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FcCkey CUCK 140 kg) ddccm diNcnsionz box ~herz tJc Gistancc Dcbccn Dnc Walls 470 distancc DaTYCCm atne walls each

Question:

FcCkey CUCK 140 kg) ddccm diNcnsionz box ~herz tJc Gistancc Dcbccn Dnc Walls 470 distancc DaTYCCm atne walls each dimension represent an infinitc square well in Cach CusC In this two-Jimensional cuse, ne state the Puck would be represented by the combinalion dimensian opposing walls standing waves 210 cuch ISSuTC thc [O (u) What the zcro-point energy cf tne puck Cacn dimcnsion? 7 OOBE-65 the first dimension tour rcsponsc dirfcrs sianiricanty trcm tnc cOmcct Jnswcr Rcwork Youl sclution trom chc jeginnino Cncct Cac7 StCp corctumy SB-64 the sec3nd dimension Your response differs significantly from the comect Jnswer Rewors YouC sclution trom che jeginnino and cnec< @ac7 stCp carefully- (b) The actual zero-point energy the sum of the zero point energy Cacm dimension Find this zero-point energy tor lhe puck in the two-dimensiona Dox.


Answers

Figure $\mathrm{P} 9.46 \mathrm{a}$ shows an overhead view of the initial configuration of two pucks of mass $m$ on frictionless ice. The pucks are tied together with a string of length $\ell$ and negligible mass. At time $t=0,$ a constant force of magnitude $F$ begins to pull to the right on the center point of the string. At time $t,$ the moving pucks strike each other and stick together. At this time, the force has moved through a distance $d,$ and the pucks have attained a speed $v(\text { Fig.

} \mathrm{P} 9.46 \mathrm{b}) .$ (a) What is $v$ in terms of $F, d, \ell,$ and $m ?$ (b) How much of the energy transferred into the system by work done by the force has been transformed to internal energy?

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Chapter four Question 11. We got a hockey puck, frictionless ice, and we give it a push of 110.25 Newtons for the 1st 2 seconds. We don't push it all from 2 to 5 seconds, and then we push it again from 5 to 7 seconds at 70.25 Newton's. So, basically, we're looking for the position of the time of two seconds, and we're looking for the so sorry. The position of the speed at two seconds and the position speed again, all the way over its seven seconds.

So there's only three parts to that motion we're pushing with 30.25 Newton's during this time interval zero Newton's during this time interval and again at 00.25 Newton's during this time interval. So we're gonna split this up into three separate parts, and these two Kinnah Matic equations that I have written right here are gonna be the key to solving this problem. Eso from 0 to 2 seconds. First, let's figure out what the acceleration is gonna be due to this force. We've gotten that forcing the extraction is equal to end times a in the extraction.

So the mass was 0.160 so 0.25 being the force and dividing by 0.160 we get an acceleration of 1.56 meters per second squared. So that's the acceleration during this time interval. And it's also the acceleration during this time interval and the acceleration when we're not applying any force is gonna be zero meters per second squared. All right, so from 0 to 2 seconds, let's do a nice little variable table, cause all of our kin O Matic equations have the same six variables X initial x final V initial vi final, the acceleration on the time. So filling in this variable table, we start at the first interval at zero meters x final.

We do not know yet V initial It does start at zero meters per second. The final we do not know yet the acceleration. We do know 1.56 meters per second squared and the time we do know too seconds. So using the Kinnah Matics First off we're gonna use this kid a Matic right here. X final ex initial zero v initial zero.

So X Final is just gonna be half of the acceleration, which we already found 1.56 times the time squared. And that gives us a final position of 3.12 meters and the final. We're gonna use the equation below on and we are gonna plug in our acceleration times or time because the initial zero so 1.56 times the time, which is to gives us a final velocity of 3.12 meters per second. So filling in our table a pair, let's use some different colors at this point right here. We're looking at a position equal to 3.12 meters and we're also looking at a velocity equal to 3.12 meters per second.

So now let's do the time and roll from 2 to 5 seconds and we're going to save your table table. This time. Ex initial is we're starting at our final position. From the last 13.12 meters X final is something we're gonna be looking for. V initial is gonna be that 3.12 from the previous and the final.

There's no acceleration here, so actually it doesn't change it all 3.12 meters per second. The acceleration is zero because there's no force acting and the time interval this time is three seconds. So the only unknown here is X final. We are going to use this kid a Matic that top one right there. There's no acceleration.

So it's just gonna be ex initial plus of the initial times t So ex initial was 3.12 plus the initial velocity 3.12 times the time. It'll three and we end up with a final position of 12. 12. My pen is not right. 12.48 meters.

So that's our position right here. X equals 12.48 The velocity is 3.12 Still, because there's no acceleration. We needed that because ultimately we wanted to find the position here at seven seconds. So moving to blue last time interval is 5 to 7 seconds. Ex initial was the one we just found.

That's 12.48 meters x final is what we're trying to find. V initial was the ending velocity of the last interval 3.12 meters per second. The VI final is another thing we're trying to find acceleration is that same as it was during the 1st 2 seconds, 1.56 meters per second squared and the time interval is two seconds. So again, using our kid Max, the 1st 1 up here that is going to be that. And we're gonna plug everything in that we've got.

So it's gonna be a 12.48 plus the velocity, which is 3.12 times the time, which was two plus 1/2 of the acceleration, which is 1.56 times the time, which is two squared playing all that indoor handed and a calculator. We end up with a final position of 21 point 84 on, then again using her second Kinnah Matic equation here view final Eagles V not plus 80 v not was 3.12 and the acceleration was 1.56 times the time, which was to throwing that into the calculator. We end up with our final answer velocity. Here at the end of all of this rigmarole is 6.24 meters per second. Who? Okay, There you have it..

So they give us on idea of our coordinate system. It's gonna be british standards. Here's positive. Why? Here's positive X. What's that right here? Plus X.

All right now, we're gonna have a puck that goes towards the wall, that's it. And it's making a 35 degree angle with the wall. And let's see it's traveling at 20 m/s. Oops. And then it leaves The angle of 25°, which would never actually happen in real life.

Yeah, 25°. uh And we is moving at 10 m/s. Okay, so first thing I need to do is um reddit velocity and unit vector notation. So a saying, okay, the initial velocity is a vector. So I need, my vector symbol is going to be the vertical component of uh I'm sorry, it's going to be the the horizontal component of this velocity, which is first going to be 20 sign.

Oh, not sim sign of 35° X. Hat is how I usually say that. And then it's going on our negative Y direction. So minus 20 a coast sign.

All right. So what we have here is a more challenging problem just because there's so many time intervals that are occurring. But if we draw really, really careful diagram and that will help us not lose track of what's going on. So um, I think it would be most useful to just use a really, really long number line for this problem because it's one dimensional or only horizontal world ex access. We don't need to draw a holograph.

We don't need to worry about vertical. Why direction? We're just going to drawn number. Hawaiian. Teo represents our X axis control. The there dot right here.

It's indicated That's our starting point. That's where the puck is and our puck starts at a location of ex not is equal to zero as well as time is able to Sarah straight all drama and blue. So see that Tina is equal to zero and we're going to travel along to three different time intervals. So the permission that we rely so far is that it starts off a Tex equal zero and Tiegel is there. I'm just running ahead.

We're going Tio cross by three other time. So they're they're two seconds. Five seconds and seven seconds. I propose naming those by t ones he to Auntie three respectively. To t one is going to be gold two seconds and just to save space, I'll not worry about the significant figures on my diagram Butty too is one to be equal to five seconds and then likewise t three is going to be equal to seven seconds on DH then we don't know Well, I kind of dropped drew them evenly spaced here not to imply that they're all going to be evenly spaced.

But these are frankly unknown distances as yet that we have to solve for throughout the problem. So we'LL just follow the subscript convention Call him X one X to and X the re So this is our schematic of what's going on in the problem on And then the reason why the puck is even moving in the first place is because there may or may not be a force acting on each of these time intervals and I'm drawing and green right here. So that means is that we have a non zero force. So that is this zero point two five Newtons s So in the first time interval the forces being applied. Then again, with that things were about sitting figures zero point five Newtons.

But in the second time, interval the forces knowing we're being applied. So that gives you a force zero. Um, but then in the third interval, the forces again turned on because the player is pushing on again. So we again have a force of zero point two five. Newton's going on here, So those are the complexities they're going on.

We can already see if the force is going to be grilled. Zero. Then Newton's second law, which is is equal to M. A, is going to tell us that the acceleration is equal to zero. Um, but the acceleration is going to be non zero in the first and the third interval.

But anyway, I'm getting a little ahead of myself. So oops, um, it's a circle that a different color. In part A. We'll be solving for this position as all of its respective lasa ti and then part B, we're going to be solving for X three, the final position and velocity. This's our schematic diagram of what is going to go down in this problem.

And so let's just go ahead and get started with a mama with party way have our very carefully, carefully crafted diagram. So actually going to be using Kinnah Matics equations he's going to tell in order to solve for the conditions and velocities. So x one is going to be equal to the initial position. X starts added to the initial velocity. The nonce on Sorry, these were supposed to be current scenes.

Um, but the time interval from T won t zero and we super explicit here. Plus one half of this acceleration from T one through T zero squared is important on the acceleration term. So you can already see here. One other piece of information that was given to us they're about to skip job is that we also know that the pup begins at arrest. So, in other words, it's initial velocity is equal to zero.

That really helps us sounds. First equation right here. Tres velocity is you will seriously at second term ghost is there. Um, and in addition, the starts, the origins, its initial position is equal to zero. That will helps out there.

So we only need to plug in numbers for the third term. So you have a one half at front on acceleration that is going to be equal to zero points. Fine. Zero by your points. One six oh times.

This time in a balti points zero zero minus zero square. Which two? Three significant figures will give us three points. One three meters. Great. So that's the first half of party, eh? I think we have enough space down here to squeeze in the velocity so we can use a simpler can Max equation.

That being the one is legal. Teo, be not simply added Teo, the product of our acceleration multiply by the time interval that were working over sense from tees here T one squared No, one half, just like before. We know that the initial velocity is zeros. The first term goes away. So we are just left with that same acceleration which is force divided by mass.

Sorry, there's our acceleration that our time interval square, anything like that. And then at the end of the day, so it's starting to look a check mark of the one is remarkably. You can see the same answer because of the coincidence that the time intervals two seconds squared before but canceled out with the one after prints. That's also a numerically through going three. But the velocity so has different units comes with meters, units of meters per sec, second and knots of anything else.

So that wraps up part a. So remember diagram. I have the have the diagram wind down. I needed a new slide, some more space. So we ultimately want to know what the position and velocity are at that third going time more.

To get there, we have to calculate the position and velocity at X to first. So luckily, there's no accelerations. So the velocity that Position two is simply going to be the same as this one, which we said earlier was through Portland three years per second. However, because we're still travelling at a constant velocity, we only to apply the kingdom attics inclusion again in order to Teo for our new position. So final velocity is equal final position and is able to initial position plus Thea velocity that initial velocity was time interval.

Um which is from T born to t to that is too one half the acceleration industry Super explicit thie. Any potential acceleration that's occurring along this time Interval squared. So you know, from our careful diagram large earlier that this acceleration is zero. So we don't need to worry about their term. However, we do have to worry about the first two terms.

So we just said in the previous part we got three point one three meters two, three, six figures, but is actually three point one two five. So this is a good instance. Where when you have intermediate calculations and problems, you should use as many significant figures as you can. Don't round off intricately answers so that it doesn't affect your final answer. So we're going to be careful and include That s our new time.

Interval is between two seconds and five seconds s. So that's going to be that. And that will give us an ex too. That's equal, Teo. Exactly twelve point five meters.

So we can underline that that's not the final answer, but it's necessary in order. Teo, get to the final answer because we have a new ah, another time interval Just wanted for an acceleration. And that's between extra next three. So this to r V. tube velocity.

Then the time intervals from Teo Teo three get the one half acceleration again in this interval. And that is still from T two t three. Mr. Terrorist Squared squared. Okay, so none of these terms are equals zero.

So I need to put in something for all three. We just found out west he was. And that is twelve point five V two. He said is the same as being one. And to forcing me from figures that was through twenty five.

And then our new time intervals from five seconds to seven seconds. That gives us a seven. All right, Fine. Then add that to one half and then remember that were cackling our acceleration by dividing the yeah force by the mass sensor of one six zero acceleration on in time Thinking that squared this time seven rights five squared pelo that all into a calculator and then two, three student If figures could get tourney horn point nine returns. This is X three, which is one of the quantities that they are asking for in party excellence today.

I didn't label that we're doing party. We're gonna party. Here we go. Okay, So we are not signing at the last thing that we need to do. So trust little box in the corner.

Walk over here now is just offer that velocity at time, interval or sorry time three. So we already know that would be too is This is It's the same as the one because there's no acceleration. But the three is going to be equal. Teo be too. Plus the acceleration a lot was time interval, which is from course two to three.

Kalou's princes. You should go back pretty quickly. The full on round answer for Bi two is three twenty five at the again to our acceleration. And then that same time interval should try to fit. It's a seven point zero zero minus five one seven zero her flute Fitz.

And if it's yes, okay uploaded onto a calculator and you get a walking from the velocity of six point two five meters per second. This is the three right, and that wraps of party. So we are older.

Eso Since Mass Mowlam is dangling, the tension in the court must be equal to the weight of the mass. So if we say attention as FFT, then that must be will do the weight of the mass and the same tension is in the other end off the card where the attention is used to maintain the circular motion off mass Big M. So on the other hand, will have FT, which is equal to which is responsible for the circular motion. And f r is the forced you to that? That's gonna be will do m times a r where air is the centripetal acceleration. And by definition we know that this is equal to B squared over R.

If we combine these two expressions, we see that mg must be equal toe begin three squared by our and from here. If we sold for V, that's it's gonna be little m g r over begin, which is, uh, what we want. Thank you.

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