Practice: A 5 kg block reaches the bottom of a long inclined plane with 20 m/s. The incline makes 53° with the horizontal, and the block-incline coefficient of friction is 0.4. What total time does it take the block to go up and back down?

Subjects

Sections | |||
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Intro to Friction | 40 mins | 0 completed | Learn |

Kinetic Friction | 39 mins | 0 completed | Learn |

Static Friction | 15 mins | 0 completed | Learn |

Inclined Planes | 34 mins | 0 completed | Learn |

Inclines with Friction | 73 mins | 0 completed | Learn |

Forces in Connected Objects (Systems) | 50 mins | 0 completed | Learn |

Intro to Springs (Hooke's Law) | 21 mins | 0 completed | Learn |

Concept #1: Intro to Incline Plane with Friction

**Transcript**

Hey guys, so now that we've talked about incline planes and we've talked about friction we're going to put it all together and we going to have problems where objects are going to be sliding up and down a plane that will have friction, OK? So, let's check it out, there's really nothing special here other than putting those two ideas together so we're going to start with an example right away, we have a 5 Kg object here so the mass is 5 it is released from rest so the initial velocity is 0 from a ramp that makes an angle of 53 degrees with the horizontal, so let me draw this over here. Here's a ramp and it makes 53 degrees with horizontal so theta equals 53, we want the angle to always be with the horizontal so that's good, the coefficients of friction are 0.4 and 0.6, remember 0.4 is my kinetic always the small one when you're given two and 0.6 is the static, so let's put the 5 Kg block in here. The first question asks us to draw the free body diagram so let's do that, let's draw the free body diagram and calculate all the forces, free body diagram F.B.D OK, so you remember how this works I have M.G. pulling down I have this is M G X down the plane always and M.G.Y into the plane always, normal opposite to M.G.Y always OK. Now which way would friction go? Well friction is always going to be in this case it's a block so we can only have kinetic friction, rubbing friction because it's going to be sliding it's going to oppose motion if there was no friction this box would fall because there is friction therefore the box the friction in going to be opposite that motion so one of the ways to identify the direction of friction is which way would it move without friction and then you know that then the friction has to be going in the opposite direction, in this situation here when you release the block friction is always going to be up the plane, OK? Now the only thing is that we don't know which type of friction but I'll talk about that shortly. let's draw a free body diagram here remember what we do is this is our YX is the new Y axis and this is the new X axis, this here is not a free body diagram technically because free body diagrams have to be just dots without the plane and without all the other stuff here so just a dot with forces on it, but I like to draw it here first as a sketch and then just basically transport it over here I'm going to put my axes first here, here is the Y which if I have this here it's easier to just kind of follow and here's the X axis right here I'm going to see that this is a positive YX because the positive YX is always in the direction of the normal force and the positive X axis depends on however you want to decide determine what the direction of positive is, in this case the block is going to slide down this way so I'm just going to choose that this is the positive X axis, now that I've drawn my dot and I drew my axes and I pick the directions of positive I can put the forces here, MG is straight down and MGX is obviously along the X axis and MGY you are in is along the Y axis, you can draw all three of them or if you wanted to you could've have drawn just those two depends on what your professors preference is, I think this is the safest version because it's got everything there Normal is going to go up against in the direction of the Y axis and Friction is going to be going this way right this is a complete free body diagram so that would have given you full credit if you were asked to draw one of those, I want to calculate all the forces let's do that real quick so MG is simply mass times gravity so mass is 5 and gravity we use 10 so it's faster and this is going to be 50. MGX. remember MGX if the angle's down here remember MGX is always going to be MG cosine of theta, I'm sorry MG sine of theta remember X goes with cosine but in the inclined plane it's the opposite it's backwards, X goes with sine just with the incline, MG is 50 and then sine of 53 and if you do this I have it here it's 30, I'm sorry this is 40 actually, yea this is 40, and then MGY is MG cosine of 53 and that is a 30 cool, so that's those forces there are more the forces in the Y axis therefore normal and MGY have to be the same so normal is 30 as well without all these forces and now we have to calculate the friction frictional force how much friction is there, the problem is at this point you don't know whether the object is moving or not you don't know if MGX is strong enough to overcome the frictional force because of that you don't know if you have static friction or kinetic friction so what we do we test it we have to calculate both and see whether or not it's going to move so let's calculate both frictions I'm going to do it over here, friction static is mu static normal, mu static is 0.6 and the normal is we have normal here is 30 so this force here is 18 Newtons. So check this out MGX is 40, friction static is 18 so this means that this block will Overcome static friction and then it will move OK, so I can say here since I am MGX is greater than friction static max because this is a 40 and this is an 18, the object moves object or block moves more specifically it accelerates so the acceleration is not 0, it breaks static friction which means that we're actually going to be going up against kinetic friction so now I know for sure that the friction I have is kinetic, let's put a kinetic there cool and I can calculate your kinetic friction, friction kinetic is 0.4 times 30 which is 12 Newtons OK, so I got a 40 here 12 here, and normal we already said it was 30, so 40, 30, 12 this here is 30 as well cool, that's part A and I already answered part B as well, part B was asking this was basically us I guess doing Part B here, Part B was asking will it move and the answer is yes the object moves why? The reason why it moves is because MGX is greater than the maximum static friction and part C. then asks us to derive an expression for its acceleration and calculate it so let's derive an expression for the acceleration, well this is a force problem I'm asking for the acceleration you going to do this using F=MA, the block will accelerate this way so when I say find me the acceleration the acceleration is the acceleration in the Y and the X axis which is along the plane because the acceleration in the Y axis is 0 in all inclined plane problems I'm really talking about the X axis so I'm going to say that the sum of all forces in the X axis equals MAx and the forces are MGx as a positive and friction kinetic as a negative going against the direction of positive and MAx. Now what we're going to do since we're deriving an expression we're not going to just plug in these numbers we have to expand these as much as possible MGx will be replaced with MG sine of theta and friction will be replaced with minus mu normal, and then we look at this once again and we realize that we can still expand it so in MGx could be written as something else so we did, friction could be rewritten so we did, MG sine of theta can't be further rewritten but normal I can replace normal with something else, normal is M G Y because those are the only two forces in the Y axis in the plane right here right? And MGy can be further rewritten as MG cosine of theta once I get to this point there's no longer nothing else I can do there so this is going to be simply M.G. sign of theta minus mu MG cosine of theta equals MA, I don't have to keep writing Ax because this was the only acceleration I have, notice at this point that the masses will cancel so the final answer the final expression becomes just G sine of theta minus mu G cosine of theta, if you want to make it look even more compact more simplified, If you want to get fancy you can factor out the G and say it's G sine of theta minus mu cosine of theta or you could have had the G in both, that's you know this is just one step further if you remember to do cool? So, this is the final expression and then it says calculate it right? So, calculate it, just plug in all the numbers, gravity is 10 this is sine of theta, sine of 53 is 0.8 minus mu, mu is the coefficient of friction which we have kinetic friction so it's point 0.4 right? Because the type of friction we've identified is actually kinetic so that's going to 0.4 times the cosine of 53 if you plug in the cosine of 53 in the calculator you get a 0.6, alright? And if you do this If you do this let me just see this real quick you're going to get.... I don't have this here but this is going to be 0.24, so 0.8 minus 0.24 is going to be 0.56 so the answer is 5.6 sorry about that I just hadn't calculated that in advance but the answer here will be 5.6 roughly 5.6 meters per second squared as the acceleration, cool? So that's it for this one hopefully you got it let's keep going.

Concept #2: Incline with Friction & Critical Angle

**Transcript**

Alright, so when we have rough inclined planes, rough inclined planes are planes that are not frictionless, we have a special angle, the angle at which an object would begin to slide is called the critical angle where the object would begin to slide so I can write this as theta crit or theta critical so it's the angle which it begins to slide so imagine if you have a surface and you put a block here and you can move the surface like this, right? and you start moving you can try this yourself with a book or something on a desk and you start moving and the block doesn't really move yet the reason why the block doesn't move is because even though it has an MGx this way, there is static friction this way that's holding, if it was a frictionless surface even a tinniest angle would cause the object accelerate but on frictional surfaces that doesn't happen and you keep doing this you keep doing eventually if you go really slowly there's going to be an angle at which the object will start to move that's the critical angle, OK? And what's special about that angle is that at that angle the block's MGx, right? So you're being pulled down this way by MGx It will exactly equal or equals exactly which means it will cancel the static friction on it, so just before it starts moving let's say it starts moving at 37 degrees that means that at 36.99 degrees right before that started happening or approximately at that angle these forces were exactly the same, Ok? friction static maximum remember the whole deal with static friction or with maximum static friction is that you have to overcome it if the friction static maximum is say 30 this M.Gx. here will increase as you do this right, as you do this the block wants to fall even more, eventually once this reaches 30 it's at the breaking point and then at the slightest additional angle means that this thing will start moving OK? so we said it exactly the same at that point so we can write it MGx is exactly equal to friction static max OK that's the critical angle, now there's another angle that's important as well and it's one that will cause an object to slide at a constant speed so if I break static friction what happens? I am now going up against kinetic friction because kinetic friction is less the second is this block starts moving this kinetic friction is less letÕs say friction kinetic drops down to 25, the only reason I started moving in the first place was because this was 30 or 30.001 and now if I keep moving I'm not going to just slide across but I'm going to accelerate because 30 was what I needed to overcome this but 30 is bigger than 25 so this block is going to accelerate now as soon as you break static friction you accelerate, the second angle which is the one that I want to talk about is an angle that allows you to instead of accelerate down the plane move down the plane with a constant speed which means that there is no acceleration, how's that possible? I just told you that if you break static friction it's going to accelerate down and it will the only way that this is possible is that you put a block at an angle that would not overcome static friction because MGx would be let's say exactly 25 it doesn't move but you're going to push on it so that you are causing static friction to be broken once you break static friction the friction against you is now 25 so all you need to keep moving is 25 OK, hopefully that makes sense it's a little weird especially at first so constant speed which means at this point MGx cancels the kinetic friction and the reason cancel is because it's the same number just opposite directions OK, so MGx equals kinetic friction so again first iteration you're not moving and you're also not touching the block this gets to a certain point and now it's going to accelerate down on the other one you're not at that point yet but someone pushes it just so that it starts moving and now it keeps moving because static friction or because kinetic friction is less than static friction OK, the setup for these is pretty straightforward if you have to the derive the equation sum of all forces we're going to call this the X axis here sum of the forces in the X axis equals MA we want to talk about the critical angle which is the angle where you don't move at this point what's happening is your MGx is the same as your friction static max, right? So, what are your forces acting on you if you are not moving and you want to find the critical angle? Well let's say......Let's call this the positive X direction so MGx is positive but friction static is negative and this will equal because the object is at equilibrium up there OK, So look what happened I move this over to this side and I have MGx equals friction that makes sense they cancel and we already mentioned that so what I'm going to do now is expand these two expressions so instead of MGx IÕm going to write MG sine of theta and instead of friction on the other side I'm going to write mu static normal. Now remember in these inclined plane problems, normal is the same as MGy as long as they are the only two forces acting on the Y axis here right which they are, so normal equals MGy and MGy is MG cosine of theta so this becomes MG sine of theta equals mu static Mg cosine of theta and look what happens here pretty neat M and G cancel on both sides and then you're left with me you mu static equals If I divide both sides by cosine to get the cosine out of the right side I get sine over cosine and that's the tangent of theta, so I started here by saying that MGx is equal to friction max and I ended up here I was able to write this using very simple F=MA depending on your professor you may or may not have to do this yourself OK, so that's cool in physics because this is a very simple answer physicists actually call this an elegant solution which is funny but it just means that it's a very simplistic answer a very simple answer with just two variables right if you want to find mu you can find by doing the tangent between the critical angle let me put a little crit here, then I can use it if I want to find the critical angle I can just take the arc tangent of both sides so this equation can be swapped around by saying theta critical is the arc tangent (it's kind of hard to fit here) of Mu static, these two equations are equivalent you get from one to the other very easily, and what's interesting is actually you can use this to figure out the coefficient of friction between a block and an incline, you can do this with like a calculator and an arm and then the sudden it starts moving you can get a protractor measure it here and once you get that angle measurement of the critical angle you just stick it in here and you're able to experimentally you might have done this in lab or you might do this in the lab figure out the static friction just by measuring that angle, cool. the second part here is very similar we're going to try to find which angle gives you a constant speed down the plane and we're going to do this by writing F=MA because I have force problems as well, sum of our forces in the X axis equals MAx and the forces now if you were moving with a constant speed it means that you're going up against kinetic friction ok, so I'm going to have MGx in the positive direction but kinetic friction in the negative direction the acceleration is still 0 because it's a constant speed. So this set up is exactly the same as this except that here it's friction static max And here is friction kinetic I'm going to move this over to the other side it's going to be MG sine of theta equals mu kinetic normal which is mu kinetic MG cosine of theta, and again the MG cancels and I'm doing this a little bit faster because it's very similar to the other one and you end up with this mu kinetic equals the tangent of theta it's the same equation, it's just that this is static mu and this is kinetic mu, this is the critical angle at which you wouldn't move at all and this is the angle that would give you constant speed OK, So one way that you can think about this is that this is the critical in go for static and this is the critical angle for kinetic, OK? So, the critical angle for static is the angle at which it would begin to move and the kinetic critical angle would be the angle at which it keeps a constant speed, OK? So those are slightly different situations and obviously here I can also do the arc tangent thing to get that theta the kinetic critical angle is arc tangent of the kinetic coefficient of friction. Two special angles here we call them critical but they're really just two special angles, one angle at which you were static constant speed and the other angle at which you would just break it, I have an example here that helps illustrate this and so you see how this would work. it says here after playing with the block on an adjustable incline you find out that the two things, one that slides down from rest for angles 37 or greater so you are doing this as soon as it hits 37 it slides down, so what does that tell you? It tells you that that 37 is your critical static angle It's 37 so when I say that it slides down from rest for these angles I'm telling you this, point 2 says it slides with a constant speed when theta is 30 so number 2 it says slides with constant speed which means the angle at which I would slide with constant speed is my critical angle my kinetic critical angle so I'm telling that theta critical kinetic is 30, OK? So, this piece of information is telling us this and then it says we're going to use this information to calculate the coefficients of friction, I'm going to know mu static and mu kinetic and at this point since we already derive these equations here we can simply use these equations because these equations are a link between critical angle and coefficient of friction, right? So, if I have this and I want this I can just use an equation that combines the two and that equation is that mu static is the tangent of theta static, right? the static critical angle in other words it's the tangent of 39 and the tangent of 37 is 0.75, same thing here mu kinetic is the tangent of the kinetic critical angle which is the critical angle which you would slide down at a constant speed so it's the tangent of 30 degrees and the answer is 0.577. So two last point here to make, one notice how this angle is greater than this angle because the idea is that the only way that the second situation happens here that you slide with a constant speed is that you are not angle which would break static friction because if you were you would accelerate but someone pushed the box and that person caused the static friction to break and now it keeps moving with kinetic friction going against it, Ok? So, it makes sense that this angle is greater and we already know that static friction is either greater or the same as kinetic so it makes sense that this is greater as well, OK? That's it for this one, a little weird I know but hopefully this makes sense let me know if you have any questions.

Example #1: Incline with Friction

**Transcript**

Hey guys, so I want to solve a few more problems of incline planes with friction. Let's check it out so I have a 3 Kg crate that is on a rough here's the coefficient of friction rough means not smooth not frictionless plane that makes 30 degrees with the horizontal, so it's an inclined plane it has a force F acting on it, an external force you're pushing all someone else is pushing on it, OK? So, I'm going to draw it this way just so that at the absence of the force it would move down and then I want this therefore to be I want the direction of positive to be to the right, you could have drawn it this way and then your MGx would be going that way that works just as well, I just like when things are pointed to the right, so 30 degrees this mass here is a 3 and the coefficient of friction is 0.8. Let's calculate these forces real quick, here I have MGx...I'm sorry MG this way MGx this way there is MGy this way, there is a normal this way, OK? And in this force situation we're saying that the force is 0 there is no force acting on this thing, there's no external force acting this thing yet all the other ones but you're not pushing on so the only other force I'm missing here is friction, without friction and without this external force this block would slide down because MGx is the only force along the block along the plane if the block would slide down without friction that means that we're going to have friction going up to try to stop the block and actually we don't know if that static friction or kinetic friction yet we don't know if the block moves or not so let's check it out first cool, so this is the basic free body diagram for this first situation the other situations will have similar free body diagrams. LetÕs calculate MGx and MGy, so MG is mass 3 gravity 10ish I'm going to round it to 10 just to make our lives easier, again you probably have to use 9.8 unless the professor lets you do 10, MGx is MG sine of theta remember X goes with cosine if the angle is in the correct place against the X axis against the horizontal but it's opposite for inclined plane X goes with sine for inclined plane it flips if you do this you get MG which is 30 sine of 30 and if you plug this is the calculator this is exactly 15. MGy is 30 cosine of 30 and if you plug this in the calculator it rounds to 26, OK? Normal I'm sorry Newtons and Newtons rounds to 26, normal in this case is the same as MGy because these are the only two forces in the Y axis so they have to exactly cancel each other so normal is 26 as well these are all the forces, so now let's calculate the acceleration in this particular situation here, to find acceleration on a force problem we're going to use F=MA remember acceleration is acceleration in the X axis because there is no acceleration the Y axis in these inclined plane problems, sum of all forces in the X axis equals MAx and the only two forces I have are MGx and friction let's actually calculate friction real quick I forgot to do that, friction is mu normal, mu is 0.8 and normal is 26 and if you doing this it's roughly 21, it's really close to 21 Newtons, alright? So MGx is 15 and friction is 21 and right away before I even try to plug in numbers I hope you realize that the system actually doesn't move at all, so don't get caught up trying to plug in numbers you would plug in let's say positive 15 plus negative 21 and if you did this you would actually find an acceleration but this is wrong because this 21 here in this case would be your static friction maximum as well as your kinetic friction because it's the same coefficient, both coefficients are the same so both forces are the same which means that there is just not enough force for this to move so I'm going to say here that since MGx is less than friction static max the acceleration of the system is 0 and block doesn't even move, OK? Doesn't move, it doesn't accelerate, right? So that's what happens there with part A you just have to realize that there's not enough force to get this block moving. Now for Part B I'm saying that there is a force that's pushing this block down the plane there's a force that's pushing this block down the plane so it's going to look something like this, I have MGx doesn't change it's 15, there's an applied force here down the plane of 15 as well OK and if both forces are down friction is going to oppose the direction of motion the direction of things would move without friction so friction's going to go this way, and friction is 21, now check it out because I have enough force these two combined for 30 because there's enough force to overcome friction, you do overcome friction you start moving which means instead of having a static friction we now have kinetic friction they happen to be the same number but it's a different type of friction, it's now kinetic friction, OK? Those are all the forces that matter, those are forces in the direction of motion the acceleration will not be 0 because you have enough forces to overcome friction and we will move, to find the acceleration we write sum of all forces is equal to MA OK. The forces are.....this is the direction of positive I have these two guys as a positive this as a negative so if you want to do this really quickly you can just say I got 15+15+-21=MA, the mass is 3 and then A is what we want, this is 30-21=9 so the acceleration is 3 meters per second squared, Alright that's it for that part we got the B down let's do C D and E and part C we're going to push up the plane with 10 Newtons. So, let's keep the block in this plane like this and I have an MGx here of 15 you don't have to recalculate MGx because the angle didn't change so MGx won't change right so we just leave it alone and I have a force this way of 10 OK. Now the direction of friction, friction has to be the last force that you put in it and the reason for that is because the depending on which way you're pushing friction might be different, friction will oppose the direction the motion would have without friction so forget friction for now just between these two this is a greater force which means this thing would go that way, OK? And I'm going to call that the direction of positive therefore it means that friction is going the other way, OK? It means friction therefore must be going up the plane over here I'll make it blue and the friction maximum is or the friction is 21, right? Is there enough force for this to move? Remember what I just said without friction the system would be going down, now we introduce friction, friction is up so the question is the 15 which used to be enough force to make this go down is that still enough force now that we introduce friction? And the answer is No right so this is actually not going to move, so since MGx which used to be the leading force was greater than 10 but it's less then F plus little F so that means we don't move at all the acceleration is 0, OK? It was not enough force to get you to move. Let's go to part D Now, part D says you're pushing with 25 up the plane OK you push it 25 up the plane let's draw that here, force is twenty five, MGx is fifty so in the previous question the stronger force was the 15 but now the stronger force is the 25 without friction it's going to move in the direction of the strongest force which means friction will come in against that OK and that's what makes this question weird and that's why I want to make sure that I talked about this so you would see it which is friction will depend on the other forces right? So, if the net force is up friction is opposing that, it's now friction is 21 this was the leading force this was a leading force, now we've added a maximum friction of 21 is the leading force sufficient to overcome all the other forces? And the answer once again is no, right? even though it's going up it's still not enough force since MGx is less I'm sorry since the leading force F is less than MGx plus little friction your acceleration is 0 Not enough force again what's tricky here is it can't just memorize there's no super easy way to do this there's no one solution one size fits all solution you have to interpret the direction of friction depending on the other forces and see if there is still enough force to cause for there to be motion. Let's do an example part E, part E is 40 and we want to know the acceleration it's 40 up the plane, OK? So first we pushed with a 10 the bigger force was MGx down but friction meant that it didn't move then we moved we pushed with a force that was bigger than MGx but it was still wasn't enough because friction now flips to the other side here I have a force of 40, 40 is certainly more than MGx of fifteen because 40 is more I'm going to call that the leading force, right? Let me get this 40 out of here and that means that friction again will be going down because friction opposes the net force and friction will be going with 21, the question here is this 40 enough to go against the other two forces? And it is because these two forces here add up to 36 so 25 was not enough, these two forces that up to 36 so 40 is enough OK, so forty is enough since F is greater than the two forces going against it the acceleration will not be 0 which means I'm going to use F=MA to find it, Sum of all the forces in the X axis equals MA we're almost done. Just for this one since we're going to be going up the plane I'm going to call that the direction of positive since that's the direction of the winning force so I'm going to say I have plus 40 and then opposite to it I have -21, -15 and the mass is 3 and then the acceleration here, so I have 40-36=3A so a 4/3 or 1.33 meters per second squared. OK so I know this was a very long question, a little painful but again I wanted to show you all the possibilities that you could have and in a weird problem like this right this literally covers everything right that's it for this one.

Example #2: Incline with Friction

**Transcript**

Alright, so here we have a block it gets to the bottom of an inclined plane with 15 meters per second, so I'm going to say that the initial velocity at the bottom here is 15, the incline makes an angle of 25 degrees with the horizontal right there and the coefficients of friction are 0.4 so that's mu kinetic is 0.4 and mu static is 0.8, we want to know the block's acceleration so this implies it is the block's acceleration as it's going up the plane so the block's going to move up the plane, and it eventually stops because it's going against gravity and it's going against kinetic friction since it's moving so it eventually going to stop so the initial velocity here is 15 and the final velocity here is 0 if you wanted to draw a sketch of...not a full free body diagram but just a quick sketch of the forces you would have MG down here, you would have MGx is always down the plane the normal up the plane, MGy into the plane and you also have friction, you friction because of the coefficients of friction that is given to you which way do you think friction would be? And I hope you're thinking that since we're moving up I'm going to call it a positive direction that friction will be opposite to that and it is, it's the kinetic friction opposite to you and to find the acceleration we're simply going to use F=MA, sum of all forces along the X axis here is a MAx or simply just A, alright? And then the forces are as you're going up there's only two forces here in the X axis which are these two, a lot of people or a good amount of people get confused with questions like this and they think that since you're moving up the plane there has to be a force going up the plane and that's just wrong you don't need a force to be pushing you up the plane for you to move up the plane you go up the plane because you had an initial velocity and because of inertia you just keep going, right? It's because at some point before you got to the bottom of the plane somebody pushed you, you got 50 meters per second and then you go up the plane but there are no forces going up OK? One way that you might detect that is that if you think there's a force going up all the sudden you can't put a finger on it you can't figure out which force that is or what number magnitude that force has and that's because it doesn't exist, OK? So, the only two forces here and I'm going to say that the forces are where the forces are negative MGx and negative friction kinetic It's kinetic friction it's a rubbing friction because you're sliding the block is sliding this equals to MA, OK? We want to find acceleration here so what I'm going to do is I'm going to expand these, now one way you could do this you could calculate MGx calculate friction and just plug in the numbers then I'm going to expand this MG sine of theta minus friction now friction mu normal and in an inclined plane normal is the same as MGy so it's mu MGy and MGy is MG cosine of theta, right? You should memorize that MGy is MG cosine of theta if the angle is down here and little F becomes all of this OK, so minus mu MG cosine of theta and just to be completely clear here, this mu is mu kinetic and that equals to MA if you do this the masses will cancel which is and I can combine the Gs, OK? There's also two negatives here so you can factor the negative G sine of theta plus mu kinetic cosine of theta that is our acceleration OK? If you are having a hard time seeing how this goes from here to here you can work it out yourself slowly and see how these two are actually the same cool? So that's sort of the expression for I can plug in numbers or again you could have just found a calculator these individually and plug it in there maybe that would be a little bit easier whatever you like OK. So A equals negative 10 sine of 25, I have it here sine of 25 is 0.42 plus mu which is 0.4 because it's mu kinetic cosine of twenty five is 0.91 and I calculated this earlier it 7.84 meters per second squared and I got a negative and it should make sense that I get a negative because I said that this is the direction of positive right here up the plane and that's arbitrary I just said well if it's going up let's call that positive direction and then therefore the acceleration is going against you because you're slowing down so it's negative, OK that's it for part A. For Part B it says once you get to the highest point will it move back down? What happens at the highest point? Well at the highest point final over here at the highest point there are our velocity will be 0, OK? I'm going to say the V top is 0 question is will it move back down? In other words, is there going to be an acceleration but instead of calculating the acceleration we just want to yes or no question and basically the idea is are there enough forces or is there enough force pulling it down to overcome static friction once you get to the top as you're moving up you're going against kinetic friction back down this way, OK? So, as you're going up you have kinetic friction once you get to the top and you stop V equals 0 now what do you have? you have an MGx pulling you down and you have static friction this way. Remember friction is almost always Anyway friction is almost always opposing motion and if the box stops again this a box it's not a ball, so let's draw a little box if the box stops at the top in now switches direction this way so friction is going to going to go this way and it's going to try to keep it from moving and to figure out whether or not it moves we have to compare the forces, OK? I can calculate MGx If you had already calculated it here it's the same number but if you calculate MGx you get the mass is 10 gravity is 10 and then it is the sine of 25 and if you do this you get a 42.3 OK? all we're doing is comparing the forces, friction static is mu static normal mu static is 0.8, normal is right here all of this here is normal MG cosine of theta so MG cosine of 25 and again feel free to solve this as slowly as you have to and if you do those the answer is 72.5 Newtons, so this block will not move down, OK? I'm going to write here since MGx is less than the static friction (the maximum static friction) the acceleration is 0 so the block does not move let me disappear here the block does not move so the answer's no it doesn't move, OK? So that's interesting as are going up the plane as you're going up the plane there is kinetic friction pulling down the second you get to the top it switches to static friction because for an instant of the second you're not moving and then you actually stay there and you don't fall because static friction is too big and MGx is not able to overcome it, alright? So that's it for this one hopefully you'll make sense let me know if you have any questions.

Practice: A 5 kg block reaches the bottom of a long inclined plane with 20 m/s. The incline makes 53° with the horizontal, and the block-incline coefficient of friction is 0.4. What total time does it take the block to go up and back down?

Practice: A 10 kg block is pushed against an inclined plane with a force F that is perpendicular to the plane, as shown. The incline makes an angle of 53° with the horizontal, and the coefficients of friction between the block and the incline are 0.5 and 0.6. What minimum force F is needed to keep the block from sliding down the incline?

Consider the arrangement shown in the figure. A box of mass m = 5.00 kg is pushed up a slope of inclination angle θ = 25° by an applied force Fapp. The coefficients of static and kinetic friction between the box and the inclined plane are μs = 0.30 and μk = 0.20 respectively. Make a force diagram showing all the forces acting on the box. Your diagram should have four forces in it. With the help of your diagram, determine the magnitude of the normal force acting on the box.

Consider the arrangement shown in the figure. A box of mass m = 5.00 kg is pushed up a slope of inclination angle θ = 25° by an applied force Fapp. The coefficients of static and kinetic friction between the box and the inclined plane are μs = 0.30 and μk = 0.20 respectively. Suppose the applied force has magnitude Fapp = F1 = 26.0 N. With the help of your diagram, determine the static frictional force acting on the box, assuming that the given applied force is not enough to start the box moving.

Consider the arrangement shown in the figure. A box of mass m = 5.00 kg is pushed up a slope of inclination angle θ = 25° by an applied force Fapp. The coefficients of static and kinetic friction between the box and the inclined plane are μs = 0.30 and μk = 0.20 respectively.
Calculate the maximum static frictional force that can exist between the box and the inclined plane. Note that your result should be larger than your answer to part b. Using your result for the maximum static frictional force, determine the minimum Fapp required to get the box moving.

Consider the arrangement shown in the figure. A box of mass m = 5.00 kg is pushed up a slope of inclination angle θ = 25° by an applied force Fapp. The coefficients of static and kinetic friction between the box and the inclined plane are μs = 0.30 and μk = 0.20 respectively. Suppose now that Fapp is increased to the value Fapp = F2 = 40.0 N. Calculate the resulting acceleration of the box up the plane.

A box is at rest on a variable incline with μs = 0.4 and μk = 0.3. If the box has a mass of 2 kg, what is the maximum incline angle of the slope so that the box doesn't slide down?
A) 23.6°
B) 66.4°
C) 21.8°
D) 16.7°

A 5.00 kg block is pulled up an incline that is inclined at 36.9° above the horizontal by a horizontal force F = 90.0 N. The coefficient of kinetic friction between the block and the surface of the incline is μk = 0.20. If the block starts from rest at the bottom of the incline, how long does it take the block to travel 14.0 m to the top of the incline?

A box on a sled slides down a long, snow-overed slope. The hill slopes at a constant angle θ. There is a small coefficient of kinetic friction of 0.16. Find an expression for the acceleration.

A 25-kg picnic cooler falls off the bed of a flatbed truck on the entrance ramp of the PG5 garage and proceeds to slide down the ramp at a constant speed. Architectural drawings of the garage show that the ramp is inclined at an angle of θ. Draw the appropriate freebody diagram and then find an algebraic expression for the coefficient of kinetic friction. Simplify the result as much as possible.

A book is at rest on an incline as shown below. A hand, in contact with the top of the book, produces a constant force Fhand vertically downward. The following figures show several attempts at drawing free-body diagrams for the book. Which figure has the correct directions for each force? The magnitudes of the forces are not necessarily drawn to scale.

A box on sled slides down a long, snow-covered slope. The hill slopes at a constant angle θ. There is a small coefficient of kinetic friction of 0.19. Find an expression for the acceleration.

The suspended 2.8 kg mass on the right is moving up, the 2 kg mass slides down the ramp, and the suspended 8 kg mass on the left is moving down. There is friction between the block and the ramp. The acceleration of gravity is 9.8 m/s2. The pulleys are massless and frictionless. What is the tension in the cord connected to the 8 kg block?
1. 42.5958
2. 33.0193
3. 45.1758
4. 39.8194
5. 46.3585
6. 36.7226
7. 43.7416
8. 35.4743
9. 38.1757
10. 42.0574

A box sits on a ramp with an adjustable inclination angle that is slowly increased. Just as the angle reaches 36°, the box begins to slide. What is the coefficient of static friction between the box and the ramp?A) 0.58B) 0.65C) 0.73D) 0.81E) It depends on the mass of the box.

A 5.00 kg block is pulled up an incline that is inclined at 36.9° above the horizontal by a horizontal force F = 90.0 N. The coefficient of kinetic friction between the block and the surface of the incline is μk = 0.20. What is the magnitude of the friction force on the block?

A mass M is initially held at rest at the top of an incline of length L, with slope forming an angle α with the horizontal (see figure). When the mass is released, it starts sliding down the incline. The coefficient of kinetic friction between the mass and the incline is μk.a) Draw a free body diagram for the mass.b) Find the acceleration of the mass.c) Find the time it take the mass to reach the bottom of the incline.Write your results in terms of M, L, α, μk, and g. Check the units/dimensions for each answer.

A block lies on a plane raised an angle from the horizontal. Three forces act upon the block:, the force of gravity; , the normal force; and , the force of friction. The coefficient of friction is large enough to prevent the block from sliding (Intro 1 figure).Now you are going to ignore the general rule (actually, a strong suggestion) that you should pick the coordinate system with the most vectors, especially unknown ones, along the coordinate axes. You will find the normal force,, using vertical coordinate system b. In these coordinates you will find the magnitude appearing in both the x and y equations, each multiplied by a trigonometric function.(A) Because the block is not moving, the sum of the y components of the forces acting on the block must be zero. Find an expression for the sum of the y components of the forces acting on the block, using coordinate system b.Express your answer in terms of some or all of the variables , , , and .(B) Because the block is not moving, the sum of the x components of the forces acting on the block must be zero. Find an expression for the sum of the x components of the forces acting on the block, using coordinate system b.Express your answer in terms of some or all of the variables , , , and .(C) To find the magnitude of the normal force, you must express in terms of since is an unknown. Using the equations you found in the two previous parts, find an expression for involving and but not .

Two blocks of unequal mass are connected by a string over a smooth massless pulley with m2>m1. The coefficient of kinetic friction between mass m2 and the incline is (mu)K and the incline makes an angle theta with respect to the horizontal. (a) Evaluate the work done by gravity in moving the blocks a distance d. (b) Evaulate the work done by the frictional force in moving the distance d traveled. Is it positive or negative? (c) Using your answers from "a" and "b" evaluate the speed of the of the blocks after they move a distance d traveled. (d) Find an expression between masses m1 and m2 such that mass m2 slides down the incline at constant velocity

A large crate filled with physics laboratory equipment must be moved up an incline onto a truck.1. The crate is at rest on the incline. What can you say about the force of friction acting on the crate?A. the frictional force points up the inclineB. the frictional force points down the inclineC. the frictional force is zero2. A physicist attempts to push the crate up the incline. The physicist senses that if he applies slightly more force the crate will move up the incline but cannot muster enough strength to get the motion started. What can you say now about the force of friction acting on the crate?A. the frictional force points up the inclineB. the frictional force points down the inclineC. the frictional force is zero3. The first physicist gets a second physicist to help. They both push on the crate, parallel to the surface of the incline, and it moves at constant speed up the incline. How does the force exerted by the two physicists on the crate compare with the force of friction on the crate?A. force of 2 physicists < force of frictionB. force of 2 physicists = force of frictionC. force of 2 physicists > force of friction

A block lies on a plane raised an angle from the horizontal. Three forces act upon the block:, the force of gravity; , the normal force; and , the force of friction. The coefficient of friction is large enough to prevent the block from sliding (Intro 1 figure).Which forces lie along the axes of the coordinate system b, in which the y-axis is vertical?

A block lies on a plane raised an angle from the horizontal. Three forces act upon the block:, the force of gravity; , the normal force; and , the force of friction. The coefficient of friction is large enough to prevent the block from sliding (Intro 1 figure).Consider coordinate system a, with the x axis along the plane. Which forces lie along the axes?

To determine the coefficients of friction between rubber and various surfaces, a student uses a rubber eraser and an incline. In one experiment, the eraser begins to slip down the incline when the angle of inclination is 36.0° and then moves down the incline with constant speed when the angle is reduced to 30.0°. From these data, determine the coefficients of (a) static friction, and (b) kinetic frictionfor this experiment.

A crate is given an initial speed of 3.0 m/s up the 28° plane shown in the figure. Assume μk = 0.16.(a) How far up the plane will it go?(b) How much time elapses before it returns to its starting point?

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