Ch 07: Work & 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 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

Concept #1: Review: Intro to Springs

Transcript

Hey guys, so in this video I want to start talking about Springs and spring forces let's check out, so when you push or pull against the spring with a force let's call this force Fa an applied force the spring pushes back this is Newton's Third Law action reaction, you push against something it pushes back against you so the force of the spring will be the negative of the force that was applied on it and again Newton's action reaction that's why the same magnitude but opposite direction so this negative here means just opposite direction, so there's nothing new there what is new is that this also equals to -Kx and I'll talk about what kx is in just a second notice that there's another negative here and again that negative just means opposite direction that negative is you see this more later is almost sort of a conceptual negative to remind you that it's opposite direction but in a lot of calculations we just get rid of that negative to make things a little bit simpler, alright? So, what is K and X? I'm going to start with X, X is the deformation that the spring will experience so the idea is that if you push or pull against the spring the spring will either be compressed which means it's now shorter or it will be stretched which means obviously that it is now longer but X is the deformation It's not the actual length but it's the change in length so in springs what matters is not how long it is but how much longer or how much shorter it gets, OK? So X you can think of it as the absolute value of the final length-the initial length the reason why it's the absolute values because if you get a negative you just think of it as a positive because the direction doesn't matter, OK? So, for example here I have this first one here I was spring attached to a wall, this green thing is sort of like a base that you can sort of drag on the spring and if the spring is just sort of left hang in there it will have no deformation it's the original length of the spring X=zero we call this the relaxed position of the spring, right? Spring is chilling basically without having been compressed, in this situation over here this red dotted line here indicates the original length of the spring and now I'm to the right and that's because we're to the right of the spring here of the original length and that's because we pushed to the right with the force Fa, if I push to the right action reaction spring pushes back this way, force of spring is this way the gap between its original length and its new length we're going to call it's (deformation) we're going to called that X, OK? Now since I have two different situations I'm going to call this X1 we're going to use X2 for the bottom here, here the bottom I was originally the spring was originally here and then I pulled it this way so my applied force is Fa like that and the spring will push back with Fs this way and the gap between its original length and its new length is deformation X which I call X2, OK? So that's how F and X kind of work. K is the spring that's the second variable here K is the spring's force constants, OK? Now there's actually multiple names for this I've seen force constant I have seen spring coefficients I've seen stiffness constant or stiffness coefficient but basically this is the coefficient of how hard it is to compress a spring, K has to do with how stiff or how hard to compress or deform in general a spring is the higher the K the harder it is to the deform a spring so for example if you've ever played I'm sure you have with one of those pens that you push on it and the tip comes out you've seen inside of it there's a small spring the allows that to happens very simple and it's a thin spring and if you've ever played with it it's really easy to squeeze and sort of stretch it out and moves very easily so that spring has a very low K. Now if you had springs for example at the bottom of an elevator shaft that design for the safety if the elevator comes down following the spring maybe tries to hold the elevator in place those springs are going to be very stiff otherwise the elevator will crush them and they wouldn't do anything, right? So that K has to do with how hard it is to deform something, real quick the units for K are Newtons per meter and you can see this from the equation, force is Newton equals KX, X is meters so K has to be Newtons per meter so that this equation is dimensionally consistence, ok? So, if you ever forget the units you can just get it from the equation by isolating by leaving the X... solving the X or the K rather leaving it by itself and seeing that the units and you do unit analysis you get Newtons/meter or you can just remember that, OK? So Force, one more point to talk about the force the force of a spring is a restoring force and it's really important it's a restoring force what does that mean? It's always going to the spring is always trying to get back to its original length if you pull the spring the spring to the right the spring will pull back left and if you post a spring to the right I'm sorry to the left it pulls back to the right and if you pull to the right it pull back to the left, you can see here that we went to the right of equilibrium and the spring is trying to pull back to the left here we are to the left and the spring is pulling to the right, so if you're on the right you can be pulled to the left and vice versa so the spring is always pulling you back the spring force is always pulling the spring back to original length X=0, it starts relaxed it likes to being relaxed and it's going to try to get back to its original length, the force is always opposite to the deformation X, OK?

Let's do a quick example here and see how this works out, I have a 1 meter long spring that is laid horizontally so very similar to these diagrams here with one of its and fixed so there's sort of a wall and you have a horizontal spring, there is usually a little bar here or sort of a platform that you can drive the spring or sort of push a box against the spring against this, Ok? So it has an initial length of 1 meter as you do this I want you to remember that what matters in springs is not the actual length but the changing length so it's not the length initially and final but the X the change in length, OK? Now you're going to come here and pull with a force of 50 which will cause the spring to stretch to a final length of 1.2 meters and remember what matters is not the length initial or final but the change in length, in this case if I go from 1 to 1.2 the change is are just 0.2 meters, OK? This is my applied force over here and I want to know what is the spring's force constant? force constant is K, your book or professor might use a different word a different terminology instead of force constant but it's the same thing, right? So what is K? Well right now we only have one equation that's deals with springs we're going to have more later but for now it's just force of the spring is the negative of the force applied or the -KX. the force applied and the force of the spring are the same, I'm just going to write a 50 here and we're going to ignore the negatives, K is always a positive so I can just ignore the negative you can think of it in terms of this negative cancels with this negative but I think it's easier to just think in terms of the negative is just a conceptual negative if I'm calculating a number I'll drop the negatives and I'll get just positives, OK? And that works better, X is 0.2 right here and now we can solve this, K is 500/0.2 and you get to 250 Newtons...Newton/meter rather, OK? Now values, typical values of K that she would see in physics problems are usually going to be in the hundreds maybe in the thousands but it's usually in the hundreds ok? It's usually not really a 2 or 5 or 10000, it's going to be a few hundred so that's generally if you remember that it might give you a little more confidence when you're solving these problems but again they can really give you whatever number want, right? Professors. So how much force is needed to compress it to 0.7? So I want to know how much force I need so what is the Fa needed to compress it to 0.7? Now really important is this word here, compressed 2.7 because I said compressed 2.7....Compress 2.7 means that my length final is 0.7 which means my X is 0.3 I started at 1 I'm going to 0.7 and the X is 0.3 had I said compress by 0.7 it would have meant that my X is 0.7 or if I said a compression of 0.7 then it would also have meant that the X is 0.7 So be very careful here, am I telling you how much it's changing by? or am I telling you what the final length is, OK? In this case we're dealing with this one here but you have to be very careful, X is 0.3 and I want to know F the equation I get is FS-FA=-KX, I want to know the force is so I'm just going to say Fa equals Kx drop the negatives. K we're talking about the same spring so K is 250 and the X is 0.3 So this is 75 Newtons, OK? 75 Newtons is the force needed to get a compression of 0.3, this should make sense, a force of 50 got me and X of 0.2, a force of 75 gets me a compression of 0.3 a little bit more force a little bit more compression or deformation technically, right? In fact this is 75 which is 50 percent more and this is 50 percent more and that should make sense because this is a linear relationship if one doubles the other one will double as well, OK? so that's it for this one I want to quickly talk about Vertical Springs So let's look at how this works if you will attach a mass to a vertical spring and let the mass come down slowly, I'm going to talk about this little detail here towards the end but basically if you put a mass what's going to happen is the weight of the mass will stretch the spring, a spring if left alone won't self-stretch, right? So you need a force to make it stretch and in a situation like this, this object has a mass it's been pulled down by the earth as a weight of MG and that MG is what's going to be responsible for pulling the spring down and it will stretch the spring until they reach equilibrium. Now remember equilibrium means acceleration=0 and that also means that the forces will cancel, OK? So here's the idea before you apply or before you put a mass here this thing is an X=0, it's relaxed because it's X=0 the force of the spring is also 0, 0 and 0, OK? Now let's go through this real quick if you have a spring here you put an object slowly let it fall, once it falls a little bit it's going to have a little bit of a compression or a deformation it's going to stretch a little bit if it stretches down the spring is going to pull back up trying to restore itself to its original length, OK? And the idea is now you get a little bit of an F... little bit of an X and you get a little bit of a force of a spring, let's play with some numbers here, let's say MG is 100, Ok? You start with the force of 0 for the spring but then you let it fall a little bit and now the force of the spring let's say is 10 and you have 100 going down and a force of spring out of 10going up, right? Now you let it fall a bit more this 10 becomes a 20 it's still weaker than MG so if you release it falls a little more and so on so forth and so this number is eventually 100, right? Why? Because this X kept growing so this F keeps growing, right? And the X will keep growing, this thing is going to keep getting pulled down until these two numbers are the same, once they both are 100 once the force of the spring reaches 100 because it's been deformed so much that it's pulling back with a full 100, now these forces are the same they cancel and if you let go it doesn't actually stretch anymore because the forces cancel, OK? So you can imagine that the if you were sort of graph of this force it would look like this and eventually you reach 100 as your deformation gets bigger and bigger, this is sort of a force deformation, here right? So once it gets 100 they will exactly cancel each other and because they were exactly cancel each other, I can say that KX which is a spring force equals Kx and MG which is the gravity will equal each other and that's the equation that you need to remember so there will reach equilibrium and I will have that KX=MG, OK? This is what I like to think of as the vertical spring equilibrium equation, OK? Now that's not the official name of the equation but I give it a name because I really want you to remember that every time you have a spring at vertical equilibrium with a little mass it's just hanging there stopped this equation will be true, now this is only true if you let the mass come down slowly so the idea is that you put a mass in here and then you hold on to the bottom of the mass and you don't just release it you sort of let it come down slowly until it's reached equilibrium once its reached equilibrium if you keep moving your hand down this thing will stay there, right? Now if you just release what's going to happen is instead of getting to equilibrium it's going to pass equilibrium and sort of go up and down it's going to oscillate around the equilibrium point that's not what we want that's a future chapter, OK? So for this equation to work you have to let the mass come down slowly. My last point is that this setup also applies if you have a mass on top of a spring, so imagine I have a spring like this and then you have a mass and top of it if I release the mass is going to do, right? But if I release slowly and just kind of level slowly compress and eventually I'm going to release and that's not going to move anymore, it works exactly the same KX=MG still applies in that situation as well because the forces are canceling just the same. Cool? So that's it for that I want you guys to do this practice problem and remember the stuff we talked about, the X is the difference in size or length not the lengths themselves, remember that this equation applies if you have vertical equilibrium and all the other stuff we talked about let's try to do this and see what you get.

Practice: A vertical spring is originally 60 cm long. When you attach a 5 kg object to it, the spring stretches to 70 cm. (a) Find the force constant on the spring. (b) You now attach an additional 10 kg to the spring. Find its new length.

Concept #2: Work by Variable Forces & Springs

Transcript

Hey guys, so so far all the work problems we've done involved constant forces then we calculate the work done by forces that were not changing over time, now is that video I want to talk about how to calculate the work done by forces that are changing over time by variable forces let's check it out.

So the work equation right here and right here the work done by a force equals force times displacement times the cosine of theta where theta is the angle between F and delta X, remember all that is applicable only if you have constant forces so everything we've had so far were constant forces now if we had variable forces it's a little bit different but it's pretty much the same thing, if we have variable forces so forces that are change over time all we have to do is use the average force instead, OK? And the simple problems will basically give us forces that are very easy to find the average of, so this equation here changes by a little bit so the work done by a force that is a variable force is the average force times delta X cosine of theta same thing or very similar, so the most important example is the spring force and if you remember the way this works the way springs work is if the spring is here at a relaxed position the compression of spring is 0 I'm going to then push this way let's say I'm going to push all the way up until here with an applied force Fa causing a compression of X or a deformation of X the spring will push back with the same force, F of spring and if you remember the relationship here is that the force of the spring is the negative of my force and it's also the negative of KX where K is the spring coefficients and X is the displacement or the deformation is the proper term here, OK? Now the thing to realize is that as I push farther into the spring I get more deformation and the spring pushes against me with a stronger force which means I have to push with a stronger force to keep going into the spring more and more and letÕs say I push and so this much of a deformation over here so what I want to do is show you is what the force or the magnitude of the force of the spring which is the same as the force that I apply just in opposite directions what the magnitude of the force would look like? I'm going to graph this against the deformation of the spring, OK? So if the deformation is 0 it's because I'm not pushing at all so I have 0,0 and if as I push more I get more of a force so as my X grows as my deformation grows the force that I have to push with and the force the spring pushes back against me with they both grow together and then you get something like this, OK? And I stopped here so this is my maximum deformation, let's make up some numbers here real quick.... LetÕs say the maximum the deformation is 2 meters and that requires a force of a 1000 meters (Newtons) so the idea is that what I can do is all what I have to do is have to get the average of this force and the average between 0 and 1000 is 500. I'm sorry this is Newtons it's a force it's Newton's so this would be 500 Newtons which by the way would give me a compression of 1 meter it's half the maximum force so it's have the maximum compression, this is my force max, this is my force minimum which is 0 and this is my force average which is just between the two numbers, OK? So I can say that the force that I'm dealing with here which is the force of the spring is KX, OK? That's the F here that I'm dealing with is KX but in this case the average force is going to be the average between the minimum and the maximum so it's (min+max)/2, the minimum force is zero the maximum force is KX/2, so the average force is KX/2 so the work done by a force that is variable is that force or the average of that force X cosine of theta, well the average of that force is right here KX/2, this is a D right here, right? A distance or delta X let's just delta X if I push and I compress the spring by 1 meter that's the compression of the spring that's deformation of the spring but it's also how much I pushed so in this case the delta X here on the work equation is the same thing as the deformation of the spring so it's (KX/2) x X cosine of theta, theta is going to be 0 because I push this way Fa and the spring compresses this way delta X so this is going to be 0 and the cosine of 0 remember is 1 so I get the work done by my pushing force in is going to be 1/2KX squared, OK? So let me rewrite this real quick, the work done by the applied force (the force that's compressing the spring or the deforming the spring) is 1/2KX squared which by the way look here that's the elastic potential energy which is the same thing as the elastic potential energy, so Uelastic and I'll talk more about this shortly, OK? And that should make sense, a spring that is not compressed has no potential energy when I compress it if I spend 100 joules of energy if I put 100 joules of work into this spring it now has a potential energy of 100, my work went into elastic potential energy so this makes sense however as I push into the spring with an Fa and the spring moves this way delta X the force of the spring is back so the spring is doing negative work against my motion it's moving that way and I'm pushing that way but the spring is pushing this way so the angle between the displacement and the force of the spring is 180 degrees and remember the cosine of 180 is -1 so I can say that the work done by the spring force you get the same exact thing here except you get a negative one right there, so it's going to be -1/2KX squared or negative the potential energy because a spring is basically fighting back against you, OK? So that's how that works I'll talk about that a little bit more and I'll give you a more generic expression here but that's just to show you how we get average of a force which is what you should be using right here, the average of the force, OK? Which is the same equation that I just showed you guys here, and in this case it's minimum maximum divide by two just the basic math average and that's how that would work and for the specific case of a spring in ends up giving you these two here, OK? The last thing I want to talk about before we go here to this other side is that graphically in other words in terms of a chart like this the work done by any force any variable force any variable or I should right here variable or constant force therefore any, right? Is the area under the FX graph so when I say FX here I mean the graph showing force over displacement or compression, right? In the case of a spring so for example if I wanted to find the work done by this...By me compressing this I could just find the area here, OK? And this case this is a triangle and I want to remind that the area of a triangle is 1/2(base)(height) right so a triangle like that, now another thing I could have done instead is....Remember the idea here is that you get the average force and you calculate this with the average force so the average force is 500 right here so I could also have found the area under the average force like this and that's a little bit easier because now it's a rectangle and effectively it's the same thing that we're doing when we simplify this equation to say it's the work done by the average force, in fact this area here is exactly the same area that area that I just erased is the same area as this, alright? So the area under a curve, let's talk about one more thing here, then we're going to do some examples of what happens if you are a spring which is already compressed here and I take it to here so this is the initial compression of the spring and I want to move it to another compression, compression final over here, well the amount of force needed to bring it to here is this much, right? You need this much force (let's draw this a little bit less steep) So I need this much force here this is the amount F initial that I need to hold at that point and then if I want to compress it even more I have to keep squeezing the spring until this point right here and it requires a force final, OK? So the amount of work that it takes to go from here to here is the potential energy so the amount of work that it takes to go from here to here is I'm going to call that potential energy 1 and the amount of work that it takes to go from here to here I'm going to call that the work needed to get to that second point is the same as...Remember it's the same as the potential energy, It's potential energy 2, OK? So let's say it takes 100 work to go from here, but it takes 300 work to go from here to here to the third point, OK? so the work to go from one to the other from initial to final here would be just the difference between these two so the work to go between 1 and 2 is you U2-U1, OK? And the conclusion here is that the work to go between those 2 points is the change in potential energy so these two equations here when I said that the work done by a force that's pushing is the potential energy and the work done by the spring is the negative of the potential energy this was kind of incomplete, this only works if you start from 0 but the more general one is that the work needed to compress is the change in potential energy, elastic potential energy and the work done by the spring in the situation is the negative of the change in potential energy OK? Now one thing that might help you remember this is that the work done by MG which you might remember the work done by gravity is the negative of the change in gravitational potential energy as I lift something gravity is pulling it down so it goes against the motion so therefore they're negative, as a I compress a spring the spring is also going against me therefore they're negative and as I lift I give something potential energy but the spring goes against me and when I compress I give something elastic potential energy but the spring goes against me, gravity goes against me, spring goes against me therefore in both cases you see that the work done by those forces is the negative of the potential energy that you gain by making that kind of motion, OK? So very very similar the idea here is very similar and if you remember the work to lift was the positive of the potential energy, here the work done to compress is the positive of the potential energy because in lifting and in compressing you are adding potential energy to the system, OK? So these are sort of the more general cases equations that's what you should remember not so much these here, OK? So let's do two examples real quick.

Example #1: Work by Variable Forces

Transcript

Alright, so here I have a box we're pushing a box along a level smooth surface it's a horizontal surface that has no friction for a total distance of 20 meters and then here it says for the first 8 meters your force grows steadily from 0 to 60 and then later on it's going to decrease steadily as well so what I'm going to do here to help us visualize this is I'm going to draw a little force as a function of position so we can show this force changing and it's a variable force and you remember that the work done by a variable force is the average force x delta X x cosine of theta or it's the area of the FX graph so by graphing FX I can simplify find the area in that graph to find the total work that your force does so the first 8 meters the total than here is 20 so let's make this a 20, right down the middle here is a 10 so for the first 8 meters let's say that's an 8 right there you're going to grow steadily from 0 to 60, so let's say that's a 60 right there 60 Newtons 0 to 60 and for the last 12 meters, here's the last 12 meters it's going to decrease slowly steadily in another words on a straight line from 60 to 0 so looks kind of like that., OK? What is the total work that you do obviously from beginning to end? So the work done by you, you're pushing the work done by this F over here that we're describing and I can do this by finding the area under the graph now in this case because I have two little triangles two different areas I'm just going to do A1+A2, this is A1 and this is A2 right here, now before I start calculating this just to be clear even though this one is going up and this one's going down they're both positive areas because they are above the X axis, negative areas (areas below the X axis) would be or areas below the negative X axis would count as negative areas right? But that's not the case here so if I were to find this work would be remember the area of a triangle is have 1/2BH So here it's 1/2 the base is 8 and the height 60 plus 1/2 for the second part the base is 12 and the height is 60 and when you do all this you get 600 joules that's the amount of work that we did here, alright let's go to the next one.

Example #2: Work by Variable Forces

Transcript

Alright so here we're hanging a 5 kilogram object from a vertical spring so let me draw a little vertical spring hanging here and when there's nothing connected to it obviously there's no deformation so I'm going to say X=0 and then when I attach an object the object's weight will pull down the spring so let's put a 5 kilograms here and its stretches this and gives a deformation of X and this question this problem says it's deformed by 10 centimeters so X =0.1 meters, I'm going to calculate the work done by gravity and by the spring so this is pretty straightforward the work done by gravity or the work done by MG is simply the negative change in potential energy that's just something you need to know, I can expand this and write this as a negative potential is MGH so the change in potential energy is MG. Delta H, right? And I have all these numbers so negative the mass is 5 gravity I'm going to us a 10 to round it and delta H=-0.1 if this thing moved down deformed by 0.1 then my delta H is the same thing as my AX here, so let me write this here, delta H is the same as X but it's going down so I have to use a negative when I do that notice that this becomes a positive number because the two negatives cancel and you get 5 joules, I got +5 joules, it should make sense that the work gravity here is positive because remember positive work is work done in the direction of motion in this case this thing is going down and gravity pulls it down so gravity does positive work and that's good, alright?

So, for Part B what is the work done by the spring? Here you should also know that the work done by the spring is much like the work done by gravity, the negative of the change in the spring potential energy or the elastic potential energy that's very...Basically the same idea here the only difference is that I can't sort of shortcut this like I did here MG. Delta H but what I can do is I can expand delta U so it's (-U final or Uelastic final-U elastic initial 0 In this case in the beginning before it started compressing there was no compression or there's no deformation so the initial potential is 0 so all you end up with is -1/2KX squared and that X is the final X so let's plug in the numbers here, 1/2 K I don't have a K actually so we have to go and get that so let me leave that alone for now and the final compression is 0.1, notice that when I put a 0.1 here I didn't put as a positive or negative it's going to squared anyway it doesn't matter. So let's find the X I hope you remember that when you have a spring at vertical equilibrium it's like that because the forces are canceling so the spring force which is KX going up cancels with the weight force which is MG going down and you're able to write KX=MG, you have to remember this, this is I call this the spring vertical equilibrium equation so I can find K here by plugging in the numbers, K is MG/X and if you plug everything here you get 500 Newtons per meter so that 500 is what's going to go in here and when you multiply all this crap you get -2.5 Joules, notice that the work done by gravity was positive but the work done by the spring here was negative and that's because gravity is helping this thing moved now but the spring is resisting as it's falling the spring is pulling it up as this thing is moving down the spring is pulling up so it's doing negative work because it's going against motion, alright? And then the last part here is I want to know what is the spring's potential energy? So obviously the potential at the beginning is 0 so we're referring to the spring's final potential energy which is just 1/2 KX squared (X final) and if you plug everything in here it's basically the same set up as this part B you get 2.5 Joules but that's 2.5 Joules is positive because the spring gained potential energy and a spring can have negative potential energy anyway, notice that one of these numbers is 2.5 negative the other one is 2.5 positive that should make sense because remember the work done by spring is the negative of the change in potential energy (elastic potential energy) that's why the numbers are the same just one is the negative of the other, Ok? You might notice that the work done by gravity was 5 positive and the work done by spring is -2.5 so there's sort of a 2.5 that's missing there and that's the work done by your hand, remember you're letting this thing falls slowly that's why it reaches equilibrium instead of oscillating around the equilibrium point so your hand basically did 2.5 Joules of energy negative, right? And if you were to add all these things the work done by weight+the work done by the spring+the work and by your hand this would be equal to 0 but that's a whole different conversation, alright? So, I want you guys to go ahead and try to do practice 1 it's very similar, we have a spring here and I want to see if you can figure this one out let's try that.

Practice: It takes 200 J of energy to compress a 1.0 m-long spring to 70 cm. How much work would it take to compress this same spring from 70 cm to 50 cm?

A 17000-kg jet takes off from an aircraft carrier via a catapult. The gases thrust out from the jets engines exert a constant force of 180 kN on the jet; the force exerted on the jet by the catapult is plotted in the figure b. Determine the work done on the jet by the gases expelled by its engines during launch of the jet.Determine the work done on the jet by the catapult during launch of the jet.
The net force exerted on a particle acts in the positive exttip{x}{x} direction. Its magnitude increases linearly from zero at x = 0, to 410 N at exttip{x}{x_2} = 3.1 m . It remains constant at 410 N from exttip{x}{x_2} = 3.1 m to exttip{x}{x_3} = 7.1 m , and then decreases linearly to zero at exttip{x}{x_4} = 13.1 m .Determine the work done to move the particle from x = 0 to exttip{x}{x_4} = 13.1 m graphically, by determining the area under the Fx versus exttip{x}{x} graph.
The force on a particle, acting along the x axis, varies as shown in the figure .Determine the work done by this force to move the particle along the x axis: from x = 0.0 to x = 10.0m.Determine the work done by this force to move the particle along the x axis: from x = 0.0 to x = 15.0m.
The figure is the force-versus-position graph for a particle moving along the x-axis. Determine the work done on the particle during each of the three intervals 0-1 m, 1-2 m, and 2-3 m.
A child applies a force F parallel to the x -axis to a 10.0-kg sled moving on the frozen surface of a small pond. As the child controls the speed of the sled, the x -component of the force she applies varies with the x -coordinate of the sled as shown in the figure .Calculate the work done by the force F when the sled moves from x=0 to x=8.0 m.Calculate the work done by the force F when the sled moves from x=8.0 m to x =12.0 m.Calculate the work done by the force F when the sled moves from x=0 to x =12.0 m. .
A force F is applied to a 2.0-kg radio-controlled model car parallel to the x-axis as it moves along a straight track. The x-component of the force varies with the x-coordinate of the car as shown in the figure .Calculate the work done by the force F when the car moves from x=0 to x=3.0 m.Calculate the work done by the force F when the car moves from x=3.0 m to x=4.0 m.Calculate the work done by the force F when the car moves from x=4.0 m to x=7.0 m.Calculate the work done by the force F when the car moves from x=0 to x=7.0 m.Calculate the work done by the force F when the car moves from x=7.0 m to x=2.0 m.
A force is applied to a 3.5-kg radio-controlled model car parallel to the x-axis as it moves along a straight track. The x-component of the force varies with the x-coordinate of the car as shown in figure .Suppose the model car is initially at rest at x=0 and F is the net force acting on it.Use the work-energy theorem to find the speed of the car at x=3.0 m.Use the work-energy theorem to find the speed of the car at x=4.0 m.Use the work-energy theorem to find the speed of the car at x=7.0 m.
A child applies a force F parallel to the x-axis to a 8.00-kg sled moving on the frozen surface of a small pond. As the child controls the speed of the sled, the x-component of the force she applies varies with the x-coordinate of the sled as shown in figure . Suppose the sled is initially at rest at x=0. You can ignore friction between the sled and the surface of the pond.You may want to review (Pages 183 - 189). For related problemsolving tips and strategies, you may want to view a Video Tutor Solution of Motion on a curved path.Use the work-energy theorem to find the speed of the sled at 5.0 m.Use the work-energy theorem to find the speed of the sled at 11.0 m.
One end of a horizontal spring with force constant 130.0 N/m is attached to a vertical wall. A 5.00-kg block sitting on the floor is placed against the spring. The coefficient of kinetic friction between the block and the floor is k = 0.400. You apply a constant force vec{F} to the block. vec{F} has magnitude 86.0 N and is directed toward the wall. The spring is compressed 80.0 cm.What is the speed of the block?What is the magnitude of the blocks acceleration?What is the direction of the blocks acceleration?
Spiderman uses his spider webs to save a runaway train (see the figure). His web stretches a few city blocks before the 2.5×104 kg train comes to a stop.Assuming the web acts like a spring, estimate the spring constant. Assume the train is moving 25 m/s , and that the distance of "a few city blocks" is perhaps about 850 meters.
A mass m is attached to a spring which is held stretched a distance exttip{x}{x} by a force F and then released. The spring compresses, pulling the mass. Assuming there is no friction, determine the speed of the mass m when the spring returns to its normal length ( x = 0 ).Determine the speed of the mass m when the spring returns to half its original extension (x/2).
Stretchable ropes are used to safely arrest the fall of rock climbers. Suppose one end of a rope with unstretched length l is anchored to a cliff and a climber of mass m is attached to the other end. When the climber is a height l above the anchor point, he slips and falls under the influence of gravity for a distance 2l, after which the rope becomes taut and stretches a distance exttip{x}{x} as it stops the climber (see the figure). Assume a stretchy rope behaves as a spring with spring constant exttip{k}{k}.Applying the work-energy principle, find x.Assuming exttip{m}{m_0} = 86 kg , exttip{l}{l_0} = 8.6 m and exttip{k}{k_0} = 870 N/m , determine x/l (the fractional stretch of the rope) at the moment the climbers fall has been stopped.Assuming exttip{m}{m_0} = 86 kg , exttip{l}{l_0} = 8.6 m and exttip{k}{k_0} = 870 N/m , determine kx/mg (the force that the rope exerts on the climber compared to his own weight) at the moment the climbers fall has been stopped.
Three identical 8.50-kg masses are hung by three identical springs . Each spring has a force constant of 7.60 kN/m and was 14.0 cm long before any masses were attached to it.How long is the bottom spring when hanging as shown? (Hint: isolate only the bottom mass.)How long is the middle spring when hanging as shown? (Hint: treat the bottom two masses as a system.)How long is the top spring when hanging as shown? (Hint: treat all three masses as a system.)Draw a free-body diagram for the top mass.Draw a free-body diagram for the middle mass.Draw a free-body diagram for the bottom mass.
To stretch a spring 8.00 cm from its unstretched length, 16.0 J of work must be done.What is the force constant of this spring?What magnitude force is needed to stretch the spring 8.00 cm from its unstretched length?How much work must be done to compress this spring 4.00 cm from its unstretched length?What force is needed to compress it this distance?
As part of your daily workout, you lie on your back and push with your feet against a platform attached to two stiff springs arranged side by side so that they are parallel to each other. When you push the platform, you compress the springs. You do an amount of work of 81.0 J when you compress the springs a distance of 0.210 m from their uncompressed length.What magnitude of force must you apply to hold the platform in this position?How much additional work must you do to move the platform a distance 0.210 m farther?What maximum force must you apply to move the platform to the position in Part B?
An air-track glider of mass 0.100 kg is attached to the end of a horizontal air track by a spring with force constant 20.0 N/m. Initially the spring is unstreched and the glider is moving at 1.50 m/s to the right. With the air track turned off, the coefficient of kinetic friction is k=0.47. It can be shown that with the air track turned off, the glider travels 8.6 cm before it stops instantaneously.How large would the coefficient of static friction s have to be to keep the glider from springing back to the left when it stops instantaneously?If the coefficient of static friction between the glider and the track is exttip{mu_{ m s}}{mu_s} = 0.75, what is the maximum initial speed v1 that the glider can be given and still remain at rest after it stops instantaneously?
The spring of a spring gun has force constant k = 400 N/m and negligible mass. The spring is compressed 6.00 cm and a ball with mass 0.0300 kg is placed in the horizontal barrel against the compressed spring. The spring is then released, and the ball is propelled out the barrel of the gun. The barrel is 6.00 cm long, so the ball leaves the barrel at the same point that it loses contact with the spring. The gun is held so the barrel is horizontal.Calculate the speed with which the ball leaves the barrel if you can ignore friction.Calculate the speed of the ball as it leaves the barrel if a constant resisting force of 6.00 N acts on the ball as it moves along the barrel.For the situation in part B, at what position along the barrel does the ball have the greatest speed? (In this case, the maximum speed does not occur at the end of the barrel.)What is that greatest speed?
A 11-cm-long spring is attached to the ceiling. When a 2.1 kg mass is hung from it, the spring stretches to a length of 16 cm . You may want to review (Pages 219 - 221).What is the spring constant k?How long is the spring when a 3.0 kg mass is suspended from it?
A 61 kg student is standing atop a spring in an elevator that is accelerating upward at 3.1 m/s2 . The spring constant is 2600 N/m . You may want to review (Pages 219 - 221).By how much is the spring compressed?
A 7.1 kg mass hanging from a spring scale is slowly lowered onto a vertical spring, as shown in the figure. You may want to review (Pages 219 - 221).What does the spring scale read just before the mass touches the lower spring?The scale reads 22 N when the lower spring has been compressed by 2.1 cm . What is the value of the spring constant for the lower spring?At what compression length will the scale read zero?
Two identical horizontal springs are attached to opposite sides of a box that sits on a frictionless table. The outer ends of the springs are clamped while the springs are at their equilibrium lengths. Then a 2.6 N force applied to the box, parallel to the springs, compresses one spring by 3.1 cm while stretching the other by the same amount.What is the spring constant of the springs?
A horizontal spring with spring constant 650 N/m is attached to a wall. An athlete presses against the free end of the spring, compressing it 7.0 cm.How hard is the athlete pushing?
The left end of a spring is attached to a wall. When Bob pulls on the right end with a 200 N force, he stretches the spring by 20 cm. The same spring is then used for a tug-of-war between Bob and Carlos. Each pulls on his end of the spring with a 200 N force.How far does the spring stretch?
How much work is done by the following force over the 12 m displacement?
If it requires 5.5 J of work to stretch a particular spring by 2.0 cm from its equilibrium length, how much more work will be required to stretch it an additional 4.1 cm?
When a 4.00-kg object is hung vertically on a certain light spring that obeys Hooke’s law, the spring stretches 2.50 cm. If the 4.00-kg object is removed, (a) how far will the spring stretch if a 1.50-kg block is hung on it? (b) How much work must an external agent do to stretch the same spring 4.00 cm from its unstretched position?
A 4.00-kg particle is subject to a net force that varies with position as shown in the figure. The particle starts from rest at x 5 0. What is its speed at (a) x = 5.00 m, (b) x = 10.0 m, and (c) x = 15.0 m?
A 38-cm-long vertical spring has one end fixed on the floor. Placing a 2.2 kg physics textbook on the spring compresses it to a length of 29 cm. What is the spring constant?
A spring has an unstretched length of 10 cm. It exerts a restoring force F when stretched to a length of 11 cm.(a) For what length of the spring is its restoring force 3F?(b) At what compressed length is the restoring force 2F?
The force acting on a particle varies as shown in Figure P7.14. Find the work done by the force on the particle as it moves (a) from x = 0 to x = 8.00 m, (b) from x = 8.00 m to x = 10.0 m, and (c) from x = 0 to x = 10.0 m.
A particle is subject to a force Fx that varies with position as shown in Figure P7.15. Find the work done by the force on the particle as it moves (a) from x = 0 to x = 5.00 m, (b) from x = 5.00 m to x = 10.0 m, and (c) from x = 10.0 m to x = 15.0 m. (d) What is the total work done by the force over the distance x = 0 to x = 15.0 m?
An archer pulls her bowstring back 0.400 m by exerting a force that increases uniformly from zero to 230 N. (a) What is the equivalent spring constant of the bow? (b) How much work does the archer do on the string in drawing the bow?