Practice: A 3-kg box is sliding on a smooth level surface with 8 m/s when it enters a 5 m rough patch. If the box has half its original speed at the end of the patch, calculate the coefficient of kinetic friction between the box and the surface.

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Intro to Energy | 15 mins | 0 completed | Learn |

Intro to Calculating Work | 36 mins | 0 completed | Learn |

Work By Gravity & Inclined Planes | 40 mins | 0 completed | Learn |

Work By Variable Forces (Springs) | 47 mins | 0 completed | Learn |

Net Work & Kinetic Energy | 39 mins | 0 completed | Learn |

More Work-Energy Problems | 39 mins | 0 completed | Learn |

Power | 20 mins | 0 completed | Learn |

Example #1: More Work-Energy #1

**Transcript**

Hey guys, let's now do some practice problems with the work energy theorem so in the first question here I have a 5 kilogram block that is initially at rest so let's draw this a 5 kilogram block initially at rest V initial = on a level meaning horizontal smooth meaning no friction surface if the block is pushed with a constant force of 20 so I'm going to draw that as a pull and I'm going to call that F=20 for the distance of 10 so distance of 10 I want to know what is its final speed? So over here it has a velocity initial of 0 I want to know after pulling it for 10, what is its final velocity, OK? You could do this using a combination of F=MA and then one of your motion equations one of the 3 motion kinematic equations or we can use the work energy theorem which is better because it's a single step so remember the work theorem says that the net work is the change in kinetic energy so remember that net work has to do or work in general has to do with F D cosine of theta and then this has to do with kinetic energy which is 1/2MV squared, obviously final initial but the idea is that I find a force and a distance and I'm looking for velocity I can use the equation, OK? So, the net work is the change in kinetic energy, alright? Now the only force that does work on this box is F I do have here MG going down and I have normal going up but remember they're perpendicular to the displacement therefore they don't do any work so the only work here is the work done by F which would be kinetic final - kinetic initial there is no kinetic initial because there is no velocity in the beginning, OK? The work done by F is F D cosine of theta equals K final which is 1/2MV final squared, V final squared is what I'm looking for and I have all the other numbers. The force is 20, the distance is 10, the angle is 0 remember that the angle here is between the force which is this way in the displacement which is also this way so the angle here is 0 degrees and the mass is 5 And that's what we have here this is going to be 200 and then when I multiply by this 2 here it's going to be 400/5=V final squared so V final ends up being 8.9 meters per second, thatÕs it let's keep going.

Example #2: More Work-Energy #1

**Transcript**

Alright so here we have how much work is needed to accelerate a 50 kilogram object from 3 to 8 so we can think of this is as box although it doesn't have to be a box it's any object moving in any direction and I want to accelerate it from 3 to 8 so if you do think it as a box, it's the box over here has an initial velocity of 3 and then I'm going to accelerate it to a final velocity of 8 and this happens you would imagine by pulling it this way since I drew the second point on the right and I want to how much work is done, notice that it's not being or it's not asking how much work is done by F just how much work period irresectable of how many forces there are so it's implied here is that we're asking for how much total work is needed? In other words what is work net? Now remember there's 3 ways you can find work net, work net could be the sum of all the individual works, it could be the work done by the net force if I can get that or the change in kinetic energy, in this problem since I'm giving you the velocities and you have the mass this is the best way and actually this one it's the only way to do this, OK? So, it's very straightforward it's just going to be K final - K initial and I have all numbers, 1/2MV squared - 1/2 MV initial square you can factor out the 1/2M if you want to it would look like this 1/2M (V final squared - V initial squared and you don't have to do that but you could, 1/2 the mass is 50 and the velocities are.... Setting up here the velocities are 8 and 3 and if you do this you get a total of 1375 Joules, so the net work is 1375 Joules just very straightforward just plug and chug into this equation that's it for this one.Alright so here we have how much work is needed to accelerate a 50 kilogram object from 3 to 8 so we can think of this is as box although it doesn't have to be a box it's any object moving in any direction and I want to accelerate it from 3 to 8 so if you do think it as a box, it's the box over here has an initial velocity of 3 and then I'm going to accelerate it to a final velocity of 8 and this happens you would imagine by pulling it this way since I drew the second point on the right and I want to how much work is done, notice that it's not being or it's not asking how much work is done by F just how much work period irresectable of how many forces there are so it's implied here is that we're asking for how much total work is needed? In other words what is work net? Now remember there's 3 ways you can find work net, work net could be the sum of all the individual works, it could be the work done by the net force if I can get that or the change in kinetic energy, in this problem since I'm giving you the velocities and you have the mass this is the best way and actually this one it's the only way to do this, OK? So, it's very straightforward it's just going to be K final - K initial and I have all numbers, 1/2MV squared - 1/2 MV initial square you can factor out the 1/2M if you want to it would look like this 1/2M (V final squared - V initial squared and you don't have to do that but you could, 1/2 the mass is 50 and the velocities are.... Setting up here the velocities are 8 and 3 and if you do this you get a total of 1375 Joules, so the net work is 1375 Joules just very straightforward just plug and chug into this equation that's it for this one.

Practice: A 3-kg box is sliding on a smooth level surface with 8 m/s when it enters a 5 m rough patch. If the box has half its original speed at the end of the patch, calculate the coefficient of kinetic friction between the box and the surface.

Example #3: More Work-Energy #2

**Transcript**

Alright, so now we're going to do a little bit more practice with the work energy theorem let's get going, so here I have I'm asking how much work is needed to throw a 200-gram ball? mass= 200 grams which is 0.200 kilograms with a horizontal 30 meters per second, how much work is needed to do that?

So notice that I'm asking how much work not how much work is done by a particular force just how much work in general is needed to do that and so the idea is that if you're throwing something you're throwing it horizontally so you're pushing it that way, now obviously there is an MG pulling you down which means part of what your hand has to do is actually also sort of hold this thing up this way in such a way that these two guys cancel but you don't have too much worry about the complexities of all the forces you just have to realize that once I say how much work is needed for something to happen but I don't specify exactly what force I want to know the work of for what this means how much total work, OK? How much total work is needed to throw a ball with the horizontal 30 meters per second? So, work net is what I'm looking for there are 3 ways I can find work net this one, this one and this one and it depends on the information I'm given, what am I given? I'm given the mass and I'm given the velocity, well delta K has a velocity in it, it has a velocity final and velocity initial So Delta K now you might be wondering what is the initial velocity? Well the initial velocity is 0, throwing implies it wasn't moving and then you threw it now it is moving so this means that the initial velocity is 0 and the final velocity is 30 if you have both initial and the final velocity you can be pretty safe there you're most likely going to have to use this equation here, this is what I want to know so all I have to calculate is Delta K, alright? So, this is K final- K initial but there is no K initial so it's just have 1/2 MV squared final and I can quickly calculate this 1/2(0.200) the final velocity is 30 squared and if you plug this all in you get 90 Joules is how much is needed for this to happen so it takes 90 Joules of energy, OK?

Part B here so that's part A... Part B here says if the ball's in contact with your hand for 0.5 meters what average force do you applied to it? Right so here's the idea the ball is here and it has a velocity of 0 now it's going to get a little bit more fancy, you're pushing the ball with an F and by the time it leaves your hand it leaves your hand with a velocity of 30, OK? As you're doing this your hand travels a distance of 0.5 meters which means that you are pushing on the ball for a distance of 0.5 meters and that tells me what the work and I can associate that with work, right? If you remember work is F D cosine of theta so whenever you see an F and a D you should be thinking of work, alright? I also have 2 velocities here, OK? And anyway, so let's get going so what is the average force? I want to know what is F? How might I find this? So, we said that this is the only force that's pushing this thing so F is effectively the net force acting on this, OK? I can say work net= 90 but I can also write another expression for work net which is F Net D cosine of theta = 90 and I want to find F Net or the throwing force, OK? So, the angle here F net is this way and the displacement is this way so the angle is 0 so I have F net is I don't know why I put a degree here this is just 90 is going to be 90 remember this is 1 / D D is 0.5 so the net force is 180 Newtons, OK? Net force is 180 Newtons so all we have to do is go back to one of these equations now it's important that you remember that work is F D cosine of theta because the first part of this problem we found W or work for the second part they gave us D and then they're asking for F so it's just a matter of tying the concepts together, I want to talk about one quick thing here is says the average force and that's because when you throw something the force is actually not average , right? Just like when there's a collision the force is not average you begin to accelerate your hand so you begin to push harder and harder and then when you're sort of almost done pushing and the object starts to sort of leave your hand the force is less or smaller so typically when throw something it looks like that but that's a little too hard for now to calculate this and in fact most of you won't have to calculate this even later on but so what we do is we can calculate the average force right there that's effectively what I got, OK? So anyway, it was just a matter of again plugging into this equation here for part B and plug it into this equation here for part A so I have now a few practice problems that I want us to continue doing so let's jump on that.

Practice: Interstate roads in the U.S. typically have a speed limit of 70 mph (31.3 m/s). How much work is needed to stop a 1,400 kg car moving at this speed? How much force is needed to accomplish this in 2 m?

Practice: To prevent a crash, a driver slams on the brakes, causing his car to skid a total distance of 80 m before stopping. The road-car coefficients of friction are 0.6 and 0.8. Calculate the car’s initial speed (the car’s mass is unknown).

Example #4: More Work-Energy #3

**Transcript**

Alright let's do a few more practice problems with the work energy theory so I have an 8-kilogram box that rests on a rough floor so let's put it here 8 kilograms rough floor so there's friction the coefficient fission of friction is 0.5 mu=0.5 and I want to know what is the work needed to move the box at a constant speed, constant speed means the acceleration is 0 means we are at equilibrium means that the forces will cancel, OK? So, this is part A, we'll do part B here on the side so the idea is if you're moving with a constant speed and there's friction let's say you're going to the right so friction will be to the left which means that you were pulling to the right this way friction kinetic, constant speed means that these two guys of the same magnitude, OK? And obviously here MG and Normal so you might be wondering if they have the same magnitude then maybe this box won't move at all, well the idea is at some point the box was already moving and now it just continues to move in a constant speed so we don't argue with how we got to where we are and what happened earlier we just go with what we're given, right? So, these two forces have to be the same it's asking what is the work needed to move at a constant speed? Constant speed according to the work energy theorem if your speed is constant remember your delta K=0 and your work net is 0 so you might be thinking the work needed to move is 0 but it's not what this question is asking is what is the work done by F, OK? What is the work done by the force? Now it's a little tricky because in other problems I told you that if the work if I'm asking you for the work done, the work needed for something to happen but I didn't specify which force you could assume that it was the net force and that's because in those problems we didn't know exactly how many forces there were so even if there was a force going against you, you just think of the whole thing at once, well as you move to the right a distance of 3 meters friction is doing negative work against you but the net work has to be 0 which means you have to do the same amount of work except positive as friction is doing negative work so that those two cancel, OK? So, what this question is asking is what is the work done by F so that you are able to accomplish constant speed, OK? I have no way of finding F if not through friction so I have to calculate friction to know what my F has to be and that's because to find the work done by F, I'm going to need to know what F is because it's F D cosine of theta, F goes this way delta X goes this way so the angle is 0 and the distance is 3, the cosine of 0 is just 1 but I do need to know the force and that force will be friction, force is friction kinetic friction is mu normal or mu normal cancels with MG so mu is 0.5 the mass is 8 and for gravity we're going to use 10 which makes this force 40 and that's exactly what goes here, so the work done by F to get this moving is a 120 joules which also means that the work done by friction was 120 Joules but in the opposite direction, OK? So here we want the work done by F to cancel out friction so that you move with a constant speed, now we are moving with a constant speed 3 meters in the vertical so if we're moving in the vertical it's because I'm pulling this way and I'm going to call F as well force going to gets here is MG, now there is a normal while it touches but as soon as it leaves the surface there is no longer a normal, OK? And that's where the problem starts when it just leaves the surface and it's going to move up with a distance this is usually D but I can call it A whatever I'm just going to use D to stay consistent with the other equations or I guess we'll call it both D and H = 3 meters, OK? Now how do I move with a constant speed, it's very similar to here if I have a constant speed my F had to equal my F K in the very beginning it was a little more to get it moving to overcome static friction but then I pulled with less force and I kept a constant force that's the same so that they cancel, same thing here to get off the ground have to pull with a little harder than MG but as soon as you pulled a little bit harder than MG and it get off the ground your force will be the same as MG and that's so that your acceleration is 0 and so you move with a constant speed, OK? And if I want to figure that out I can say the work done by F is. Force D cosine of theta, here the force is this way the displacement is this way these two guys are parallel to each other so the angle is 0 and the force that I need is MG, OK? The force is MG and for I'm going to call instead of D I'm going to call this H and the work needed is MGH which make a lot of sense this is the potential energy, you're giving this thing potential energy, another way that you could have thought about this is the work done by F is the work to lift so you might remember that the work to lift is MG Delta H, now H here is 0 so this H here implies H means H final so these two guys here are in this case the same exact thing, OK? So those are two ways you could have done this. The force is...Let's just plug in from this equation here so M is 8 gravity is 10 and the height is 3 so the answer is 240 joules of energy, what you might see real quick is that it takes more energy to lift this than it takes to move it horizontally, OK? And that's always going to be the case let's keep going.

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 2-kg block is originally at rest at the bottom of a smooth inclined plane that makes 37° with the horizontal. You push the block with a constant 20 N directed up the plane. Find the block’s speed after you push it 5 m up the plane.

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Example #1: More Work-Energy #1

Example #2: More Work-Energy #1

Practice #1: More Work-Energy #1

Example #3: More Work-Energy #2

Practice #2: More Work-Energy #2

Practice #3: More Work-Energy #2

Example #4: More Work-Energy #3

Practice #4: More Work-Energy #3

Practice #5: More Work-Energy #3

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The force acting on a particle varies as in the figure below. (The x axis is marked in increments of 1.50 m.) Find the work done by the force as the particle moves across the following distances. from x = 0 m to x = 12.0 m

The force acting on a particle varies as in the figure below. (The x axis is marked in increments of 1.50 m.)Find the work done by the force as the particle moves from x = 12.0 m to x = 18.0 m

The force acting on a particle varies as in the figure below. (The x axis is marked in increments of 1.50 m.) Find the work done by the force as the particle moves across the following distances. from x = 0 m to x = 18.0 m

A system of two objects has ΔKtot = 7 J and ΔUint = -3 J.Part AHow much work is done by interaction forces?Part BHow much work is done by external forces?

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