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A small 8.00-kg rocket burns fuel that exerts a timevarying upward force on the rocket (assume constant mass) as the rocket moves upward from the launch pad. This force obeys the equation $F = A + Bt^2$. Measurements show that at $t$ = 0, the force is 100.0 N, and at the end of the first 2.00 s, it is 150.0 N. (a) Find the constants $A$ and $B$, including their SI units. (b) Find the $net$ force on this rocket and its acceleration (i) the instant after the fuel ignites and (ii) 3.00 s after the fuel ignites. (c) Suppose that you were using this rocket in outer space, far from all gravity. What would its acceleration be 3.00 s after fuel ignition?

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Physics 101 Mechanics

Chapter 4

Newton's Laws of Motion

Section 3

Newton's Second Law

Applying Newton's Laws

University of Michigan - Ann Arbor

Simon Fraser University

University of Winnipeg

Lectures

03:28

Newton's Laws of Motion are three physical laws that, laid the foundation for classical mechanics. They describe the relationship between a body and the forces acting upon it, and its motion in response to those forces. These three laws have been expressed in several ways, over nearly three centuries, and can be summarised as follows: In his 1687 "Philosophiæ Naturalis Principia Mathematica" ("Mathematical Principles of Natural Philosophy"), Isaac Newton set out three laws of motion. The first law defines the force F, the second law defines the mass m, and the third law defines the acceleration a. The first law states that if the net force acting upon a body is zero, its velocity will not change; the second law states that the acceleration of a body is proportional to the net force acting upon it, and the third law states that for every action there is an equal and opposite reaction.

03:43

In physics, dynamics is the branch of physics concerned with the study of forces and their effect on matter, commonly in the context of motion. In everyday usage, "dynamics" usually refers to a set of laws that describe the motion of bodies under the action of a system of forces. The motion of a body is described by its position and its velocity as the time value varies. The science of dynamics can be subdivided into, Dynamics of a rigid body, which deals with the motion of a rigid body in the frame of reference where it is considered to be a rigid body. Dynamics of a continuum, which deals with the motion of a continuous system, in the frame of reference where the system is considered to be a continuum.

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A small 8.00-kg rocket bur…

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A small $8.00 \mathrm{~kg}…

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A small 8.00 -kg rocket bu…

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A small $8.00-\mathrm{kg}$…

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A small $5.00 \mathrm{~kg}…

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A 2540-kg test rocket is l…

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A 2540 -kg test rocket is …

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A $2540 \mathrm{~kg}$ test…

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A small rocket with mass $…

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A rocket accelerates by bu…

Okay, So what we have here is a Newton's second law problem, but one with a time varying force. So first important key piece of information is that we know what the mass of our rocket is, and that is a kilograms. And we know the functional form of our time during force. Actually, go on, step further, Teo, explicitly remind ourselves that this is a function of times. I'm going to write it in the traditional way as a function of tea is eagle to some constant officer A plus, a different constant be being multiplied by time squared. So this means graphically is that we are going to get, um, half a parabola. It's what this means they have for graphing our forces a function of time every time on the X access and force on the Y access. What this means is that we're going to get a problem with some sort of positive offset and the amounts of that positive offset. It's going to be a sorry too terrible aim on the reason why a is this value right here. It's because when time was able to zero, well, then that second term vanishes and you're on Lee left, eh? So that is the graphical intuition we're going Teo using here. Let's go ahead and I'm rights into part, eh? Which is, too. Find the constants of a bee and any wacky units that they might come back. So I think the easiest way to do that is to just use the information that's given to us. So now that we have are explicit function, we can say that that the value of the force at time is equal to zero. It's going to be equal to a plus being where we plug in that value for tea, which is equal to zero every square. Senior Lazar scored a zero multiplied baby is going to make the entire value cool deserve. So we are on ly left with a no only that we are told that the value of this force is one hundred point zero Newton so can actually already go ahead and put a box around that because we successfully solved for that first constant aim on that we can do the same thing by plugging in the value for what's happening at two seconds. Then I won't include all this, and you think years, just for sure hand. But remember, though two seconds there's the same constant pain add it to be, but plugging in two seconds with all of its glorious, significant players and the entire constantly squared that is going, we're saying we want to assert now that is equal to one hundred fifty Newtons. But before we even get there, I want t o go ahead and directly plug in. Our answer from before, which is to sit at a constant, is equal to a hundred point zero Newtons. Um, we still have that constant being. And if we evaluate the square that's occurring here, that gives us four point zero zero. I remember that the units also get squared in addition to numerical value, so that has units of second sward. And then again, the giving piece of information to us is that the left hand side is equal to one hundred and fifty point zero notice so we can go ahead team after this of track one hundred new tins from both sides divide the signs by force have been squared and then at the end of the day, two three six figures get value of twelve point five on the end is gonna have work units. Newton's her square seconds. Okay, so that is part of a And just because the Texas problem so long we're starting to run in the rooms, we're gonna insert new slide to keep all that in mind as we move on to part B, where we now and then B is divided into two parts even further apart from tomorrow, I remember one. So now we want to know what the net force on a rocket is as well as the acceleration at time is equal to zero s o. The subtlety here is that you know, we have a rocket that were launching from the surface of the earth. So there are two forces acting on this rocket. That's the force that we've just described and solve for R penny ante, probable force. But there's also the force of gravity because we're dealing with a rocket. So I'm going to go ahead and subtract the force of gravity because it is working downward or as our other forces fortune upward. Um, so we'll just go ahead and keep yes, Force zero, and then the force of gravity is be the mass of our rocket, which you sit is eight kilograms multiplied by the graph to show acceleration your surface of earth. So just fuse and memories Rover from part A well, calculated that it was given to us rather that the force at times zero is one hundred new tins and then our mess is eight quite serious. Zero kilograms. And lastly, we have ah guard official exploration to three significant fears of nine point eight one meters per square seconds. Plug that into the calculator and then you will get a whoops. Sorry about that. You will get a force of twenty one point five mutants. But this problem is not only asking for the Air Force, but also in the exploration. So we're gonna have to take that one step further and apply needs second law one more times. The acceleration at time is zero can be found simply by dividing the force. That would just solve for it for us that time. Equal Teo zero by the mass. So that means is that we are twenty one court five new tunes on divide, but by through by eight point zero zero kilograms. Plug that into the calculator and then get that Arnett force is equal to two point six nine meters per square second. Cool. Uh, so that was part of the Roman world. I of being. So let's go ahead and sure ourselves in imaginary barrier andan, continue with Roman. Your world too. Andan this part. We want to hide the net force after three seconds instead of zero. So that will look, Lex grammatically is thie. Sorry. Time. It was terrible. Start over. Something helped force after three seconds. Thomas Guns vehicle. Teo Force We found earlier also after three seconds subtracting off the force due to gravity. His gravity is always the same s so well that look like in terms of functional form. Somebody moved from equal sign over a little bit just because we're running out of space, that's going to be a plus. Being with three seconds tow three sniffing big years. Oops. School. Where'd tracking off that same force of gravity? Twenty two law. That's mass time's gravitational acceleration on. Then don't mind me my evil side over even a little bit more sewing. Other units will work out so well, Mitt units. But that gives us one hundred. Add on twelve point fine square the three to nine point zero zero and then take Ah, a point zero zero. What? Supplied by nine points. What? There we go put it all into a calculator and gets that the that force, after three seconds is going to be equal to one hundred and thirty for new chintz, right? Of course, we're not done yet because we also need to find the acceleration. But as we saw earlier, this is just one more small calculation. Come in service. That's three seconds is going to be equal to applying needs in second law, the F Nets Seconds mass going Here s we just found it. And that was one hundred and thirty four. Technically, it was one hundred thirty four point zero two. Andan mass is still a point zero zero. So I think the day and then we'LL be able to type things out or not. No force. It's going to be equal. Teo. Sorry. That force, specifically, at three seconds, it's going to be equal to twenty six point six b. I'm sorry. I'm jumping ahead of myself. That's the numerical answer for part. See, absolute slips sixteen point eight meters per second squared. Crites was long party had four unnecessary answers to Okay, so what's internalize that? And then, lastly, do part c on a third slide. So this is part seeing, like probably write bigger since for the last part, all rights on. There's no Net sub script on this one because we're in the middle of space and were not subject to gravity. So on ly getting the acceleration of the force with that time dependent force we found earlier. So that's going to be what's new, occurring at three seconds. That's what being closed, parentheses and divide through by Mass. So what goes on top Constant, eh? Plus be well supplied by the moment in time that were at which is three seconds entire quantity square. And then the mass being carried over the bottom hurt. It's our constant A, which is one hundred had into our constant being, which is twelve point five and then evaluating the square to nine point zero zero and the mass on the bottom being eight point zero zero. Okay, so here goes Thie answer that I put it out earlier, which is the acceleration in the middle of space after three seconds being equal to twenty six points. Six meters per square. Second big box around that. Okay, now we're done.

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