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Read the following description very carefully and then create and fully label diagram of the situation: Once you have diagram; please determine the amount that the ...

Question

Read the following description very carefully and then create and fully label diagram of the situation: Once you have diagram; please determine the amount that the spring will compress in order to stop the crate from moving:Working on a completely frictionless surface, physicist decides to have some fun: The physicist sets up 5 kg bowling ball on level ground, directly in front of our favorite crate full of fruit: The stationary; 31kg crate is connected to a long; horizontal spring with k 250 N/

Read the following description very carefully and then create and fully label diagram of the situation: Once you have diagram; please determine the amount that the spring will compress in order to stop the crate from moving: Working on a completely frictionless surface, physicist decides to have some fun: The physicist sets up 5 kg bowling ball on level ground, directly in front of our favorite crate full of fruit: The stationary; 31kg crate is connected to a long; horizontal spring with k 250 N/m and the other end of the spring is connected to wall: The physicist decides to exert 2000 N horizontal force for 0.1m,at the center of the stationary bowling ball: The bowling ball was allowed to slide into the crate, creating an elastic collision with the crate and ending up with velocity of 1.44 mls in the opposite direction: The crate leaves the collision with a velocity that will eventually g0 to zero,as the spring accelerates the crate torest: Extra Credit: If this was normal surface;with friction; would the spring stretch more or less? In order to receive any credit on this part; YOU must justify your answer with conceptual and/or mathematical support A diagram wouldn't be= bad idea;either_. MacBool



Answers

This problem extends the reasoning of Problem 59 in Chapter $9 .$ Two gliders are set in motion on an
air track. Glider 1 has mass $m_{1}=0.240 \mathrm{kg}$ and moves to the right with speed 0.740 $\mathrm{m} / \mathrm{s}$ . It will have a rear-end collision with glider $2,$ of mass $m_{2}=0.360 \mathrm{kg}$ , which initially moves to the right with speed 0.120 $\mathrm{m} / \mathrm{s}$ . A light spring of force constant 45.0 $\mathrm{N} / \mathrm{m}$ is attached to the back end of glider 2 as shown in Figure $\mathrm{P} 9.59 .$ When glider 1 touches the spring, superglue instantly and permanently makes it stick to its end of the spring. (a) Find the common speed the
two gliders have when the spring is at maximum compres-sion. (b) Find the maximum spring compression distance. The motion after the gliders become attachession of the cencombination of $(1)$ the constant-velocity motion of the center of mass of the two-glider system found in part (a) and
(2) simple harmonic motion of the gliders relative to the center of mass. (c) Find the energy of the center-of-mass motion. (d) Find the energy of the oscillation.

No discussion. We start off with two separate messes and one and two that are traveling on a frictionless uh, a trick. No, the M one is traveling at a faster speed. It was M two. So eventually even meet with him to which has a spring attached to its end. And you spring itself has a supergroup. So and anyone touches into a Mesquita, uh, prominently connected to each other. So with regards to this motion, we wanna find for Sovaldi speak that the two fighters were half when the spring Is it maximum compression? So when it is set maximum compression instead, messes are no longer moving with respect to each other, right? So audio at maximum distance. Yeah, of course. It's possible distance with each other because the strongest, most compressed and what this means is that they are moving with the same collonville istea. Let me just call this V A at this particular point and the way they were gonna find this is by conservation off momento. So every start we have total momentum and one speed one plus and two times B two. But the final speech the momentum is in one plus two times common velocity V. So I commend fee one is, uh, 2.7 full. Just a second for me to this is what's your point? 12 to spook sickened. So we just have to substitute in abuse, right? So m one issue and that's your 10.24 Get a grams itis by 2.74 plus mess off into to a 0.36 for the by 2.24 plus your 0.36 My energy should get the common Flossie to be your point tree. Six it you just for a second. Next, we want to find, uh, the motion off. So the maximum compression distance and maybe that we can do that is by looking at conservation off energy. So it is but a good 0.20 this maximum the compressed dean. A moment off compassion that's just considered as X, Then the amount of energy start in the spring. ISS half k x squared. This is de energy story this spring, so the total energy would be the energy storing your spring plastic kinetic energy off the blocks. This must be equal to the total in your show energy, which is the just eka net energy off m one plus m to it. It's not so half one if you want, square us off into the two square speak list to off and want us to times be a square plus the energy d spring. So once again, we substitute in the various values. So we have one one into a P two. We have feet A and we have the spring constant. Okay, Just 45 turns for meter. I saw rearranging this equation. Just doing some energy break manipulation. We should get X to be close to so point you 351 beat us. No, we want to fight the and she off the center mass motion, which is simply the motion off the entire system. No, we already followed it this particle case when this maximum compression the overthrow speed off the system. It is the common Rossi v a This must be d philosophy off the central mess. True entire motion since, uh, you know, there's no additional forces exciting on this to mess us already. So the said no mess must be constantly moving at this fianc? E So the energy associated with this motion is just huff M one plus two times feei square. And this would give us your point too. Force your six juice when you have a hand for the oscillation. No, the energy of the oscillation This associated with the energy starting in the spring, right? Basically et elastic potential energy just given us half cake square. And we used the extension. There was privacy form T. U s question points you treat 51 script. This will give us your points. Your 277 juice.

And this problem where reminded that we usually ignore the kinetic energy of the moving coils of a spring. But let's try to get a reasonable approximation of this. So consider our spring of mass m like l not and forced constant K the work done to stretch or compress the spring by a distance. L is 1/2 k x squared where X is l minus l not. Or the change in Maine. Consider a spring as fragged abroad that has one end fixed and the other end moving with speed V Assume that the speed of points along the length of the spring varies linearly with the distance l from the fixed end and assume also that the mass of the spring is distributed uniformly along its length. Okay, so we can draw a little schematic of a spring here. I've used a coordinate X here. I guess they called it little l. But same thing. And one end of this spring is fixed. Although we could actually say that this was the not and generalize this a bit. But this case were just considered at one end of spring is grounded. The other end has a velocity V sub l. And we have, um, uniform density along the spring, and it's we're gonna assume that the velocity varies linearly. So at this point, we have a velocity zero here and here we have a velocity of Visa. Bell had a x equals l. And so over linear velocity changes. Is this line here? So the velocity grows linearly to, um, this point here. Now what we need to do is we need Teoh. Figure out how much kinetic energy is stored in this spring with the linear velocity change along its length. So what we can do is again we know that, um the spring is uniform, so weaken determine a density density per unit length or lineal density. So that's the mass of the spring divided by the length of the spring. So that's the mass per unit length along with spring. And it's a constant. And so what? We actually want to do what we're trying to look for here. Is this integral? Um, 1/2 B squared the over the whole body. Okay, so that's the end agro that we want to do because that's what have MV squared. But this is if the If the velocity is changing over the body, then we need to do an integral here. And so what we could do is we can say that the differential mass element here is the differential. Um is the density times differential length so we can substitute in that the m equals wrote L the X. And now we have a coordinate the X that we can integrate along. So all right, our kinetic energy integral here reduces down to this, which is an integral can. We can do easily. So what Basic. Just adding up all of these little mass elements. We're selling them up. Each one has a differential kinetic energy, and we're going to send them all up and get the total kinetic energy over the length of this spring. So we get 1/2 the integral from zero to L Rosa Bell V squared DX. And we have, um we have an equation here for V as a function of X, and that is Visa Bell over L um, Times X. So you plug that in and we square and so that all weekend we now we have an integral is a function of X over D X, and we can work with that. Yeah, um, a lot of stuff comes out. So this all of this stuff here is Constance, and we can pull it out, and then we get X squared the X from zero to l. And that's just l cubed over three. Okay, um, so and I have against substituted row in here as m over l. And so we get 1/2 m V. L squared over L Cube. Sometimes l cubed over three. The l cubes obviously cancel. And so you get 1/2 m over three every time V l. The velocity of the end of the spring squared. And so you can say here is that we basically have an effective mass of 1/3 that of the spring. So if we wanted to, we could. It's taking real spring. That has mass. Let's see who Strada thick spring here that has masked and instead and make it a thin spring here with no mass again. These are all gonna be fixed at this end for this formula to work. And we can love all of its mass out here at the end in a turn in a block that has a mass of the third of the mass of the thick spring, Um, and dump it here and then whatever. Have whatever else is attached to this spring act attached here. So it's a way of taking this what we call a distributed system that has a stiff as a distributed mass. So it's distributed Ambridge Energy and turning it into a lump system where all the stiffness is in all the energy kinetic potential energy is in one element, and all the kinetic energy is in a different element. So now we need to, um, given that, given an example where this is important. So we have a spring gun. A spring of Mass. 0.243 kilograms on the force constants of 3200 Newtons per meter is compressed 2.5 centimeters from its own stress length. When the trigger is pulled, the spring pushes horizontally on a 0.5 feet kilogram ball, and the work done by friction is negligible. Um, getting assume I worked on by any all areas of sense of to what I calculate the ball speed when the spring stretches because it's un compressed like by either my first ignoring the Mass of the spring and then including it and then seeing how much the kinetic energy is divided between the two. So here's our parameters we were given. Here's the mass of the ball, the mass of the spring and we can figure out that the initial energy when the spring is compressed but nothing is moving is 1/2 k. L not square. And if you plug the numbers and that just turns out to be one tool and then if we the final energy, um, the spring is un stretched in the state, so there's no lasting potential energy. So we just have kinetic energy of the ball. So we have one F M V squared, and that has to be one drool since where no energy is being dissipated out of this system. So we wind up with the squared is 37.7 meters squared per second squared, or V, is 6.14 meters per second. Now we can, um, take a look at the other case where we include the mass of the of the spring, and actually we should get quite a bit of a different answer because most of the time the mass of the spring is negligible compared to whatever connected to. But in this case, the mass of the spring is about five times almost five times the mass of what it's connected to. So clearly we should get a different answer here. Um, so in this case again, we have the same initial convinced energy state because just that nothing is moving. But now, in the final state, we have the kinetic energy of the ball and kinetic energy from the spring. So the kinetic energy that ball is this in the kinetic energy is of the spring. Is this from our previous calculation? And again, that still needs to be one tool. So you can see that our velocity here is gonna be smaller because we have a bigger mass. And so if we plug in our numbers, you get that V is now 3.86 which is substantially less than 6.14 So we have a much lower velocity here because we've included this extra mass of the spring that that, um um has kinetic energy. And so if we actually divide this figure out If we plug this number into these two terms here and divide, figure out which what fraction of this energy is in each of the thing. In each of the elements. We have 0.40 jewels for 40% of the energy in the ball and 0.6 year old jewels or 60% and vanity in the spring. And we can expect that because that there's more energy here, because this is more than three times that.

So our question tells us that we have a spring that's lying on a table had that is connected to a box. The box has a mass m appointing kilograms, and spring has a spring constant K of 59 Newtons per meter. And there is friction between the box and the surface that it's setting on that friction constant abusive at 0.74 And we're asked to figure out how far, uh, how far out can we pull the box away from the wall that it's attached to? So I have a little diagram here where the spring is in red, in the boxes and green. How far out can we pull the box and have it sit there? So essentially, what this is asking is at what point is the force of friction that's keeping the box in its position That doesn't want the box to move? At what point is that equal to the force of this spring? That's trying to pull the box back to equilibrium because springs want to stay in equilibrium. So if I pull something away from it and wants to pull it back, if I push it in, it wants to push it away. So we're pulling it out and we want to figure out when the force of the screen of some mess is equal to the force of friction. FC DEATH FULL This force of a spring comes from hooks law, and it's equal to K Delta X mean okay and Del Taxes What we want to find when the force of friction is equal to cowfish. Coefficient of friction, music s times, the mass, um times acceleration due to gravity. Okay, so gravity here is 9.8 meters per second squared. So we know everything and we simply just need to solve for Delta X. When we find that Delta X is equal to Musa Bess times in times g divided by Okay, So plugging all these values into this expression we find that Delta X here is equal to zero 0.98 when the units here would be meters mike, um, boxes and is their solution to the question

Okay, So, uh, this problem we have the set up showing here and in part A We want to Ah, let me see. So in part we want to find out Senator, regional, blah gay. At the moment it is released from the rest from the initial position, so we know that it is compressed by one over four times our way. So according to the Hawks law before is acting on the block is the Well, we're four times l a times K this sequel, sometimes a right. So the initial acceleration is ah ah que el over full Em and Inbee say's that Ah, we want Thio. Ah. So if the V winds have lost a block one before impact showed at a velocity of ah, the velocity book one just after impact is 1/2 B one. So because according to a problem, we know that the only half of the kind neck energy is transferred into block too. Right? So, uh, we see that Ah ah, I'm sorry. 0/2 off. The kind of energy is transferred into heat so only half off the continental energy is transferred into the mechanical energy. Right? So we have half times have Ah and be one squid. So this means half off the initial kind of energy. It kuo of envy One prime squared plus half and the two prime squid. Excuse me. So if you want me to me, is the velocity off? Ah, block one block to after the collision. So this is Ah, this is the energy Conservation Hall And then we have the moment of provisional. We have m times the one equal Sometimes we own prime plus end times made to prime. Right? So we have this too quick. We have this two questions and only at well, nonverbal is the view on prime the way to prime. So if you solved this two equations, we will You will pretend that the view in front equal v to prime equal have everyone all right and in part See, we want to show that to the also Asia attitude after the cliche So before the collision we know that Ah, we have a relation half k time's a squid which is the end which is the potential energy stored in the spring Equal half and be one squared. So this is the before percolation. Okay, So you will attend that, eh? Which is the attitude? A call? Sorry. Ah, no, we did not obtain from here Way, Captain. The attitude, the way up to an aptitude from the problem himself. So it's well over four times out, Ray and ah, after the collision and we have relation. Okay, A prime squid, which is the amplitude after the after the collision. A cool half times ofem be one Squitieri. So because hobbled energy is transferred into heat only Ah, half of the energy is kept in the kind of in a modicum mechanical energy. So we have to That's a prime equal. One over squared two times A. Alright, so this gives you what? Over full. You know, I score two comes out and d want to know the period after location. So we know the period only depends on it Has the expression and over case where route times to pirate. So you can say that to the only depends on the mass of the block and, uh, the end of the okay. Sorry and the this bring constant ray and ah, because ah ah yeah, Because because the mess does not change before and after the collision. So we see that Ah ah t prime after the collision, actually equal to note. All right. Actually, equal toe the period before the collision, and e Ah. So anyway, we want to find out. Ah, we want to find out that Ah, greasy the expression of our in terms off. Ah h k o m g. So, uh, Ari's, uh are you so this does That's right. So, uh, we can, firstly, find out the total time off the flight. So we have half gt squared, equal edge. So we pretend that at the time of flight t equal square root to H over G. Okay. So we can tend the displacement horizontal, which is Ah ah v one sort of half the one. Because as we obtained a part B, we have tended the veto. Prime is also happy. One, Marie. So how the wise the is the loss of your block to after collision? Ah, half the one ah times. T people are right, and we plug in expression for the tea and also wish you're plugging the expression full V one. All right, so, to attend the expression of you want winning the utilize this expression we have off. Okay? A squid equal half m times the once great ray. So this gives us V one equals square root k over m times, eh? Or it becomes times went over four times. L. Okay, so Ah, the total time started. Art becomes half times V one times. T it call half times square root k over an times over full. L and T is Ah, he's right here is go to H over G square, root to age her to a J or a G. So this is a funny expression for the are Okay, so this equal one over a thing and ah times l and square roots two K eight over mg, okay.


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