questions-1-2-and-3-are-short-free-response-questions-that-require-about-13-minutes-to-answer-and--5

Question

Questions 1, 2, and 3 are short free-response questions that require about 13 minutes to answer and are worth 8 points. Questions 4 and 5 are long free-response questions that require about 25 minutes each to answer and are worth 13 points each. Show your work for each part in the space provided after that part.
(DIAGRAM IS NOT AVAILABLE TO COPY)
A horizontal spring with a force constant of 40 $\mathrm{N} / \mathrm{m}$ is attached a 0.1 $\mathrm{kg}$ block.
(a) If the block is pulled to a distance of 0.5 m and released, what is the maximum speed of the block?
(b) What is the frequency of the oscillations?
(c) If the spring were flipped vertically and attached to the ground with the block placed on top, how would the natural length of the spring change?
(d) How does the frequency of the oscillations of the vertical spring block oscillator compare with that when it was placed horizontally?

Linda Winkler · Accepted Answer

here we are considering a simple spring past system to look at some of the quantities that are involved when a mass starts oscillating on a spring. So we&#x27;re going to start off with a mass oscillating horizontally. We are told its mass like one kg and it spring constant for the spring is 40 newtons per meter. And what I suggest is we immediately know the frequency with which this mass will oscillate. We&#x27;ll pretend it&#x27;s on a frictionless surface so there is no friction. Um But the angular frequency omega is simply equal to the square root of K over em. So those two parameters govern how quickly it oscillates if we want that in terms of frequency in cycles per second. There&#x27;s a simple relationship. Um And if we use the numbers given 40.1, What we get basically is 20 radiance per second for the angular frequency. And that means are uh frequency in cycles per second Is just that divided by two pi. So that would be 10 over by cycles per second. So roughly three cycles per second. So we know already how this thing will oscillate wants to get set into motion but we can figure out another parameter. Let&#x27;s assume that we stretch the block immediately to the right by an amount of five centimeters. It&#x27;s a reasonable amount to stretch that spring. But we can now calculate is not only the frequency of oscillation, but how fast it is going to oscillate back and forth. We can simply use conservation of energy that the original elastic energy that&#x27;s in that system. When you stretch it gets transferred all into kinetic energy when that mass goes through the equilibrium position. So we&#x27;ll consider that the point of highest velocity through the equilibrium position where the spring is neither stretched nor compressed. Um And of course kinetic energy is 1/2. V. Squared. So we could figure out the velocity of this mess. Yeah. Mhm. Some things will cancel their nicely and the velocity final will be equal to the square root of K over em. Um And then the amount of stretch so notice that that final velocity is going to be related to the omega, the angular frequency and that&#x27;s kind of interesting but you need a stretch in order to turn this into a velocity. Um And we knew that that Initial Omega was 20 radiance per second And the s. is five cm Or 5.5 m. So that will translate into a final velocity Of 1.0 m/s. Okay, finally we can ask what if instead of um oscillating the mass horizontally, what if we put it on top of the spring? So we&#x27;ll assume there&#x27;s some sort of little platform of something that we can put the mass on to load the spring down vertically. The thing to note is that you really haven&#x27;t changed the stiffness or the mass. Hopefully compressing it will not um do anything to the spring constant. Um Ideal Spring would not work that way. So the Omega is still 20 rads per second. And about all that has happened is you&#x27;ve changed the equilibrium position. Hook&#x27;s law tells us that okay times the amount of compression, so just sitting on their the mass is going to compress the spring. But according to Hook&#x27;s law, that compression will change the length of the spring KS equals M. J. In equilibrium. And if we saw for S it&#x27;s roughly 0.1 times 10 over 40 meters, 4.025 m. So the result of putting the mass on the top is to shorten the spring bye. A couple of centimeters. And hopefully that will not change the spring constant, not for an ideal spring.

Kristela Garcia · Answer

in this problem, we&#x27;re asked to draw a free body diagram and calculate the acceleration for two masses on an incline plane. So we&#x27;re gonna start with our diagram and these messes air connected over a frictionless pulley. And this has a massive three M, and this has a mess of em. And so for part A, we just want to draw the free body diagram. So we have normal force for particular tour surface mg straight down the ramp, straight down, we have a tension going up the ramp and a friction down the room before mess em. And then for mastery M, we have the attention going up and if three mg going down. And so for part B, when we&#x27;re asked us offering its acceleration, this is just gonna be in terms of variables, and we&#x27;re actually gonna treat this is the system. So we&#x27;re gonna take all the all of the forces that would affect its motion. So if you look at M G, it&#x27;s gonna have two components. It&#x27;s gonna have a component that&#x27;s adjacent and a component that is opposite. And these components MG signed data friction. These tensions and three mg are all going to affect the acceleration. So we&#x27;re gonna take this direction down for three. We&#x27;re gonna take that direction is positive. Sobering to say three mg minus T plus T minus F minus mg signed data. And this is Newton&#x27;s second law. The sum of the forces? No, we&#x27;re just treating it as a system. So I&#x27;m gonna write some of the forces on the system equals mass times, exhilaration, and our mass. Here is our total mass. So that&#x27;s three em plus m. And then the acceleration is going to be for the system. And here you can notice those those tensions cancel out, and friction can be replaced with the coefficient of friction times my normal force. Now, in this case, my normal force is going to be equal to m g co sign. So friction can be replaced with the coefficient of kinetic friction times, mg coast and data. So replacing all of my variables. With that, we have three m g minus mu m geico sank data minus mg sign fatum all divided by three m plus m, which is four em equals a And my accelerations here can actually counsel. I&#x27;m sorry, my masses here can actually counsel. And I&#x27;m love with three G minus mu. Geico signed Tatum minus She signed data all divided by four. And then part see asks Well, if we have a coefficient of kinetic friction given as 0.3 and an angle of 45 degrees, how far are we going to travel in three seconds? So for that we can assault for exoneration, given our values. So using 10 for gravity, I end up with three times 10 minus point three times 10 times. He co signed a 45 minus 10 times. He signed a 45 divided by four gives me an acceleration of 5.56 meters per second squared and then, using a cinematic equation, I can assume that the initial velocity was zero since it started from rest and then our distance traveled will be 5.56 times when half times our time squared of about 25 meters

Ze-Han Lee · Answer

So in this problem, Initially, we have a blanc spring system. So the mass of the blood has kept two am and Ah, so we wanted to know the velocity of the block when the position is that half can&#x27;t say we&#x27;re raised amplitude. Okay, so we see that. Ah, According to the energy conservation role, we have half times k times, uh, have a squid plus of envious Quit quote times K Time&#x27;s a square, right? Because ah, business the Polina chip assist them. So this gives us the velocity off this Welsh at this moment. So this equal three or four k over an times, eh? All right. And in part B ah, we want So there is a new situation. So there&#x27;s the spring. This is the block. Mismatch them. So there is a ball with Miss Moore M drop the front dropped about this dropped above this. Ah, this block and ah, we want to know their velocity when it keeps this Ah, block. So, uh, we simply utilize the moment the moment, um, conservation law so that I gave that gives us and plus m times of the one equal m times feet. Right? So uh, this weekend soulful. Be one according to this equation and the V one equal and Bob, I am close to him. Times Square roots three over four K over M k over capital. I am sorry. This is the kept him and time&#x27;s a All right. So this is the new velocity when the boys, when the bull is getting on the under on the block and in Parsi want to know the new period of this? Ah, after the two combined together. So in this case, equal two pints. Beirut M class M. You are by K all right. And in party. Uh, so we want to know the new amplitude. So according to the energy conservation law, we have half k times half a squid, which is the pitcher energy stored in the spring, plus the energy off the ball block system, which is 1/2 and plus m times of anyone squared equal to the total energy. Have k a prime squid. It&#x27;s a problem. Here is Ah, the a prime here is that new attitude. Okay. And the wine is the one that we found previously, so we can&#x27;t in. That&#x27;s a prime equal square roots One over four plus three over four times come to empty by by a small implants kept om and time say All right, so this is the new attitude and in part E So if we heat if the bulky to the block at different, at different position, like x equal while we&#x27;re full, eh? So clearly the pure itself does not depend on the position, no matter where the ball hits. Ah, this ah book. The period is always to pi square root m plus m the Rabbi Kay All right, appear it is ah is the same. However ever, if we&#x27;re looking at two, the amplitude empty to actually actually changes. So again, if exit poll want more full A in this case, then we actually get a prime equal square root won over 16 class 15 over 16 times and they buy by m class m. I&#x27;m c All right. So come bit compared to this one compared to the result that we that we obtained here, clearly they&#x27;re different. So if we if the book it&#x27;s blowing a different position, the temperature which Inge Yeah,

Salamat Ali · Answer

Hey off the problem, the maximum can take. Energy is equal to maximum potential energy. From here we write equations for boat for spring we have can take energy. One word to every maximum square is equal to one over to K employees shoots square potential energy for spring from here. Then we&#x27;ll cancel out one or 21 or two on both sides and soulful. Do we? Then we were here. We get here is square root off. Okay, square. So it will come out here then Kay derided by him here. So? So I&#x27;m starting the value for amplitude, which is zero point 05 meter square, wrote off a K we have here is a 40 Newton per meter derided by mass which is a report one then this gives us they were lost You if one meter per second maximum velocity you bought me We were asked to find a frequency frequency busy pulled over on a spring Is one over two pi r squared off this spring Constant okay, divided by the mass. Then we can substitute the values for a K and them And uh for the two pi K we have is a 40 divided by again as a 0.1 mask. Cagey. That gives us the frequency off the 3.2 hearts. 3.2 hearts. So this was part Be in part. See, um the forces will be called forth. Restoring force will be equal to the weight off the right off this mass attached to a spring. And then we can replace restoring force by K X That is a cool toe mg from here x will be called to moss. Damn she divided by K then substituting the values we get. Tau your reproof press G 10 meters per second squared, divided by four d. This gives us the ex increase in lengthy zero point ah to five you major the natural lent off the spring would decrease by 0.25 Indy slipping the spring water curly the if the frequency remains unchanged

Dading Chen · Answer

Okay, so first we have to determine the spring Constant spring constantly sequel to everyone over acts and we know at the beginning there&#x27;s only gravity Will block was applying to the spring as distance was stretched by about 0.70 meters. So therefore the spring constant this evil NGO s was equaled 0.10 Kilburn test 9.8 meters per second squared over 0.70 meter which would give us the spring constants about 14 Newton commuter. And here this is everybody that were true for this question. So there&#x27;s a restoring force. It was point upward when we ah pull the spring in a downward position. So in the Donald position, there should be gravity and the force that we apply, which is F two here. So therefore, after he&#x27;s able to restoring horse minus embassy when they&#x27;re restoring for the three news and NGC was 0.10 grams trans 9.8 meters per second square which would give us the force that we apply is 2.0 to Newton. So therefore will stretch a certain amount This in here, OK, which is the the place of the position, then a. There&#x27;s a block costar bouncing while it one and then the block gold. So we have have toe over Kate here, which is 2.0. So Newton over 14 years and premiere, which will give us 0.14 from year. So the block will star somewhere which is below the equilibrium position. Okay, so start at 0.1 whole meter below the equilibrium position. Why let it go? And for the next question, it was asking us the work doing stretching process. So the work of they&#x27;re in the stretch process to people want, have times K and has extras work because we stretch this amount distance. Okay, so there for all the work down should be good. 0.1 qualifies you and questions. He was asking us the total energy of assistant. The total energy of assistance you wouldn&#x27;t want have case where and I m still here should be the initial stress distance, which is when the gravel block applied to the spring. And then plus the design we stretch, which is extra here. So therefore have used with won&#x27;t have Kate has x one plus x two square, which is we won&#x27;t have times 14 centimeter and then times your point, uh, 070 meter plus through a 700.1 point. Peter Square. What is it with a 0.2? It would you So that other energies people that you&#x27;re pointing three told you and we&#x27;ll question t well, remember, with stretch about 0.144 meter. So So the distance of a block that will be travel. After I releases, she&#x27;ll be two times the distance I stretch OK, because you bounce up and down like a oscillation. Okay, so their voice two times, x two, which is two times 3.14 from year, which is your point to a meter. And for the last question, this is the energy bar charts I drew for his question. So the energy already will be reduced as ah, tangles. OK, so the question was saying that the reparations last 4 50 cycles. Okay. And we know there&#x27;s a resistance involved during this process. So which means that this will be sorry Energy lost due to the air resistance. So has tangos. Theology really will be reduced. So our lab, this is the and energy A cycle one and then on the rise thesis the energy as cycle 25 and ah, these are the answers. What is question?

Averell Hause · Answer

today the system This will essentially be this mass, Uh, the spring and finally the earth and, uh, they interact the, uh the spring force, you can say the normal force and of course, gravitational force. So I would be your answer for party for part B. The points of interest are we can say the points of interest. Ah, points of interest O R. Point where mass is released from rest. And this is essentially a X equaling 6.0 centimeters. And of course, um, point of equilibrium or the equilibrium position. And here, that, of course, is ex equaling zero centimeters. So those two points or where what we&#x27;re interested in now, for part C, the energy stored in a spring would be the elastic potential energy. This would be equaling 1/2 k the spring constant times. X squared X would, of course, be the compression or the stretching of the spring. And so we can say that the potential energy of the spring would be equaling 1/2 times 850 Newtons per meter. And then we&#x27;re evaluating this at six centimeters or 60.6 meters quantity squared, and this is equaling 1.53 Jules, This would be our final answer for part C and then at the equilibrium position 1/2 times 850 Newtons per meter. But then, of course, it&#x27;ll just be multiplied by zero meters squared, giving us zero. Jules, this would be our second answer. This would be again for the equilibrium position, which makes sense. If the string isn&#x27;t stretched or compressed, it will have no associated potential Energy. Four Part D. Now we know that the only force doing work on the masses the conservative Spring Force Um, here, of course, the normal force and the gravitational force are of course, existent. However, they act perpendicular to the displacement and so they&#x27;re not performing any work. And so we can say that then, applying the conservation of mechanical energy, we can say that 1/2 and the final squared plus 1/2 K x final squared equals 1/2 and the initial squared plus 1/2 Okay, ex initial squared and so essentially solving for the final speed, the final speed is gonna be equal in the square root of the initial velocity squared, plus the spring constant divided by the mass M multiplied by ex initial squared minus X final squared. And so this would be our formula for the final velocity or the final speed? No, If the final position is the equilibrium position and the objects has started from rest, this means that here again, the object starts from rest. So we&#x27;re eliminating this term here and then it&#x27;s, um final position would be the equilibrium position so that X final is also eliminated. And so we can finally say that here than the final would be equaling the square root of K over M multiplied by accidental. And so this would be equaling weaken, say 0.6 meters or six centimeters multiplied by the square root of 850 newtons per meter. This would be divided by 1.0 kilograms and we find our final speed to be 1.75 meters per second. This would be our final answer for final speed. This is again given that X final is equaling zero centimeters and that the initial is equaling zero meters per second. Those are the constraints or rather the the variables for that answer Now four part E the object is halfway between the release point and the equilibrium position. So now, although RV initial is still zero, we can say that VI final is now gonna be equaling the square root of K over Ama multiplied by X initial squared minus X final square. So this is gonna be equal in the square root of 850 newtons per meter. This would be divided by 1.0 kilograms multiplied by 0.6 meters quantity squared minus point 03 meters. That would be halfway between the release point and the equilibrium position, quantity squared. Extend the square root and we find that the final velocity is then equaling 1.51 meters per second. This would be our answer. And as you can see here, um, this is not half the speed this we can say this is not half thes speed from Part D, because final velocity is not a linear function of position. So that would be our second final answer. Four part E. That is the end of the solution. Thank you for watching

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Questions 1, 2, and 3 are short free-response que…

Problem 3 Easy Difficulty

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