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A block with mass m = 16.3 kg is moving along horizontal plane at constant velocity. The block then starts moving Up the incline (0 23.59) with an initial velocity ...

Question

A block with mass m = 16.3 kg is moving along horizontal plane at constant velocity. The block then starts moving Up the incline (0 23.59) with an initial velocity of V1 = 12.23 m/s, at that moment; force F (30 _ 2.lt) N (where is in seconds) is applied to the block The force acts in the direction of motion parallel to the incline. as represented in the image. The coefficient of friction between the block and the incline is ut = 0.37. Determine the time necessary for the block t0 attain velocily

A block with mass m = 16.3 kg is moving along horizontal plane at constant velocity. The block then starts moving Up the incline (0 23.59) with an initial velocity of V1 = 12.23 m/s, at that moment; force F (30 _ 2.lt) N (where is in seconds) is applied to the block The force acts in the direction of motion parallel to the incline. as represented in the image. The coefficient of friction between the block and the incline is ut = 0.37. Determine the time necessary for the block t0 attain velocily of zero 0 m/s) aller slarts moving Up the incline Solve the problem using the theory corresponding "Chapter 15: Kinetics of particle: Impulse and momentum



Answers

A force $\mathrm{F}$ is applied as shown on a block of mass $\mathrm{M}=100 \mathrm{~kg}$, kept on a fixed, rough incline of angle $\theta=37^{\circ}$ with horizontal. The force is gradually increased and when $\mathrm{F}=840 \mathrm{~N}$, the block starts moving up and covers a distance of $7.5 \mathrm{~m}$ with an average velocity of $1.5 \mathrm{~m} \mathrm{~s}^{-1} \cdot\left(\mathrm{g}=10 \mathrm{~m} \mathrm{~s}^{-2}\right)$ (a) The coefficient of static friction between block and incline is $\mu_{\mathrm{s}}=0.3$. (b) The coefficient of dynamic friction between block and incline is $\mu_{k}=0.28$ (c) The coefficient of dynamic friction between block and incline is $\mu_{k}=0.225$ (d) The minimum acceleration with which M will move is $0.6 \mathrm{~m} \mathrm{~s}^{-2}$.

Here, somebody has propelled a block down an incline with a speed of 1.6 m per second. And the block slides a distance of 1.1 oh meters before coming to a stop. And the question is, what is the coefficient of kinetic friction on that surface? So to answer that, we are going to need both Kinnah Matics and Newton's second law. So a reminder of what those are. And then we'll get into the kingdom attics, which usually is a little bit easier to do. But Newton's second law is the sum of the forces equals mass times acceleration. And there are three equations usually that you use to work with constant acceleration. I am going to concentrate on the one equation that does not involve time because I see that no time is given in the problem. So I'm going to use the final squared minus v. Initial squared is equal to twice the acceleration times the distance. And we're going to pick a cord, that system with why pointing up and X pointing down along the incline some and with that as our situation we have the final squared zero and of course the initial is 1.60 m per second, which we will square is to times the unknown acceleration times the distance of sliding 1.1 oh meters is down the incline. So what this will provide for us is that the acceleration is negative and specifically it is in the minus X. Direction and kind of working that out. I it winds up to be 1.60 squared with a negative in front divided by twice 1.1 oh. And let me sub label that with an X. So I realize that that is an exhilaration along the incline, hence force dubbed the extraction. And that comes out to be minus 1.16 m per second squared. And why I did that first is knowing that Newton's second law is going to involve the acceleration. And typically what you want to do is break this into two equations, one for the X. Direction and one for the Y direction. So decompose your vectors and you really can't use Newton's second law until you have a force diagram and then you can start working with the components. So let's create a force diagram. So I have my mass, which I know and I'm going to create a weight downwards, that's always good place to start. And I'm going to show that that weight ex at an angle. So I'm going to kind of show my axes in there. I have an X. Axis and a Y axis. Let's make that a different color. And my weight actually makes an angle of 30 degrees to my Y. Axis. That is it becomes zero if I am flat on a horizontal, completely horizontal um oriented surface. So that's the easy one. The other easy one to draw is the normal force is straight along the y. Axis. And finally there is a force of fiction which is going to act opposite the motion and it is sliding friction. So we will call that F. Sub K. And now I can decompose things into components and set up my Newton's second law. So I usually make a column for my some of Fx and another column for my some of fy. And then I take each force separately and work with it sometimes starting with the easy ones. Um So the force of friction for example we know is completely along the X. Direction. It's up the incline which is in the negative direction. And knowing what I do about friction, It is U. K. Times the force of the normal, Let's be consistent. We call that normal voice. Good idea and now I'm happy I see my unknown there um and it has no Y. Component so whoops. Yeah no why component good? And my normal force is again the easy an easy one to draw out. It is no zero, no X. Component and all lying along the Y. Direction. And finally the weight is the one that has components but I also know a lot about it. We know the mass, we know 9.8 m per second squared. Um And the X. Component goes with the sign of 30 degrees and the why component is downwards and goes with the co sign of 30 degrees and I'll give myself a little bit of space to work. It's always a good idea um to actually work out numbers because we do need numbers. Okay, so the X component working it out with, the sign of 30 degrees is 12.25 newtons. And the Y component working out with the co sign is minus yeah, 21.2. And then at the bottom we know everything has to add up to zero so we can write an equation in our bottom a table element where we've added everything together and set it equal to zero. I'm sorry though, it's equal to zero in the Y direction but it's equal to M A X in the extraction. Yeah, so that will produce two equations. Um And hopefully only two unknowns and I can identify my unknowns fairly readily that they are going to be the coefficient of friction and the normal force and the algebra is not going to be too terrible, we hope. Um so what we have is minus U sub k times normal force plus 12.25 buttons is equal to M A X. And we'll put in a X. Because we do know that quantity on the other side, we have fn minus 21.2. Newtons equals to zero. And as promised, the algebra is not too difficult to do um equation number two. There can be solved for fN fairly readily. 21.2 newtons, we can use it in our first equation and solve for the coefficient of friction. Yeah. Mhm minus UK times 21.2 newtons. I'll leave off the units just so things don't look so complicated and all that has to equal the mass times the acceleration which we found above. Mhm Okay. And I'll show one. An intermediate step are actual, Yeah, just show one intermediate step. And so we have a minus U. K. Yeah. That doesn't even dane. Another step. We can turn all our negatives into a positive and we get a coefficient of friction fairly readily from that arithmetic.

Question. What stoop? I've two blocks are with Aram like this. This is a poorly and one block. It'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'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's sliding downwards, then we can die And don'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'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

Mhm. Okay. So in this video we're trying to we're looking at this apparatus so we have object on a friction full, there is friction on this uh ramp, So that someone attached to a string over a pulley attached to another mass. M2, that is just dangling over the edge uh the box. And we want to know three things. First, we want to know the mass of em too. That is necessary to move M. one up the plane at a constant speed. Mm. Then we want to know em to to move M. One down the plane at a constant speed. And then what range of M. Two S. Will the blocks remain at rest? So, first let's draw our FDD. So for and one. So just to make a little bit easier drawing until ting it. Yeah, So, into the ramp actually, let's keep it keep it. Let's withdraw that, sorry. So and one we got F. G. Straight down towards the center of the earth. Yeah, we got normal force perpendicular, we've got tension, pulling it up the ramp and we got the force of friction opposite motion. That's what's sliding up the ramp towards the frictions opposite. Now, here we have a problem F. G. We really can't use like that because it's in between directions. So we have to break F. G up. So let's get rid of this. We're gonna break it up into its components and we're gonna do that is figure out the perpendicular component, which is just gonna be MG coast data and the parallel components, which is M. G. Side potato. And let's also draw our fPd for our other box. That's too this means a lot easier. We got M two G. Let's actually go back and let's make sure we put in those ones and tubes. These are both, and one's in our previous equations. This one We're going to be right. The force of gravity or in other words, mass to gravity and tension. So essentially, you know, everything is moving at a constant velocity. That means that all of the forces have to be balanced. So for our M two we can write T equals M two G. And for our other box for M one we can write two equations. T equals M one G. Sign. Alfa. You go back and correct that earlier. Yeah. Right. Mhm. We're using alphas instead of fears of the culture. So M one G sine alpha plus force of friction. And we also know that normal force is equal to M one G co sign outside. Okay, we're getting somewhere and one more thing you need, the force of friction when it's moving is equal to mute. Okay? Yeah. Times. FN which in this case are force of friction is going to be um U. K. Times M1G face alpha. Mhm. So if we mash this all together, I'm going to substitute the M2G in for attention. I'm gonna substitute this expression for force of friction. So I'm gonna end up with and to keep Equals M1G sign Alison bus. You okay? M one G coats house? Thanks. So now we can get rid of Gs. So let's cancel those out, be cancels out everywhere and I'm left with and two equals and one side alpha plus UK and one looks oh yeah, especially. Okay, so we got our first part done. Mhm. Yeah. In part B now we're looking at the box sliding down the plane at a constant velocity. So let's redraw our F Bds. So the FBD for M. two, okay, hasn't changed. So you got M two G and T. And they are still equal to each other and for one this time I would have brought the components in already. So we got Tennessee and yeah, yeah. Yeah, force of friction up the plane. Mhm. Yeah. And one she signed alpha. So here we go. Normal force and And one G coats. Yeah. Already. So now let's write our equation for are in one box. Now I'm gonna skip the F N. And M one G close alpha part because that's already gonna be baked into the force of friction. So now we have M1G sign alpha equals T plus UK M one G Coast Alfa. And a little bit room. Let's re write that down alone. I got M one G sign up for equals T plus UK. M Y G cooks. No, mm no we gotta get now we got a substitute in for tea. So that's M1G sign alpha equals. And to G what in the UK? Yeah. M one G coast alpha. Yeah. Again we can get rid of all of our G's. Each one is a G. Some common factor cancels out. So you must subtract the UK and one goes out to the other side. So now I have I am too because and one side alpha minus UK. M one coats. No. Yes. So notice the only thing that changed was in sliding up the ramp. It's N one side Alpha plus U K. M one coast alpha. And in the slide down the ramp it's minus. Yeah. And the real nice thing about going to static friction. This is our last part part seat. What is the conditions for the blocks term made at rest? Two things change. We switch the equal sign to less than or right. Less than or equal Tucson. And we also switch arm us or our location for you guys. So for static friction it only balances out less than or equal to. So the only thing that changes from our two expressions earlier is that I am too has to be less than or equal to and one sign also plus U. S. M one coast Elsa and M two has to be greater than or equal to the reason that is is any lighter then it's going to accelerate down the ramp heavier so it keeps it from sliding down the ramp right? Everyone is almost winning here and trust we agree with an equal to M one side. Alpha made this us 10 1 pulse Alfa. Yeah.

This question covered the concept of Newton's second law, that's all the free world diagram for each of the block. So for someone the forces acting on The bloggers, the weight, which rejecting what degree downward that is M1G. And the normal election force that is n. one and attention force team. Okay, so let's say I'm too is moving downward. So the friction falls and one if I'm always moving downward, everyone will move up along the plane and the friction falls will act down along the plane. And that is new tents. And Okay. And for mass M. two the forces and to G. And that the chinese tea vertically. So From Newton's 2nd law along the from the Newton's second law form us and to you can write I am to G minus T. That is a cost to zero. Okay, so this is the question. one and four M. One. We can. Right and one minus M. one GT Car Sofa. That is a cost to zero. Okay Oregon. Right. And when is the cuba into? And one G cause of let's say this the question too. Now we can right along the perpendicular to the plane, sorry, along the, along the plane G minus mealtimes and win minus And one G Sine Alpha. That is a question zero. Or we can write t minus mealtimes. Mhm And one day course alpha -M one G Sign α.. That is because 20. Yes. Yeah. So let's say this is the question three. So from the question one and 3 we can right I am to G minus you. Okay two M one Geico sulfa -M one G sign alpha. That is the question zero or from this we get and two is I'm one G and two. You okay course alpha less sign. Alfa a bond cheap. Rm two es I'm one in 2. Okay, causal, fabulous sign. So this is the mass of em too. I did a maximum mass of em too that can move the block and one with a constant speed up to the plane. So now when the block moved down to the plane the friction falls the change direction. and by doing this, the mass M2 is I'm one and 2 sign Alfa my s okay. Of course, alpha. So this is the minimum mass of the M two so that someone will move down to the place. So this is minimum mass and this is maximum, yeah. Now, for the third case, the range of M2 for which it will begin to slide. The coefficient of friction will come into place. Uh The static kinetic uh static friction coefficient will come into play and it will replace the coffee set of kind of exception. So and two must be greater than I'm, one sign alpha minus us cause alpha, and it must be less than and one sign alpha plus Mewes, because so this is that and separate part C. Okay, This is what part B and this is what part.


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