Ch 08: Conservation of EnergyWorksheetSee all chapters
All Chapters
Ch 01: Units & Vectors
Ch 02: 1D Motion (Kinematics)
Ch 03: 2D Motion (Projectile Motion)
Ch 04: Intro to Forces (Dynamics)
Ch 04: Motion in Two and Three Dimensions
Ch 05: Friction, Inclines, Systems
Ch 06: Centripetal Forces & Gravitation
Ch 07: Work & Energy
Ch 08: Conservation of Energy
Ch 09: Momentum & Impulse
Ch 10: Rotational Kinematics
Ch 11: Rotational Inertia & Energy
Ch 12: Torque & Rotational Dynamics
Ch 13: Rotational Equilibrium
Ch 14: Angular Momentum
Ch 15: Periodic Motion (NEW)
Ch 15: Periodic Motion (Oscillations)
Ch 16: Waves & Sound
Ch 17: Fluid Mechanics
Ch 18: Heat and Temperature
Ch 19: Kinetic Theory of Ideal Gasses
Ch 20: The First Law of Thermodynamics
Ch 21: The Second Law of Thermodynamics
Ch 22: Electric Force & Field; Gauss' Law
Ch 23: Electric Potential
Ch 24: Capacitors & Dielectrics
Ch 25: Resistors & DC Circuits
Ch 26: Magnetic Fields and Forces
Ch 27: Sources of Magnetic Field
Ch 28: Induction and Inductance
Ch 29: Alternating Current
Ch 30: Electromagnetic Waves
Ch 31: Geometric Optics
Ch 32: Wave Optics
Ch 34: Special Relativity
Ch 35: Particle-Wave Duality
Ch 36: Atomic Structure
Ch 37: Nuclear Physics
Ch 38: Quantum Mechanics

Example #1: Energy in Connected Objects

Transcript

Hey guys so in this video I want to show you how to use the conservation of energy equation to solve problems where we're going to have multiple objects moving together as a system let's check it out, so something like this here I have multiple objects moving together and it says here when we're dealing with multiple objects we'll still we use only one equation if you remember when we did this with forces F=MA if you had two objects you would write F=MA twice, here we're going to use just one equation the big energy equation but you have to consider the energy of each object individual independently so let's check it out.

Here I have a system that is released from rest so the initial velocity is 0, the 5-kilogram object is initially 3 meters above the floor so this gap here it's going to be released it's going to fall so I'm going to do this, 3 meters let's call this object A let's call this object B, OK? So, if the 5 kilograms is initially 3 meters above the floor calculate its speed just before it hits the ground so I want to know what is the final speed of object B (Vb final)? But remember this is a system and in systems when you have a connected system like this all the objects move with the same speed and they also have the same acceleration so when I ask for the final velocity of the 5 kilogram I could have just as well have asked for the final velocity of the 4 kilogram or the final speed of the 4 kilogram it would have been the same and I could have also asked for the final speed of this system so it's all the same I'm just going to think of this as V, they're all going to have the same V final, OK? Now once the in 5 kilogram drops 3 meters this guy has to go up it's going to go up the same amount if this drops 3 this goes up 3 so I'm going to draw this over here as well, 3 meters so the block B starts here so the initial height of block B is 3 and the final height of block B is 0 and for block 4 block A it's the opposite the initial height starts here so the initial height is 0 and the final height is 3 so they basically swap places, cool. Let's write the energy equation, Kinetic initial + potential initial + work non-conservative = kinetic final + potential final, the reason why we're going to use the energy equation is because I'm being asked for a final velocity or final speed and the energy equation deals with that, it deals with changes in speeds with changes in heights, right? So it's a good use of the energy equation here but I have to consider every object individually I'm going to first write the long version of this but don't scared you don't have to do it every time but I just want to show you all the different elements you have, there are 2 objects so when I talk about kinetic initial, it's kinetic initial of A plus kinetic initial of B, Potential energy here is going to be just the gravitational potential energy MGH but I have two objects so it's Ua U initial A and U initial B and then there's the work done by non-conservative forces either one of the two objects which here will be simply 0 because there's no friction it says here that we can disregard friction and there is no external force, right? So I'm going to have kinetic and potential so it's the same thing K final A + K final B + U final A + U final B, these are all the possible energies there are 8 of these things but again as you get better with this you don't have to write all of them because you're immediately going to recognize some of them are 0 for example the system starts from rest which means nothing is moving so there is no initial kinetic energy you could have very well have just crossed this out, OK? Now the potential energy in the beginning this guy starts at 3 and this guy starts at 0 so there's no potential is for block A only for Block B at the end you get the opposite of the inverse of that they flip places so at the end the A has potential energy but the B has no potential energy so you can U final B is 0 so I have initial B and final A, OK? Initial B and final A, alright? So now I have kinetic energy of both of these guys because they both have speeds after this thing moves, right? So, I can start writing this out here this is going to be MbGH initial this is the initial height of B so if you want to put all the little letters equals 1/2MaVa final 1/2MbVb final and they're all squares plus this guy here MaGHa final those are all the letters. one thing to realize is that this V here is the same so instead of calling it I'm going to say same V instead of calling it Va final or Vb final I'm just going to call it V final which means that if these two are the same I'm going to be able to combine them, OK? So, let's start plugging in some numbers here, the mass at B is right there 5 gravity 'm going to use 10, the initial height is 3, OK? This is going to be half the mass of A is 4V final squared this is what I'm looking for + 1/2 mass of B which is 5 V final squared this is what I'm looking for, + mass of A 4(10) final height is 3 as well notice that the initial heights of one is the same as the final height of the other because they swap places, now all we have to do is solve for V final in this equation so I have here this is going to be 150 this whole thing here is going to be 120 so it's going to come here as a negative and then I can combine these guys I get a 2V final from the first plus 2.5V final from the second this adds up to very simple 4.5 V final squared so V final will be this is a 30 over here divided by 4.5 and then you got to take the square root of both sides when you plug this into your calculator you get approximately 2.58 meters per second and that is your V final for both objects again it didn't matter that I asked you for the final speed for one it's the same for both of them, OK? Now one last point I want to make here is I want to talk about the kinetic final right here IÕm going to do a different color I want to talk about kinetic final right here which becomes this which then eventually becomes this, OK? Kinetic final you can think of it as 1/2MaV final + 1/2MbV final, right? You can think of it that way or notice what happens the V's are the same so really what's going on is you end up with something like this 1/2Vfinal squared (Ma + Mb). Notice that the masses combine into a single velocity so you can think of it this way the kinetic final in the system is just 1/2 big M V final squared where big M is just the sum of all the masses M1 + M2 + all the masses etc. OK? So, form now to make life a little bit simpler we can just write kinetic final as treating all objects or the entire system as just one big mass and you can do that because all the masses move together instead of writing this and then taking the bit longer to combine everything, cool? So here in this case big M would have been 4+5=9, alright? So I have another example here let's jump into that

Example #2: Energy in Connected Objects

Transcript

Alright so here I have a 3 kilogram block right here 2 meters above the floor so this gap between the initial position and the final position is 2 meters I like to draw this little arrow and I like to indicate that the initial height is 2 and that the final height is 0 so that when I plug this stuff into my energy equation I already have some of these numbers figured out, now if this block moves 2 this way it means that this block on top has to move 2 this way, right? So, this guy is going to move 2 meters like this, OK? And the problem says that there's a coefficient of friction here, coefficient of friction here is 0.5 I only give you on coefficient so that means it's both kinetic and static friction we don't have to worry about it so what happens here? Well remember the way we think about these problems is what would happen without friction? Without friction this would certainly go this way because there is no force opposing it therefore friction is going to be going this way, OK? And we're going to use the since I'm being asked for the final speed or the speed just before I hit the floor I have changes in height I have changes in speed and I get a friction I'm going to use the conservation of energy equation which means I can solve this much faster, an alternative way to solve this by the way would have been to write F=M.A for both blocks combine those two equations like we did back in the day and find the acceleration of system and then I would be able to use one of the motion equations to figure out how fast this thing is so I could have done one written F=MA twice to find my A and then two I could have used motion equation to find the final velocity but what's better is I can just do this in one step using the big energy equation and that's much better that's what we're going to do here, alright? I just want to point that out you could do this in other ways. Kinetic initial + potential initial + work non-conservative = kinetic final + potential final, right away here the system starts from rest, right? It's released from rest which means the initial kinetic energy for both objects, the potential energy there are two objects let's treat them separately potential energy we're going to be a bit careful I'm going to call this guy A and I'm going to call this guy B so it's Ua initial + Ub initial, the work done by non-conservative forces is the work done by friction because here we have friction but we don't have any external forces, right? And remember non-conservative is friction plus the external so the only one here is the work done by friction, the kinetic energy we talked about this in the previous video kinetic energy I'm going to treat it as an entire system of one 1/2 of the total mass times the final velocity of the system and this is what we're looking for, this total M here is going to be Ma+MB which is 3+4=7, OK? Plus So we're going to treat the kinetic energy as if you have just one body because they move together so they have the same velocity they're able to do that it's just a mass simplification there and the potential have to treat them separately because they have different heights, Ua final I'm sorry Ub final let's go through it and see if anything else is 0, so I'll go 4 kilogram certainly has a height above the floor but it has no height relative to the table and even more importantly its height doesn't change so because the floor doesn't change heights it doesn't even matter what its height is if it doesn't change heights it doesn't change potential energy so we're going to just cancel out the potential energies, so look at Ua here whatever number this is would be the same number here so we can just cancel the two of them out because they have the same height so I want to do a little thing here to sort of so you have it your notes the change in height for A is 0 therefore I can eliminate both potential energy it doesn't matter what the potential energy is it's going to be the same in both places so we just get rid of it make our lives simpler. Is there a potential energy for Block B? well in the beginning it does it has a height but not at the end, OK? So, this simplifies a lot and I have just these types of energies continuing here this is MbGHb initial so that's just MGH with all the subscript don't get lost in all of the letters, the work done by friction this is kinetic friction because the block is sliding so it can be -FD, OK? And I can calculate this friction is mu normal the little side thing here and normal in this case is MG because I got this block here it's being pulled this way by tension it's being pulled this way by kinetic friction it's been pulled down my MG there are no other forces in the Y axis except for normal and MG therefore they have to exactly cancel and normal = MG so I can write that friction is mu MG and it's the mass of the block A so it's a 4-kilogram I can calculate this real quick friction, friction mu is 0.5, mass is a 4 gravity we're going to around that as a 10 so friction is simply 20 newtons and I'm ready to plug this in here for now let me just continue here FD 1/2 Big M V final squared, OK? I have all these numbers so let's start plugging them in, mass of B is right there 3 make sure you be careful be very careful with the letter so you don't pick out the wrong numbers, gravity is 10 height of B is a 2, friction is 20 what is the distance? So again, if you move 2 down this block over here is moving 2 sideways so this distance here happens to be the same as your drop because they're connected so they have to move the same distance so this D Here is a 2 1/2 the combined mass is 4+3=7Vfinal is what we're looking for, alright? This number is a 60 this number is a 40, 60-40 equals this is 3.5 V final squared so Vfinal=20/3.5 and we take the square root of that 20 divided by 3.5 we get 2.4 meters per second and that's the answer, now one last point I want to make here is in both of these problems we calculate the same thing we asked for the final speed obviously you can get a different question you can have a problem where I give you the final speed but then I ask you for the mass, right? So, there's all kinds of combinations of things you can do even though I'm asking in both of these for the final speed I could have given you that is taken away and asked you for a different number the process would have been almost the same thing, OK? That's it for this one let me know if you guys have any questions.

Practice: The 4-kg block is 1 m above the floor, and the surface-block coefficient of friction for the 3-kg block is 0.4. If the system is released from rest, find its speed just before the 4-kg block hits the floor. (The string and pulley are massless)

Practice: The system below is released from rest. Calculate the speed of the system after the hanging block has moved 1 meter. (The string and pulley are massless, and you may disregard any effects due to friction)

Use work and energy to find an expression for the speed of the block in the following figure just before it hits the floor.Find an expression for the speed of the block if the coefficient of kinetic friction for the block on the table is k.Find an expression for the speed of the block if the table is frictionless.
Two blocks are connected by a very light string passing over a massless and frictionless pulley (Figure ). The 20.0-N block moves 75.0 cm to the right and the 12.0-N block moves 75.0 cm downward.Find the total work done on 20.0-N block if there is no friction between the table and the 20.0-N block.Find the total work done on 20.0-N block if s=0.500 and k=0.325 between the table and the 20.0-N block.Find the total work done on 12.0-N block if there is no friction between the table and the 20.0-N block.Find the total work done on 12.0-N block if s = 0.500 and k = 0.325 between the table and the 20.0-N block.
Consider the system shown in the figure . The rope and pulley have negligible mass, and the pulley is frictionless. Initially the 6.00-kg block is moving downward and the 8.00-kg block is moving to the right, both with a speed of 0.600 m/s . The blocks come to rest after moving 6.00 m .Use the work-energy theorem to calculate the coefficient of kinetic friction between the 8.00-kg block and the tabletop.
At a construction site, a 66.0 kg bucket of concrete hangs from a light (but strong) cable that passes over a light friction-free pulley and is connected to an 83.0 kg box on a horizontal roof (see the figure ). The cable pulls horizontally on the box, and a 48.0 kg bag of gravel rests on top of the box. The coefficients of friction between the box and roof are shown. The system is not moving.Find the friction force on the bag of gravel.Suddenly a worker picks up the bag of gravel. Use energy conservation to find the speed of the bucket after it has descended 2.10 m , from rest. (You can check your answer by solving this problem using Newtons laws.)Find the friction force on the box.
In the following figure, the pulley doesn’t rotate without friction, so it limits how fast the system can move. In this particular case, the 15 kg mass can only drop at a maximum speed of 10 m/s. At this speed, how much work does the pulley need to do every meter the 15 kg mass drops?
Two objects are connected by a light string passing over a light, frictionless pulley as shown in the figure. The object of mass m1 = 5.00 kg is released from rest at a height h = 4.00 m above the table. Using the isolated system model,(a) determine the speed of the object of mass m2 = 3.00 kg just as the 5.00-kg object hits the table(b) find the maximum height above the table to which the 3.00-kg object rises
A block of mass m1 = 20.0 kg is connected to a block of mass m2 = 30.0 kg by a massless string that passes over a light, frictionless pulley. The 30.0-kg block is connected to a spring that has negligible mass and a force constant of k = 250 N/m as shown in the figure. The spring is unstretched when the system is as shown in the figure, and the incline is frictionless. The 20.0-kg block is pulled a distance h = 20.0 cm down the incline of angle θ = 40.0° and released from rest. Find the speed of each block when the spring is again unstretched.
The coefficient of friction between the block of mass m1 = 3.00 kg and the surface in the figure is µk = 0.400. The system starts from rest. What is the speed of the ball of mass m2 = 5.00 kg when it has fallen a distance h = 1.50 m?
Two masses are connected by a string as shown in the figure. mA exttip{m_{ m A}}{m_A}= 3.6 kg rests on a frictionless inclined plane, while mB exttip{m_{ m B}}{m_B}= 4.6 kg is initially held at a height of h exttip{h}{h}= 0.75 m above the floor.(a) If mB is allowed to fall, what will be the resulting acceleration of the masses?(b) If the masses were initially at rest, use the kinematic equations to find their velocity just before mB hits the floor.(c) Use conservation of energy to find the velocity of the masses just before mB hits the floor.
A system of two paint buckets connected by a lightweight rope is released from rest with the 12.0-kg bucket 2.00 m above the floor. Use the principle of conservation of energy to find the speed with which this bucket strikes the floor. You can ignore friction and the mass of the pulley.
As shown in Figure P8.46, a light string that does not stretch changes from horizontal to vertical as it passes over the edge of a table. The string connects m1, a 3.50-kg block originally at rest on the horizontal table at a height h = 1.20 m above the floor, to m2, a hanging 1.90-kg block originally a distance d = 0.900 m above the floor. Neither the surface of the table nor its edge exerts a force of kinetic friction. The blocks start to move from rest. The sliding block m1 is projected horizontally after reaching the edge of the table. The hanging block m2 stops without bouncing when it strikes the floor. Consider the two blocks plus the Earth as the system.(a) Find the speed at which m1 leaves the edge of the table.(b) Find the impact speed of m1 on the floor.(c) What is the shortest length of the string so that it does not go taut while m1 is in flight?(d) Is the energy of the system when it is released from rest equal to the energy of the system just before m1 strikes the ground?(e) Why or why not?
Two objects are connected by a light string passing over a light, frictionless pulley as shown in Figure P8.7. The object of mass m1 is released from rest at height h above the table. Using the isolated system model, (a) determine the speed of m2 just as m1 hits the table and(b) find the maximum height above the table to which m2 rises.
The system shown in Figure P8.11 consists of a light, inextensible cord, light, frictionless pulleys, and blocks of equal mass. Notice that block B is attached to one of the pulleys. The system is initially held at rest so that the blocks are at the same height above the ground. The blocks are then released. Find the speed of block A at the moment the vertical separation of the blocks is h.