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41s3-kgbock restsonahorizonta table andis attiched to oneendot a massless horirontal spring By pulling horizantallyonthe other endof the spring someone Causes the b...

Question

41s3-kgbock restsonahorizonta table andis attiched to oneendot a massless horirontal spring By pulling horizantallyonthe other endof the spring someone Causes the block to arcekerate uniformly And reacha spcrd 433mkIn L34 t Inthc proccss Iho spiret stretchcd by 0212m The blork # then rulled at # constant prcdot 4 33rVs durins whichtimn Lpnnecanichedtn '0.0632 m Find (a} the 50"| constaniotthc prinpand '[b) Uhc cocffcicntot kinatic friction Ectwcen Ihc block tnd thc tabl Numbct 64

41s3-kgbock restsonahorizonta table andis attiched to oneendot a massless horirontal spring By pulling horizantallyonthe other endof the spring someone Causes the block to arcekerate uniformly And reacha spcrd 433mkIn L34 t Inthc proccss Iho spiret stretchcd by 0212m The blork # then rulled at # constant prcdot 4 33rVs durins whichtimn Lpnnecanichedtn '0.0632 m Find (a} the 50"| constaniotthc prinpand '[b) Uhc cocffcicntot kinatic friction Ectwcen Ihc block tnd thc tabl Numbct 64047 Unts Nuemlcg elea Nouil



Answers

A 15.0 -kg block rests on a horizontal table and is attached to one end of a massless, horizontal spring. By pulling horizontally on the other end of the spring, someone causes the block to accelerate uniformly and reach a speed of 5.00 $\mathrm{m} / \mathrm{s}$ in 0.500 $\mathrm{s}$ . In the process, the spring is stretched by 0.200 $\mathrm{m}$ . The block is then pulled at a constant speed of 5.00 $\mathrm{m} / \mathrm{s}$ , during which time the spring is stretched by only 0.0500 $\mathrm{m} .$
Find $(\mathrm{a})$ the spring constant of the spring and $(\mathrm{b})$ the coefficient of
kinetic friction between the block and the table.

Hello and welcome to this video solution of enumerate. This question is based on simple harmonic motion principles. So if we have got a block or 1.5 Kg that is addressed on a table top And it is connected to a horizontal spring having forced constant 19.6 newtons per meter. Okay, let us tablet given parameters here, you know, we about mass m Equal to 1.5 kgs. And next we have the force constant of our spring as equal to 19.6 newton per meter. The spring is initially announced. Reached a constant force of 20 newton is acting on the object That causes a string spring to stretch. So we have the external force acting on it f equal to 20 newtons. I we have to determine the speed of the block after it has moved .3 m from the equilibrium position. And assuming that the self is between the block and these table doses tabletop is fiction less. So the distance movies X equal to 0.3 m. Oh we have to find out the speed of the block. Now here we can actually apply the conservation of energy principles for that. Now look there is a block and there is a spring right now of course if he's acting on it, making it compressed by that same distance X. Which is three m .3 m. Thus we can see the work done by the force is F. X. Is the wagon right? That will be equal to the energy required to compress the spring which is half Okay, X squared plus the net total kinetic energy of the block. Yeah, So let us plug in all these values here. So 20 times x. .3 equal to half K. is 19.6 Annex. We have .3 squared plus half Time masses. 1.5 G Geez times we square right now here we have to deal with the philosophy which is equal to let us calculate and see how much we're getting. Uh huh. That's 2.6 m/s. So this is the answer to question party. No, we have got a separate part B where we have got a certain amount of friction between this glock and the tabletop. So the amount of friction is new equal to 0.2. Right now we have to calculate the same velocity of the block after it has more .3 m. So now look the 4th external force here for some will also act upon buyer reverse force which is the frictional force. Right? Thus we can say the net force. That's the F net. Let me write it in the blue itself. Net force. FN which is equal to f minus friction. Right? This is equal to The external force that was acting was 20 minus um um G Right. Your game is the word of the block Which is equal to 20- new. Here is .2 Times the masses out. Here is 1.5 mm. James name branded. So here we have, let us see 17 Newton's. Now we have to apply the same equation here. Just we have to plug in ah This 20 with 17. And let us calculate people. Right? Yeah. So the velocity here is coming around to point seven m/s. Okay, so we have got that two velocities, one with the surfaces frictionless and another one when the surface is with friction. I hope this is clear to you and have a very good rest of the

In this exercise, we have spring that has a spring constant K and and then stretched length l zero that's connected to a block A that has a mass m E which is close to a block B that has a blood mont Mass and beat Now block be suppressed against block A, such that the spring is compressed by a distance deep has shown here in figure and then they are released are going to show that in order for block a block B to be separated after some time, we need that d be greater than two times the sum of the masses of the blocks times um UK Times G divided by K And actually, I'm sorry, there shouldn't be a d here. And also we need to find what is the distance that the blocks slide on the surface before they they separate. And the formula for D UK is the kinetic coefficient. Friction. Okay, uh, can connect kinetic friction coefficient. Okay, what we need to know in order to solve the exercise that the elastic force f is equal to minus k times Delta X no tax is the difference between the on stretched length off the the spring and the compressed length or the stretched length. And the minus sign is to show that when the spring is compressed, the force points in the opposite direction as the compression. When when the spring is stretched, the forest points in the opposite direction as well. Okay, so what we need to do first is to draw the free body diagram on the blocks. Yeah, so let's start by block A like a subject to the forces off the the elastic force. Have we also have the friction Force F F onda? We also have the force of block B exerts a block A which I'm gonna call in. Okay. And according to Newton's second law, we have that F minus and minus F f physical to the massive block A time is the acceleration. Now the force, which now is magnitude off the elastic force is K times D minus X where X is the distance traveled by the block. So let me draw here the spring that has an unscratched length zero. It's compressed by This is D. Okay, this is block a. Now consider that actually block a traveled a certain distance. X after being compressed. So, like a traveled this distance here. Okay, notice that this distance is equal to D minus X so that the force that block a experiences is equal to K times D minus X. Yeah. So going back to the equation we have came times D minus X minus and minus a frictional force, which is, um, u K times the normal force that the surface exerts on the block A and the normal forces just m a g in this situation. Okay, so this is the first equation that we obtain, and now we can carry on to block B. Yeah. So for block B, we have the force, and we also have the frictional force. And that is all we have. So we have an minus. F f is equal to M B G. So an minus m k m the G is equal to actually MBA. I'm sorry. Okay, So I have the two equations that I've highlighted in green. So when When this is some of these equations. But let me first just write it here. So we have, um okay. Times d minus X minus and minus mu k m A G is which, um a B I'm sorry. A they don't have an minus. Mu k m b g is equal to m B A and I'm going to some of the equations. So the ends canceled out. E gets k times d minus X minus mu k g m a person B is equal to m a plus m b times e So a Is it what you k d minus x divided by m a US and be minus mu K G. Hey, this is the acceleration. E can take this exploration. Go back to the previous equation. So this one here, for example, and find what the force and is. And I'm going to do that because when the force and is equal to zero, then the two blocks start to separate so that n is equal to m b times a plus m u k g. So when we have an energy which can be times K d minus x divided by m A plus and be, I noticed that the other two terms cancel out since the next term would be and be my M minus and B u K G. And the other term plus and B McKay g So we have it End is this thing here? And this is equal to zero when the easy go to x. Okay, so the block the blocks need to travel a distance d before separating. Okay, so now we are in a position to calculate to show that the inequality that the exercise gave us actually holds. So remember, remember that the exploration is equal to D v D t. So we have a eight times easy go to V times DVT, and I'm gonna write V as the x p t. We have a DX DT is v d v d t. Now the DT's cancel out. So adx right is vdv. You don't have to integrate it and I'm gonna integrate it from vehicle zero to V and from X equals zero too deep. You know that the exploration is equal to K D minus x divided by a purpose and be minus m u k g time is d X from zero to d and this is equal to B squared over two. Okay, eso we have a k times the squared minus D squared over two divided by AM A plus and be minus mu K G D is equal to B squared over two. Okay, so we need the speech. Be greater than zero if the blocks are are going to separate at this point D here. This is the upper limit of the integration, and we know that the blocks will separate at this point. So if we want the block to separate, then V must be greater than zero. So we have k d squared over two m able ism be must be greater than um u k g d one of these cancel logs. So you have d is greater than two mu k g. I am a person even be the vital backache. This which concludes our exercise.

In this problem. We see there's a spring that has been compressed and a box is attached to the spring whenever the boxes released thus or yeah, whenever the boxes released the spring D compresses and pushes the box towards the right side. So since the surfaces friction surface has friction, that's why the box stops at a certain point. Ah, because the friction is being acted on the box on the opposite direction to the displacement. So if we called this point as 0.0.1, where the spring has been attached to the box and call, this point is point to where the box finally stops. We can say that at this point the box had initial velocity of zero. Um, it's going to be one because we're wanting that as one, and we tow or the final velocity will be here as well, because due to friction, the box has stopped. Now, before you use the question 17.16 from the textbook, we can see the following question. Ah, let me explain what this equation means. So K one our king karma is corresponding to the kinetic energy of the system, or, for example, ifyou're concerning the box, then this is the kind of the energy of the box, and the one here is denoting that it's in 0.1. Then you want is the potential energy at 0.1 and w other is some other form of energy. So in our case, it will be the friction and ah ah, energy due to the friction for So let's call it F s, which is acting this way. That's why the boxes stopping it here because it's kind of in against the direction of motion of the box and that point, too. We have the frantic energy and the potential energy, but we don't have this other form of energy because the box is stationary here. So let's try to evaluate each town one by one. So when the system is a 0.1, we see that the system is at rest. That means there is no candidate energy, since the very one term is here is zero, and we know that kind of thing. Energies have in peace grade, so if easier than we can get rid of our candidate energy term, so let's make that zero. Similarly, Kato will be here as well because we see that Vito is zero here. So let's make that zero as well. For the potential energy. We see that the spring has been compressed, so there would be potential. It is you do to that compression and we can write that as half. Okay, X one squared. Now we should not be confused with this K on DH this K, this is big hair which is denoting the kind of energy. And there's a small cave which is denoting the spring constant. So at this point, we'll have the potential energy and this other form of energy. Let me discuss living. Come back to that in a moment. But let's talk about the you do first. So you two is the and pretension. And at this point, we can see that since this is stationary. And if we fix this as our origin, Yes. So, like, if this is this is our origin, then in the white direction of motion, there is no motion. That means we don't have to use gravitational potential energy here, since the height is not changing. And then there's no other form of off energy that's present here so we can get it off this YouTube harm here. So now let's come back to this w other time. So this is just the word than due to some force so we can write, worked, and due to some force as, um w f in our case because this is the work of the two frictional force. And we can write that as f k call sign fi times s. Now, if you have done the previous problems, we I know that, um, Wharton is nothing but force times the displacement Dying's co signed five. Where Fei is the angle between the force and the displacement on DA again essence the displacement and forces thie in darkest forces. F k. So we What? What do we see about the angle Fei? Here we see that the displacement is on the right. Although the friction forces acting on the left that means the angle between the direction of the displacement and that the frictional force will be 1 80 degrees So we can set that as our fight. So let me right w if one more time we'll have f s. Oh, are we have f k? It is a fictional force. Then we have co sign five times s. And as we say, that fire is 1 80 degrees. Also. What? What can you say about the frictional force? So we know that frictional forces mieux que times the normal force and anarchist normal force will be m times g. Hey, so let's call it in. That means new K times and will be the frictional force and let substitute and by writing mg. So finally our wk becomes, um yeah, okay, mg Then let's call the displacement s. And for cause I in 1 80 we have caused 1 80 Please, We know that co sign one. It is negative one So we can write that as I'm Yuki m g s Right. So let's combine all the questions together so we know that this is no negative. This one is no negative But then the other times are all zero. So we'll have you won plus w other are in our cases affection. No, we're going to do frictional force will be zero So you want is half k X one squared Plus we mentioned w f s and negative new K MDs So it's negative Mut and d s. It is equal to zero. And from here, if you dare aerosol form Yuki, we'll have nuclear. It is equal to half off Key X one squared, divided by sometimes ci times s So if we plug in the numbers now will have half. Then K is 100 newton per meter. Yeah, excellent is 0.2 meters and then there's a square term and is 0.5 kg g is 9.8 meters per second. Squared and s is the displacement of these 1.0 meter using all the terms of the numbers. Together we get new K as zero point for one now is that there is no uniformity and we can see clearly. If we do, the calculation on the units will see that all the units get cancelled and we have Ah, we have any unit this quantity over here. Thank you.

In this question. We have a block or five kg mass. Then it's attached to a spring on being pulled by a false Um, continue this to the right. Okay. The initial equilibrium line of the spring is 0.6 m and that is being put 25 addressed at this incident shown. So there are two parts in this question. We had a we want to find the speed off the block. Are that point B k eso in this case will be using conservation of energy king. Then the system is spring block system. Then the semen for conservation of energy is changing total energy. It goes to net work done. So that's changing. Told energy is half k X square plus half MV square. Okay, the network down will be the force times the displacement king. So to find a speed, um, and we just rearrange the equation. Okay, so this is what we get. Okay. Two times. 20 time. 0.25 minus 14 times 0.25 square. Uh, right by 0.5 and then take the square root. Okay. And then this is equal to 3.87 m. per second. Okay, so this is the answer for part day. Mhm in part B. Um, you want to find, uh, the distance between the block and the wall, The closest distance between the block and the wall when the block goes back, goes back to the left. And he so, uh, be okay so that we know that the total energy off the system he is equal toe times x, which is 20 times 0.25 Okay, so this is five Jews. Yeah, so? So the 20 new turn force waas apply until it travels a distance off 0.25 m. Yeah. Okay. So since this is the total energy off the system and then we know that its maximum compression me so half que ex Mex square equals to five so we can actually calculate X makes S o X max is whatever. Two times total. Bye bye. Katie Scott roots See, two times five divided by 40. Square it. And this is 0.5 m. Okay, so the the sense between the war and block, that's next compression. Okay, we'll be 0.6 minus 0.5. Goes to 0.1. Yes. Okay. So, um, so this is the answer for party, and that's all for this question.


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