Practice: Your hand moves a horizontal distance of 1.6 meter while you throw a 0.140-kg baseball horizontally. If the ball leaves your hand at 35 m/s, calculate: (a) the work done by you, and (b) the average force you exert on the ball.

Subjects

Sections | |||
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Intro to Conservation of Energy | 52 mins | 0 completed | Learn |

Energy with Non-Conservative Forces | 45 mins | 0 completed | Learn |

Escape Velocity | 12 mins | 0 completed | Learn |

Conservative Forces & Inclined Planes | 50 mins | 0 completed | Learn |

Motion Along Curved Paths | 107 mins | 0 completed | Learn |

Energy in Connected Objects (Systems) | 31 mins | 0 completed | Learn |

Solving Projectile Motion Using Energy | 33 mins | 0 completed | Learn |

Springs & Elastic Potential Energy | 63 mins | 0 completed | Learn |

Force & Potential Energy | 22 mins | 0 completed | Learn |

Example #1: Non-Conservative Problems

*Calculation Error

d=75m

**Transcript**

Hey guys, we're now going to do some energy problems where there will be work done by non conservative forces, let's check it out. So, if you remember, if either friction or an applied force or an external force where I refer to as the force done by you, if either one of those two forces do work on an object mechanical energy will not be conserved then you can see this from the conservation of energy equation, this is initial mechanical energy, this is the final mechanical energy and if this is 0 the energies are the same but if this is not 0 then the energies will be different, okay? So, let's check this out. So, even though the mechanical energy will not be conserved you can still use the conservation of energy equation. So, I put an asterisk here because even though we refer to that as the conservation of energy equation it works for problems where energy is not conserved either, okay? So, basically irrespective of whether or not energy is conserved, we can always use the main equation as a starting point, some of the simplest problems we're going to have or just horizontal motion problems and we'll do a few. So, I can explain I can show you guys how this stuff works, let's check it out. So, I have a 10 kilogram block initially at rest on a level smooth surface. So, let's put a little 10 block here, it's initially at rest, so the initial velocity 0, on a level horizontal smooth no friction surface, okay? The block is pushed with a constant horizontal force so the force is horizontal and it's a constant 20 and you do this for a distance of 5 meters, okay? So, it starts here, I push it for a distance of 5 meters and until it gets over here. Now, when I calculate its final speed, in other words, how fast it will be moving after 5 meters. So, if this is initial this here 5 meters later is final and I want to know what is V final, okay? So, we can use the energy, conservation of energy equation to do this, there's an object moving and we can do that. So, kinetic initial plus potential initial Plus work non-conservative equals kinetic final plus potential final, is there a kinetic energy in the beginning? There isn't because the object is not moving, is there potential energy in the beginning? There isn't because the object is on the floor so the height is 0, is there work done by non conservative forces? Well, the work done by non conservative forces is the work done by you, which is an apply or external force plus the work done by friction, there is no friction because we're talking about a smooth surface but you are pushing on the object over a distance. So, there is work done by you. Remember, when you want to calculate the work done by an external force you can use the general work equation, which is F, D, cosine of theta. So, let's use that in here, the work done by you is just F, D, cosine of theta, we know the force, that's a 20, we Know the distance it's a 5, and remember theta here is the angle between the displacement and the force, in this case we're pushing, we're pushing to the right horizontally and then this object moves to the right so the angle between these two is 0, okay? So, and the cosine of 0 is simply 1. So, this becomes F, D. Remember, when the force is perpendicular to the displacement the work simplifies to F, D, and that's what's going to go right here. So, the work done my non-conservative forces is F, D, do I have kinetic energy at the end? I do because this object is moving and I want to know how fast it's moving. So, we have that, let's replace that with half m, V final squared and there is no potential energy because the object is still on the ground level, okay? So, this is the final equation you get here or the equation that you get after you cross out all your energies and leave just the ones that exist, okay? And we're solving for V final and we have all the other numbers, so the force is 20, the distance is 5, half 10 v final squared, if I move around all the numbers I'm going to get, this is 100 times 2 is going to be 200 divided by 10 is going to be 20. So, V final squared equals 20. So, V final is the square root of 20, which is roughly 4.5 meters per second but the most important part is obviously the setup, once you get here, the rest is just algebra, okay? So, that's it for the first example.

I'm going to jump in a second example here and we're going to solve something similar, so it says here, a block of unknown mass. So, we don't know the mass but that's okay because as you've seen with a lot of these energy questions the mass will cancel anyway, it's sliding on a flat surface edge or a flat surface with 30. So, he has an initial velocity of 30, when it enters a long rough patch, okay? So, let's draw its a rough patch, which means there is friction and it's going to kinetic friction. So, we're going to draw it like this to represent friction, if the coefficient of friction is 0.6 calculate the block's stopping distance, in other words, how long will it move along this path until it stops, if I'm asking for stopping distance it implies that it stops. So, I'm just going to put that somewhere here, let's say the final velocity will be 0 and I want to know what the stopping distance is, if my velocity is 30 before I enter the friction path it means that it's going to be 30 all the way up until right there so this is sort of my initial .point, okay? So, I can say that it's 30 before it enters, 30 just as it enters and this is the stopping distance that I want to know, from here to here what is the distance, d or Delta x, and I know that in this interval the coefficient of friction is 0.6, okay? Again, we can do, we can solve this with the work, with the energy equation, with the longer energy equation and the reason we can do that is because is because, when I write the energy equation, let me put here, this guy here, work non conservative is made up of two parts, work non conservative is made up of the work done by you plus the work done by friction, since this is a problem of changing velocities due to friction, velocities are kinetic energies and friction is part of this, this guy here, then we can solve this, okay? So, all the elements of this problem are in the energy equation. So, let's start, is there kinetic energy in the beginning? there is because there's a velocity, is there potential energy in the beginning? There is not because if this thing is on the on the ground level, is there work done by non conservative forces? So, the work done by you is 0 because you're not doing anything, you're just watching this thing move, the work done by friction does exist because as these objects moving to the right friction is pushing it to the left. So, friction is slowing the block down it's consuming its energy, right? So, there is work done by friction and the work done by friction, if you remember the work done by kinetic friction is negative friction distance, okay? So, we do have this, I'm going to replace this here, half m, v squared initial plus negative friction distance, let's keep going here to the other side, is there kinetic energy at the end? The answer is no because it comes to a stop, is there potential at the end? There's no potential at the end because you're still in ground level. So, you get this interesting thing where there's a bunch of stuff on the left and then the right is just is 0, that might seem weird at first but I can move this to the right and then I have positive equals positive. So, half m, Vi squared equals friction d, we're looking for, we're trying to solve for d, one thing I can do is calculate friction using the friction equation and then just plug in the numbers, I have all these numbers and I have all the necessary numbers to find friction, friction if you remember, is mu normal. So, I could just calculate this number here and put it there or I can plug in the variables for it, I'll show you. So, I have m, G is this way, normal is this way, because there are no other forces and the y-axis and they're only forces in the x axis with this object moving sideways, since there's no force in the y axis except for normal and m, g they have to cancel each other. So, they equal each other. Remember, that. So, mu k, normal becomes mu k, m, g, if you do that and you plug that in here you see that the masses will cancel, which in this problem you have to do this because you have an unknown mass. So, there's no way around it, you have to write out the entire expression for friction, plug it into the equation so that the masses could actually cancel. So, let's do that half m, vi squared equals, friction is mu k, m, g, distance is what I'm looking for, the masses cancel. So, we're almost done. Now, we just have to solve for d, d will be v initial squared divided by 2, I'm going to divide both sides by mu k, m, g, so mu, m, g, the initial velocity is 30, friction is 0.6, the coefficient of friction is 0.6, gravity we're going to round that to 10 and when you plug all of this into the calculator you get 45 meters, 45 meters. Alright, that's it for this one.

Example #2: Non-Conservative Problems

**Transcript**

Alright, so I have one more point to make and one more example to show you guys. So, if you're being asked to find the work done by a force, we did a lot of this earlier, you're typically going to use the work equation, that's what you should do most of the time, right? If you get a work question you should go straight for the work equation and try to plug in numbers but sometimes you're not going to have enough information to be able to calculate this you need f d and theta, in some cases you're not going to have all that and that's because you're not supposed to use this equation instead you're supposed to be using the conservation of energy equation. So, I'd like to think in the conservation of energy equation also as a backup for to calculate work in cases where the work equation involves work no pun intended. So, let's try this example here and you'll see what I mean. So, here we're pushing a 5 kilogram block, initially at rest. So, here's a 5 kilogram block it's, initially at rest and you're going to push it along a rough horizontal surface. So, surface is horizontal and rough. Notice how it doesn't say whether you push it horizontally or at an angle or whatever but, let me just draw it this way though I don't know exactly, which way you're pushing, which is the first problem here, we don't really know the angle and what you're pushing this thing but you do this in the horizontal surface since it reach, until it reaches a velocity of 20. So, I'm going to say v final equals 20 and it says if the kinetic energy dissipates 200 joules of energy, how much work did you do. So, if kinetic energy or kinetic friction rather dissipates 200 joules of energy, if friction dissipates energy it means that the work done by friction kinetic is negative 200 joules, friction is slowing the box down or at least wasting some of the mechanic's work out of the box and then changing it into heat. So, if the work done by friction is negative 200, how much work did you do, your first thing might be and I think it's it's a good instinct to go straight for the work equation F, d, cosine of theta, that's what I would do, right? Try this first, it's simple, more straightforward, but then we realize we don't know F, we don't know the distance and we don't know the angle, you don't know anything, so the only other thing you could do at this point is use the conservation of energy equation, okay? So. Notice how the conservation of energy equation can be used for new problems but also to solve work problems that we were doing in a different way earlier, okay? So, k initial plus q initial plus work non-conservative equals k final plus U final is their kinetic energy in the beginning? There is not because this thing is not moving, what about potential energy? It's on a flat surface it has no height so the potential is a 0, what about the work done by non conservative forces? The work done by non conservative forces, I'm going to expand it over here is the work done by you plus the work done by friction, this is what I want and this is given to us. So, plug that in, in just a second their kinetic energy at the end? There is kinetic energy at the end because we have a velocity. So, I'm going to write half m, V final squared here and there is no potential energy because we're still on the ground, okay? The work done by U is what we're looking for, the work done by friction is negative 200 half, the mass is 5 and the velocity is 200. So, almost done, if you calculate this you get 400 divided by 2, that's 200 times 5, that's 1,000 so the work done by U is 1,000 plus the 200, on the other side the work done by U is 1,200 joules, in other words, if this box went from here to here and achieve the final velocity of 20 while friction was going against it, it must have taken you this much work to be able to accomplish that, okay? Again very, very nice that we get to use one big equation to solve so many of these problems. Alright, that's it for this one.

Practice: Your hand moves a horizontal distance of 1.6 meter while you throw a 0.140-kg baseball horizontally. If the ball leaves your hand at 35 m/s, calculate: (a) the work done by you, and (b) the average force you exert on the ball.

Practice: A 500-kg load is originally at rest on the floor. A crane pulls the load vertically up on the box with a constant 7,500 N until it reaches a height of 20 m. Calculate the speed of the load once it reaches 20 m.

Practice: A 800-kg car leaves a skid mark of 90 m in stopping from 30 m/s. Calculate the car-road coefficient of friction.

Example #3: Energy with Resistive Forces

**Transcript**

Hey guys. So, now that we've seen some basic examples of using the conservation of energy equations to solve problems I want to show you how this works in situations where we have the resistive forces, these are basically friction, a different types of friction such as air friction or wire resistance stuff like that, let's check out. So, we can use the conservation of energy equation to solve problems with resistive forces such as air resistance or water resistance, many, most physics problems ignore air resistance and most physics problems don't deal with water at all but we can use energy to solve some of these basic problems, right? And resistive force is going to work just like kinetic friction, if you have a box sliding to the right kinetic friction is pulling the object, the box to the left. So, these are resistive forces, it's in the name, they resist motion. So, they will always oppose motion, their direction will be opposite to the direction of your speed or velocity. The work then by kinetic friction, if you remember, we could have simplify that as the negative friction d and we can do the same thing for all these resistive forces. So, it's just going to be negative F, D, just the same, okay? So, the work done by any resistive force is negative f, d where d is the distance. So, let's do an example here, we have a 2 kilogram object released from a height of 80 and it reaches the floor with 30. So, let's draw that, 2 kilogram object and it was released from a height of 80, let's just say, this is 80 meters and it reaches the bottom over here, with a final speed of 30. So, I'm going to put a V final equals 30, just draw a little floor there, the initial velocity since this is released, its released from rest and obviously follow that way. Alright, now if you want to calculate this and we can do this using the energy equation you would see that objects release from 80 at the absence of air resistance will actually get to the bottom with about 40 meters per second but this one is getting there a little slower and that's because there's some air resistance and it says, what is the amount of work done by air resistance, so the first thing I want to know is what is the work done by air resistance. So, we can think of this as, what is the work done by the friction of air or air friction or air resistance, okay? And we can do, we can use the work, I'm sorry, the energy equation to solve this, the big long energy equation, k initial, u initial, work non-conservative, k final, u final, and one way you might be able to identify or realize you're supposed to use this equation is that work non conservative is made up of the work done by u, some sort of external force plus the work done by friction. So, that's what we want right here, Alright, let's go to the equation here, there's no kinetic energy in beginning because there's no velocity, it's at rest, is there potential energy in the beginning? There is, there's potential range because I have a height, is there work non conservative? Well, the work done by you is 0 because you're just watching this thing fall, from the top to the bottom you don't do anything, there is the work done by friction on, because when I asked you to find the work done by air resistance, it's implied that there is some work done by air resistance, okay? That's, what we're looking for at the end is there kinetic energy? There is because just before hitting the ground you have a speed, in fact, that speed is 30 and is there potential energy at the end? And there is no potential is at the end because the height is 0 just before you hit the bottom, okay? So, we have these, these three terms here, one, two, three and let's write them all out, this is m, g, h, this is going to be replaced by just this guy, work of friction, okay? And, this is half m, v squared final. Now, usually you'd replace work of friction with negative f, d but we want this number here, we want to find the work done by friction. So, you leave it alone you don't expand it out otherwise you're going to be calculating something else that's not the work done by friction. So, leave it alone and all we got to do now, is move this to the other side, so the work done by friction is half, mass is 2, velocity at the bottom or speed at the bottom it's 30 minus m, g, h. So, m is 2, g we're going to use 10 just to make it a little faster and h is 80, if you do this you're going to get a negative 700, which at first you might be confused why we're getting a negative but hopefully you remember that the work done by resistive forces will always, always going to be negative, that's why we can simplify and say it's negative f, d. So, there should be a negative, okay? That's good that this thing is a negative, it means that friction is taking energy from the system.

Now, let's go for the second part here, we're done with the first one where's the average force of air resistance, okay? Now, the force of air resistance is variable but we're not going to get into that. So, in these problems really all you can do is find the average force of air resistance and that's because we have this equation here and the force of air resistance is this one. So, we can write out this equation and Part B is much easier to calculate this. So, I'm asking basically what is f. Well, the work done by f right here, we have it, it's negative 700 equals negative f and d is the distance that you fall because friction was acting on you throughout the entire fall. So, it's just the distance that you move, which in this case is your height with these 80 meters. So, this negative and this negative cancel, they become positive and you just have that friction will be 700 divided by 80 and I have that here, that's 8.75, okay? I might be wondering isn't friction negative? Well, friction is opposite to motion, right? But this number here ends up giving you the magnitude of friction and that's because the negative is already included here, okay? So, don't worry about the fact that this is positive, it should give you a positive. Alright, that's it for this one, I have a practice problem that's similar to this, it's another very typical example, we have a block being shot into a bullet, being shot into a block of wood or some sort of wooden wall and then it stops. So, you can use a very similar process, I want you guys to give that a shot.

Practice: A 10-g bullet hits a wooden wall with a horizontal 300 m/s. If the bullet penetrates the wall by 5 cm, calculate:

(a) the amount of energy lost by the bullet.

(b) the average frictional force that stops the bullet.

Example #4: Energy with Resistive Forces

**Transcript**

Alright, so here we have a diver who's jumping into the water and he enters the water with 10 meters per second and then eventually it stops, right? So, let's say there's a little platform here or whatever, we don't know how high the platform is, here's a jumper, doesn't really matter if he jumps up or down all that matters is that he eventually hits the water with a velocity of 10 meters per second, we know the average resistive force of water on the diver is 2,000, okay? So, the idea is you hit water with 10. So, you go down until you eventually stop, okay? Until you eventually stop throughout this distance here you have a frictional force, acting against you, of 2000 and I want to know how deep underwater will he reach, how deep underwater will he reach, so the idea that eventually you stop, it's implied in the question so the final velocity is 0, initial velocity is 10 and I want to know this distance here, what's this entire distance, okay? We're going to use energy as well, So, how you know to use energy, well velocities are changing, heights are changing and there's friction, right? Friction is part of the energy equation as well. So, we want to know what is the distance underwater. So, kinetic energy, kinetic potential, kinetic initial potential, initial work, non-conservative kinetic final plus potential final, what is their natural kinetic energy? we're going to go from here to here, what happened before it doesn't matter, okay? Is there kinetic energy at the beginning? there is. Alright, we know that the velocity is 10, is there potential in the beginning? Well, there is, this is the lowest point. So, I'm going to call this, I'm going to say that the height here is 0 okay, if the height here is 0 and there's a distance of d between these two points then the height here at the top of the water is d, okay? So, for example, real quick, we don't have to draw this, let's say this gap is 10 meters, if this height is 0 then this height here is 10, right? If the lowest one is 0 then the highest one is just the same as the gap between the two. So, but we don't know that number is, that's actually what I'm looking for. So, that's d is a work non-conservative, work non-conservative is the worth done by you, you're not doing anything you're just sort of watching this guy fall into the water but there is work done by friction, okay? The kinetic energy at the end of 0 because you reach final velocity of 0 and you go to the bottom, it might not be the bottom of the pool but it's the lowest point that you're at and we're going to call that 0, okay? So, we have these three energies on the left side of the equation, let's expand this, half m, V initial squared plus m, g, h initial plus the work done by friction. Now, remember that the work done by friction is negative F, d. So, if I want I can just plug in negative F, d and in this case I should do that because I'm not looking for the work done by friction, I'm looking for d, and if I'm looking for d, I have to make the show up in the equation, okay? So, I'm going to go like this. Alright, and instead of just going ahead and plug in a bunch of numbers, I'm going to actually first try to solve for the variables, one of the things we have to do here is remember, the initial height was actually d. So, I have to change that because we don't know that initial height, that is my unknown. So, it's m, g, d and just to be clear, I'm going to write here, h initial was d. Now, if you notice here, what happens is that this equation with two D's in it and the way to solve for d, I'm going to have to combine those two terms, I'm going to have to combine this term with this term and then move everything to the other side. So, let's start doing that, m, g, d, minus FD equals negative half m, V initial squared, if I'm combining the DS, I have to factor it out, d parenthesis m, g minus f okay, if you work backwards you see that that's exactly the same thing as this, just got to do it carefully equals to this and I just have to divide both sides by m, g minus f. So, d equals negative m V initial squared over m, g minus f. Now, just a matter of plug in the numbers, mass is 70, initial velocity is 10 that gets squared, the mass 70, gravity I'm going around that to 10 just to make things a little bit faster and friction will be 2000, it's the average frictional force, or resistive force on you, interesting is that down here you're going to have a negative and that's okay because it needs to cancel out with this negative so that you end up getting a positive, if you put all of this in a calculator, work it out carefully, you get a distance of 2.7 meters, which is pretty reasonable, it's about a little little less than 10 feet, you go in underwater pretty much, a good distance but that's a reasonable number, okay? So, that's it for this one, let me know if you guys have any questions.

0 of 8 completed

Example #1: Non-Conservative Problems

Example #2: Non-Conservative Problems

Practice #1: Non-Conservative

Practice #2: Non-Conservative

Practice #3: Non-Conservative

Example #3: Energy with Resistive Forces

Practice #4: Energy with Resistive Forces

Example #4: Energy with Resistive Forces

A 30 g object is dropped from a height of 50 cm and bounces off the ground. If after the bounce, the ball leaves the ground with 50% of the speed it hit with, how high will the ball bounce?

A small rock with mass 0.22 kg is released from rest at point exttip{A}{A}, which is at the top edge of a large, hemispherical bowl with radius exttip{R}{R} = 0.52 m (the figure ). Assume that the size of the rock is small compared to R, so that the rock can be treated as a particle, and assume that the rock slides rather than rolls. The work done by friction on the rock when it moves from point exttip{A}{A} to point exttip{B}{B} at the bottom of the bowl has magnitude 0.22 J.Between points exttip{A}{A} and exttip{B}{B}, how much work is done on the rock by the normal force?What is the speed of the rock as it reaches point exttip{B}{B}?Of the three forces acting on the rock as it slides down the bowl, which (if any) are constant and which are not? Explain.Just as the rock reaches point exttip{B}{B}, what is the normal force on it due to the bottom of the bowl?Between points exttip{A}{A} and exttip{B}{B}, how much work is done on the rock by gravity?

In an auto accident, a car hit a pedestrian and the driver then slammed on the brakes to stop the car. During the subsequent trial, the drivers lawyer claimed that he was obeying the posted 40 mi/h speed limit, but that the legal speed was too high to allow him to see and react to the pedestrian in time. You have been called in as the states expert witness. Your investigation of the accident found that the skid marks made while the brakes were applied were 280 ft long, and the tread on the tires produced a coefficient of kinetic friction of 0.30 with the road.In your testimony in court, will you say that the driver was obeying the posted speed? You must be able to back up your conclusion with clear reasoning because one of the lawyers will surely cross-examine you.If the drivers speeding ticket were $10 for each mile per hour he was driving above the posted speed limit, would he have to pay a fine?
If so, how much would it be?

Two identical arrows, one with twice the speed of the other, are fired into a bale of hay.Assuming the hay exerts a constant "frictional" force on the arrows, the faster arrow will penetrate how much farther than the slower arrow?

A baggage handler throws a 15 kg suitcase along the floor of an airplane luggage compartment with a speed of 1.2 m/s. The suitcase slides
2.1 m
before stopping.Use work and energy to
find the suitcase’s coefficient of kinetic friction on the floor.

In a truck-loading station at a post office, a small 0.200-kg package is released from rest at point exttip{A}{A} on a track that is one-quarter of a circle with radius 1.60 m (the figure ). The size of the package is much less than 1.60 m, so the package can be treated as a particle. It slides down the track and reaches point exttip{B}{B} with a speed of 4.60 m/s . From point exttip{B}{B}, it slides on a level surface a distance of 3.00 m to point C, where it comes to rest.What is the coefficient of kinetic friction on the horizontal surface?How much work is done on the package by friction as it slides down the circular arc from exttip{A}{A} to exttip{B}{B}?

A sled of mass m is given a kick on a frozen pond. The kick imparts to the sled an initial speed of v. The coefficient of kinetic friction between sled and ice is µk. Use energy considerations to find the distance the sled moves before it stops.

A 11.0 kg box is pulled by a horizontal wire in a circle on a rough horizontal surface for which the coefficient of kinetic friction is 0.300.(a) Calculate the work done by friction during one complete circular trip if the radius is 2.00 m.(b) Calculate the work done by friction during one complete circular trip if the radius is 4.00 m.(c) On the basis of the results you just obtained, would you say that friction is a conservative or nonconservative force?

A 62.0-kg skier is moving at 6.60 m/s on a frictionless, horizontal, snow-covered plateau when she encounters a rough patch 4.60 m long. The coefficient of kinetic friction between this patch and her skis is 0.300. After crossing the rough patch and returning to friction-free snow, she skis down an icy, frictionless hill 2.50 m high.(a) How fast is the skier moving when she gets
to the bottom of the hill?(b) How much internal energy was
generated in crossing the rough patch?

You drop a ball from a height of 2.1 m, and it bounces back to a height of 1.3 m.(a) What fraction of its initial energy is lost during the bounce?(b) What is the balls speed just before the bounce?(c) Where did the energy go?(d) What is the balls speed just after the bounce?

A 2.3 kg piece of wood slides on the surface shown in the figure. The curved sides are perfectly smooth, but the rough horizontal bottom is 33 m long and has a kinetic friction coefficient of 0.26 with the wood. The piece of wood starts from rest 4.0 m above the rough bottom.(a) Where will this wood eventually come to rest?(b) For the motion from the initial release until the piece of wood comes to rest, what is the total amount of work done by friction?

A small block with mass 0.0425 kg slides in a vertical circle of radius 0.525 m on the inside of a circular track. During one of the revolutions of the block, when the block is at the bottom of its path, point A, the magnitude of the normal force exerted on the block by the track has magnitude 3.90 N. In this same revolution, when the block reaches the top of its path, point B, the magnitude of the normal force exerted on the block has magnitude 0.680 N. How much work was done on the block by friction during the motion of the block from point A to point B?

A smooth circular hoop with a radius of 0.500 m is placed flat on the floor. A 0.400-kg particle slides around the inside edge of the hoop. The particle is given an initial speed of 8.00 m/s. After one revolution, its speed has dropped to 6.00 m/s because of friction with the floor.(a) Find the energy transformed from mechanical to internal in the particle–hoop– floor system as a result of friction in one revolution.(b) What is the total number of revolutions the particle makes before stopping? Assume the friction force remains constant during the entire motion.

A sled of mass m is given a kick on a frozen pond. The kick imparts to the sled an initial speed of 2.00 m/s. The coefficient of kinetic friction between sled and ice is 0.100. Use energy considerations to find the distance the sled moves before it stops.

A uniform board of length L is sliding along a smooth, frictionless, horizontal plane as shown in Figure P8.79a. The board then slides across the boundary with a rough horizontal surface. The coefficient of kinetic friction between the board and the second surface is µk. (a) Find the acceleration of the board at the moment its front end has traveled a distance x beyond the boundary. (b) The board stops at the moment its back end reaches the boundary as shown in Figure P8.79b. Find the initial speed v of the board.

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