an incline. So we have block was massive. Three m. And it is Attention was Ah ah, boy. So and there&#x27;s another block to M in this English data. Okay, so we want to find out. Except regional. Persistent. So basically, the net for is acting on the system is pointing down. Went into this direction, and the magnitude is a two mg time Santa. Right? So there&#x27;s an air force acting on this hold on the whole system and put a master&#x27;s five. So this gives you two or five g time. Send data, okay?

Averell Hause · Answer

be a quick free body diagram for the system. And here we have Newton&#x27;s We&#x27;re gonna ply Newton&#x27;s second law for the X and Y axis of the first block and then on Lee the Y axis of the second block. And then we have tea minus m sub one g sign of data equaling. I&#x27;m someone Times acceleration, eh? The normal force is gonna be This will be minus m sub one g co sign of data. And this would be going too zero. And then finally, we have themselves to G minus t equaling EMS up to hey! And so we can then four part A. We&#x27;re going to add the first and third equations. And we have m sub too. She minus m someone g sign Fada. This would be going to em someone, eh? Plus, I&#x27;m so up to a And so the acceleration a would be equaling two m sub two minus. I&#x27;m someone sign of fada. Multiply this bite g divided by the sum of the masses. And so the acceleration A is going to be 2.30 kilograms minus 3.70 kilograms multiplied by sine of 30 degrees multiplied by 9.8 meters per second squared divided by 2.3 plus 3.76 point 00 kilograms. So the acceleration is found to be equaling 0.735 meters per second squared. This would be our answer for part II and we have the four b. The results to a is greater than zero positive, indicating that block one block ones acceleration is up the incline and we can say blocks. We&#x27;re blocked twos. Acceleration is downwards so directly downwards that were answer for part B and then for part C quickly. We can say that the tension in the court t equaling the mass of one times acceleration plus m sub one times g sign of theater. So this would be equaling 23.70 kilograms multiplied by the acceleration of 0.7 three five meters per second squared plus 9.80 meters per second squared sign of 30 degrees. And we find that the tension forced t is gonna be equaling 2 20 0.8 Newtons. This would be our final answer for part C. That is the end of the solution. Thank you for watching

Averell Hause · Answer

Okay, so let&#x27;s first draw the incline with free body diagram of the incline. And we have a mass A lying on this incline. Forced normal always always acts perpendicular to the surface of contact. So this would be forced, normal and always, always going straight down would be the weight. Ah, we have an angle here, Fada. And now we can&#x27;t evaluate. We want to find the, um total X displacement s so we can say that the sum of forces in the X direction would be equal to M g sign of data. This would be the X component of gravity, and this would equal the mass times acceleration. So the acceleration would be equal to G sign of data if the masses here will cancel out. And so this is gonna equal a 9.8 meters per second squared sine of 22 degrees. And so we find that their acceleration is going to be Ah, let&#x27;s actually keep it 9.8. Yeah, because we don&#x27;t really need to evaluate. We can actually evaluate at the very end. So when we want to find the change in displacement in the extraction, we confined, We can use rather to de Cana matics, so Velocity Final squared would be a true equal to velocity. Initial squared plus two a Delta X and so here. Delta X would essentially be equal to negative the initial squared divided by two times a so Delta X would be equal to negative negative 4.5 meters per second quantity squared, and then this would be divided by two times acceleration of 9.8 sign of 22 degrees. And so this is giving us negative 2.758 meters. Now, the fact that is negative means that this would be up the plane. So essentially you&#x27;re giving an initial velocity, and you just simply want to see how far up the planet goes given that the incline restriction lists and ah, in this case, gravity would be the only thing that would be stopping you stopping the block. So this would be your answer for a part A. And then for part B, we want to find the round trip time. So for Part B, we need to say that the final equals V initial plus 80. However, if we&#x27;re trying to find the round trip time for tea. Ah, we can say that T is gonna equal velocity final minus Flossie Initial divided by a. However, in this case, because we&#x27;re looking at the round trip time, the final velocity would actually be positive 4.5 meters per second because it&#x27;s going up the incline stopping and then coming back down. The and the block will have the same mass initially as it will as it. Okay, so the initial velocity, it&#x27;ll if it has a certain initial velocity, it will stop. Ah, again, 2.758 meters off the plane. However, once it once it reaches back to you, it will have that same inertial velocity. So that&#x27;s why the final velocity is positive. Four points, five meters per second and then will be negative. A rather minus negative, 4.5 meters per second. This would account for the initial in the initial velocity of the block and then again divided by 9.8 times sign of 22 degrees. And so are our hour round trip time will be 2.452 seconds and this will be our final answer. That is the end of the solution. Thank you for watching

Ze-Han Lee · Answer

So this problem had the model like this. So the angle dating clients data and we have block over here. So it is a pulley we have blocked there. Right? So this is m one and business them too. And suppose the freak, the surfaces Friction ist. Then the total force acting on two blocks is, um, one g, which is the gravity of m one. Lina&#x27;s them to ji han sine theta some two g time Sunday A is the gravity. Ah, is the ah ah, power component of the gravity. Okay, so this is the This is the net force acting on this. Ah, the whole the whole system and develop divided by the total mass, which is m one plus him too. All right, so this gives you the total acceleration of the system.

Guilherme Barros · Answer

in this problem, we&#x27;re gonna talk about inclined planes. So consider that we have a block that is lying on an inclined plane that makes an angle Alfa with the horizontal. And can you compose the force? The gravitational force that acts on the object? Remember that the gravitational force just point upwards and has a magnet. I&#x27;m sorry Downwards and has a magnitude of M G. And we can decompose it into components, uh, that are perpendicular and parallel to the plane. Okay, I&#x27;m gonna call what I call them Y and X. So f in the Y direction is equal to minus m g co sign off Alfa and F in the X direction is mg sign of awful. Okay, on in our problem what we have in a situation very similar to this one where awesome is equal to 22 degrees and the mess of the block is equal to seven. Killer were okay, so in question A, our goal is to find what is the acceleration of the block as it slides downwards. About the only force that acts on the block in the X direction is the, um, gravitational force and the acceleration in the X direction times the mass. So let me write it because the mass times the exploration is equal to the mass of the object times G time Sign off 22 degrees. Since this is the only force in the X direction, the EMS cancel out and the acceleration in the X direction is just G. That&#x27;s 9.8 m per second squared times design have 22 weeks. The exploration in the X direction is equal to 3.671 All right, let me round it down to 27 meters per second squared. Okay, that question be, we have to suppose that&#x27;s the lane off. The inclined plane is 12 m and that the block starts at the top of the inclined plane and then slides downwards. And at the top of incline, playing the block is a breast. Eso. What I&#x27;m gonna do is use Torricelli&#x27;s formula in order to find what is the value of the velocity of the speed. I&#x27;m sorry of the block at the bottom of the inclined plane. Remember that their churches formula tells us that we squared is equal to be zero square was to time the exploration times the attacks in our case v 00 So he is just a squared up to a The attacks of the acceleration as we found in the X direction is equal Thio three point 67 m per second squared while Delta X is 12 m. What is that? Both the exploration and the attacks are in the same direction. That&#x27;s important. So V is equal to 9.39 m per second. This is the final speed of the block.

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