directions-questions-1-2-and-3-are-short-free-response-questions-that-require-about-13-minutes-to--7

Question

Directions: 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.

(Figure is not available to copy)

An experiment is designed to calculate the spring constant $k$ of a vertical spring for a jumping toy. The toy is compressed a distance of $x$ from its natural length of $L_{0}$ , as shown on the left in the diagram, and then released. When the toy is released, the top of the toy reaches a height of $h$ in comparison to its previous height and the spring reaches its maximum extension. The experiment is repeated multiple times and replaced with different masses $m$ attached to the spring. The spring itself has negligible mass.
(a) Derive an expression for the height $h$ in terms of $m, x, k,$ and any other constants provided in the formula sheet.
(b) To standardize the experiment, the compressed distance x is set to 0.020 m. The following table shows the data for different values of m.

(Table is not available to copy)

(i) What quantities should be graphed so that the slope of a best-fit straight line through the data points can help us calculate the spring constant k?
(ii) Fill in the blank column in the table above with calculated values. Also include a header with units.
(c) On the axes below, plot the data and draw the best-fit straight line. Label the axes and indicate scale.

(Table is not available to copy)

(d) Using your best-fit line, calculate the numerical value of the spring constant.
(e) Describe an experimental procedure that determines the height h in the experiment, given that the toy is only momentarily at that maximum height. You may include a labeled diagram of your setup to help in your description.

Linda Winkler · Accepted Answer

many examples of conservation of mechanical energy involve spring potential energy getting converted into other forms. So typical examples might be clowns getting shot out of spring cannons during a circus event, um Or toy guns that can shoot marbles upwards. Um And in these cases you&#x27;re converting spring potential energy into gravitational potential energy or and or kinetic energy. But let&#x27;s take a concrete example, let&#x27;s suppose you have one of those little spring toys, they usually look like chickens or rabbits, but we&#x27;ll pretend it it&#x27;s just um a mass. And will also say that you can load different masses onto the toy. The spring starts off relaxed with a relaxed length L. Zero, which will change as you load the toy down. But we&#x27;ll have uh some way of measuring that, we&#x27;ll say. So you squeeze the toy down, press it down a little bit. Let&#x27;s say you always maintain the same distance through which you&#x27;re going to compress that spring. We&#x27;ll call that distance X. And what you&#x27;re doing with your experiment is you&#x27;re trying to determine the spring constant K. Of that spring. And it&#x27;s a little bit too hard to do sort of the vertical hanging experiment. So what you&#x27;re going to do instead is watch the little toy leap up into the air and it will leap up to a certain height after the spring completely relaxes and leaves the surface. Um And that height will be measured relative to where it started with compression by delta X. Or by X. We&#x27;ll call it. And the question is, how can you experimentally determine the spring constant from these experiments? Well, this is conservation of mechanical energy. And the form that you&#x27;re starting with initially is the um spring potential stored in the spring. And that&#x27;s an amount one half K. X. Squared. And the final state of energy is gravitational potential. MGH Provided you measure that h um upstairs from the low point of compression. Okay so that&#x27;s just equal to MGH. Ah um Now you could just take one data point uh and solve it for the spring constant knowing all the other parameters. But of course it&#x27;s better to do an experiment. So the experiment is going to have us load different masses onto the toy. Um And for each mass you&#x27;re going to somehow determine how high the toy grows with that mass loaded on there. And a way to do that is you would want to um put a ruler a meter stick, ruler depends how high it goes behind the toy and perhaps record with your phone so that you can go look at the videos later and catch at what marking the mass rose. And you can play through those videos slowly so you can see what that marking is more clearly. Um But in any event let&#x27;s suppose you do that experiment and you gathered some data. Um So I&#x27;ll share some data. Um But what you would like to do is to make a plot of some sort in order to make a straight line and determine the spring constant from the slope of that line. So let&#x27;s see in your experiment will say that you&#x27;re controlled variable is the mass that you&#x27;re loading on. So that&#x27;s what&#x27;s called the independent variable. Um And H. Is the dependent variable. It&#x27;s the result of the experiment. Um So you basically want to get H. As a function of em. And so let&#x27;s divide both sides of that equation by M. M. I. G. And you get Kay X squared over two MG. Is equal to age. Now that won&#x27;t make a very good straight line. Um Remember you&#x27;re holding the x. constant and of course G is just 9.8. Um But because amazon the denominator that&#x27;s an inverse relationship. So the plot that you actually want to make to get a straight line for your experiment would be H versus one over M. Uh And then we expect as the mass gets larger the height will get smaller and vice versa and we would expect a nice straight line relationship. Um So I&#x27;m going to share a table of data and I&#x27;ve made the corresponding plot of the height versus one over the load mass. And let&#x27;s see uh what we get from that, I&#x27;ll share that. Right? And I&#x27;ve grafted using Excel, I do like Excel. Yeah. Uh And if you don&#x27;t have Excel, one of the things you can do is use google sheets, which is freeware if you have uh Uh huh. A google account um you should be able to use google sheets. So indeed when I plot the height in meters vs. One over the mass and inverse kilograms. Indeed, there&#x27;s a straight line um the data table is shown above. Um And what I can do with Excel, which you can also do my hand is I can fit what&#x27;s called the best fit straight line, which tries to get as close to as much of the data points as it can. So some of the data will fall a little bit below the line and some a little bit above the line. Um I actually forced my graph to go through 00 as well. Um So that is one of the other points on there. And I find that my slope is 0.0101 with units of meters um per Uh one over kg, so that&#x27;s a little bit cumbersome. Okay, but the slope From my graph is 0.0101 beaters times kilograms, we&#x27;ll call it that&#x27;s meters times kilograms. Um And now we want to solve this for the spring constant, but we can see that in the formula above the quantities that make up the slope, R k, X squared and to G. And that has to equal 0.0101 um giving us the K and S. I. Units. Um So we simply have to cross multiply and know what the initial compression was. So let&#x27;s say we were starting with the compression of .02 m and we tried to control that, that should not have been too hard to do. Um And so we can solve for that unknown spring constant from slope um and X squared and twice times 9.8. And because everything is in units of S. I meters, kilograms and seconds, the units for k should be newtons per meter. Um and putting all that together, let&#x27;s see what we get. Uh huh. It looks like it&#x27;s going to be surprisingly large, surprisingly stiff um spring constant And I&#x27;m getting about close to 500 newtons per meter. Um I&#x27;ll write it down with all its decimal Figures for 95 Newtons Per meter. Probably not that precise, but um very close to 500 newtons per meter, so it&#x27;s a pretty stiff spring. And if you&#x27;ve ever tried to punch these little toys down, that may not be too surprising.

Eric Mockensturm · Answer

We&#x27;re told that in a physics lab experiment, one under the horizontal spring that a raise hook&#x27;s law So the forces linear, proportional to displacement or through the change in length is attached to a while. The spring has compressed 0.4 meters and they blocked with mass. 0.3 kilograms is attached to it. The sprinters and released on the block moves along a horizontal surface. Sensors measure the speed of the block after it has traveled a distance. Steve from its position, um, that measure values are listed. The data shows that the speed V of the block increases and decreases has a spring. We courage to its on stretch length. Explain why this happens in terms of worked on a block by forces that that Okay, well, Teoh, start off here. So as the spring he compresses, it does work on the block. Once the block passes the point where the spring is neither compressed or stretched, the black will start doing work on the spring. But friction will always be taking energy from the block acting to slow it down. So what we have here is a If if Thea, if there was no friction than we would actually reach a peak speed that 0.4. Okay, so it would go like this instead. We have something that goes like this, and why that is is because of the friction. The friction is actually his slowing it down is taking a lot of energy from the system. And in fact, it looks like even if we continue down, we&#x27;re taking data that it would actually stop when it got too got to 0.4, which is at the equilibrium. So you basically what you did, you compressed the spring, and then the block this law moved started moving pretty fairly fast, and then it started slowing down and basically was gonna just come to rest rate about, come at theon stretch, length of spring. Because if there was no velocity at that point, which it looks like there was very little, um, there&#x27;s no there&#x27;s no energy to then transfer back from the block into the screen to then stretch it. So it&#x27;s basically just gonna come to rest at the point where the spring is at its unstrapped length. So then we&#x27;re asked, um, let&#x27;s see here to use the work energy theorem to derive an expression for B squared in terms of D. Okay, so we have the potential energy in the spring. Okay, The knot is gonna be are under stretch length with which is 0.4 meters. And then we have the rate of work done by the the rate of work done on the system by the friction force. Okay, so this is the power that&#x27;s being dissipated by the pissed the friction force. Um, so then we know that the kinetic energy of the block, And so then we have that the power, the rate of energy change in time, rate of change, of energy, my secret, the power flow out of the power into or out of the system. Um, we can integrate this because efforts a constant. And then he just becomes the So we get that, um, e is some constant initial energy. Some of the initial energy of the spring and the block times, um, the this should be a d times Thea. Their work, um, dissipated by the friction force. And then that equals, um So there&#x27;s though initial kinetic energy in the spring. Now in the in the block and the spring has a connection. Initial kinetic energy. I want half k d not squared the energy at a get at any given time. Then is, um, 1/2 k times the change in length of the spring squared. So the energy stored in the square in the spring at it some given time and then a the potential energy in the block So we can then saw for V squared. OK, so we go back to buy am multiplied by two. And we get this expression here where K over AM times the quantity d not squared minus thes squared minus. Do you not squared all that and minus two times the friction force divided by m times? The now you can obviously simplify that a little bit and begin it down to a form that looks like so there&#x27;s a d. I&#x27;m multiplying the whole thing. And then in here you have this expression here that also depends on D and D, not on the friction in the massive spring Cost it. So we were asked to then a lot, um, the data as the fund has b squared as a function of d. So I did that here. And so here is our the data points and then I just have lines connecting them. And so this is V Squared versus D, And then I then took this function here substituted in the value for D squared or D Not where this is to be not And then I had the computer fit that this function to this data, I think at least squares fit and we find that then and basically saying that, OK, d is our, um variable and then K over em and ff ever Amaar unknown fitting quantities, fitting parameters And what we get from that fit is that K over M is about 40 and FF over m is about 7.83 If we then plug these back into here, then we get V squared as a function of D. And then we complied That and you can see this Prabal here that actually goes fits the data very well. Then we can maximize that and we find that the squared max is 1.67 which means that V max is 1.29 meters per second and then that occurs at 0.24 mirrors, so, you know, right? A little bit after here. Okay, so then there we were asked to figure out what, um, what Spring Carson is and what the coefficient of friction is. So we could do that first, If we know the mass of the block, then K is 40 meters per second squared times the mass of the block. And that gives us, um um 12.0 newtons per meter. So that is the spring constant. And then if we look at our friction, you can write our friction force as the kinetic, uh, kinetic friction coefficient times the normal force from the table on the block, which is mg. So the M cancels out and we get, um, UK Times T equals the value from our fit 7.83 And so we get you que is about 0.80

Jincy M Saji · Answer

e no position of the peace of the spring position as and so any quarter zero, it is a fixed point, huh? Any legal capital is the moving end of the spring. How do we understand if something is there? One spring is attached to the site. So one scientists next on the other side is moving. So here the length a legal zero. And here the length Caldwell capital. This is how that is Define No, the velocity off the point corresponding to end The velocity is in ordered by you Now you off. L equal to be led by it That this when the spring is moving l b l b the function off time, then spring moves Then what happens? L a will be the function off time. What&#x27;s it is dependent on time? Therefore, you is an implicit function off thing day. So in the A bob gm equal to capital m upon el into deal Therefore DK equals half d m you square because it was too half capital M b squared upon a que into elsewhere, do you? And then okay will do integration off decayed which is equals to em. These where a phone Tu que integration from the limit of zero to capital and square D l, which is equal to mm. These square once is I missed it. No, the beauty part. We have MB Devi upon DT last k x dx upon DT equal to zero or begin right and me last k X equal to zero. This is what we have. We&#x27;ll be long to the sea Buck Emmis replaced by M by three Now therefore, neither will become Rudolph three k a one him there n dash will do and like

Shital Rijal · Answer

Okay, so let us first Ystad with Christian seven b on that say&#x27;s t equals two twice by gender, Mikey City goes to group I returned or m by Kate Now from this equation If we started, it&#x27;s where on both sides. So this would be de squired is equal to for by squired by Ki times So if you compare these with a question of a straight line you know that&#x27;ll slow pass x Okay, let&#x27;s see Then you can see that Okay, so you can see that this is the question one. This shows that the question of one is straight nine straight. So from the from this, how can keep Christian and slope So if you compare so the slope will be called for by a square by kill Now you remind the key. What we would do is just divide full bias Where the snow so easy, right? Okay, So where the lions were intercepted acceptably Obviously the nines wind is equal to zero because there is no see parked here. The windows have locked in this equation So c cold studio so would have to know brought d c square versus m So with me I have plotted this in Emma&#x27;s excel. Okay, so in the wax, it is G square on the XX. There is em from the data on extra. The the question why is it called a 0.1278 x plus Every want 3 88 reason balls to the straight line through this. So the question up, this prison of this is straight line No would remain the stoop. So from here. So this is part of this was this was a party, so no, for part B, we have plotted on Dhe knew tremendous stops is still pass is equal to, you know, 0.1 through 78 Right. So nearly court to 0.13 This is the soap on DDE. A wider set. So that is going to see. Seems to be so much. 0.13 88 Okay. Okay, so we have been part of the problem. Now let&#x27;s move on to park Si. I&#x27;m Parsi, a nun in show that Alondra light so you can experience practically if rather than using. Okay, So we had initially expected from our question that this y two sips of Voyager, But if if you were bought it. But when we parted the experimental results so it&#x27;s sort off. Winer said Right. So this is like a twist. This is a president. So how do we do it this So if we consider the mass m to be m plus and nor so the estimable intercepts A let&#x27;s check it. So let&#x27;s again start with the aggression. Second be so t is equal to to buy roots under and bless. I&#x27;m not upon key, right? So proud of this. You can write a thesis choir is called you or by squire by Ki terms Mm less for vice. Why by Ki times I&#x27;m not right now. These resembles read Why is equal to a max. See where a musical too. So I don&#x27;t want to use I&#x27;m here because we have already used and for mass. So I really use us for slope and seek our intercept. Okay, so this question in this aggression the slope is four fights Where upon gay and who understood is full price for bikey. I&#x27;m not Okay, so So we need to find out, okay? And I&#x27;m not from the stop. Right? So to find out Kay This is lies, Katie. Close to pull pi squared by S. I squired by smoke. This is equal to four party squire divided by a slope in 0.13 So this gives meaning the vendor off gate to be 308 or in mine. Nugent for a major. So we got cake on dhe the well off. I&#x27;m not to find out the value off. I&#x27;m not. This isn&#x27;t good. You compare this to so I&#x27;m not. Seems to be like so I&#x27;m not. Seems to be c upon for price choir biking. Right now. We know everything. So this cans out to be a general this chance to be one point 1.8 eat g g. We know that I love s three other band. All right, so just plug bringing Losing It&#x27;s me. I&#x27;m not. It runs your 30.1 point zero. Agent Casey, this is where the party of the promised you were part of. What is what is an officer here? It&#x27;s pretty simple because as in the Austin, it&#x27;d system not only the mass Arcelor&#x27;s, but also the spring right? So obviously I&#x27;m not. Should be on the effective mass off this spring. Okay, so we&#x27;re done with the solution of this problem. Uh, thank you for watching

Khoobchandra Agrawal · Answer

even this is the problem is if I spring having the mouth and here it is David. I spring off Sam and length and not It&#x27;s a bull and it&#x27;s fixed on the other end is free to move. We have to fight. It&#x27;s kind of particularly if the velocity of a point where is linearly with distance from the extent. So here the velocity is zero. Then it gives increasing and becomes Maxim at the other end. Then we have to find that kind of technology and second part if mass image has to it. So by considering the energy concept, we have to find its angular velocity. If spring having the mask, then what will be the spring Anguilla relatively off oscillation that we have to calculate. Let us start doing it. Starting with part A. I&#x27;m drawing the diagram again for upon the friends. This is all this is a at all and zero. So velocity is zero. Well, the city is increasingly nearly velocity is deflection off. Can Britain has we? L upon and is not. It&#x27;s total length of this unstitched length of this week, Matt. Unique, partly distributed over their spring so mass per unit length will be, um, upon and not or simply, l If it is given l so I should take and only let us start solving it. If we consider a very small element, the l distance from it, then it&#x27;s a mass will be I am upon Elder on it is moving with velocity use culture the and upon. So he element off having that kind of technology You can find that kind of technology d m into you square How, um upon l deal and to you will square that is the square Ellie square upon elsewhere. So what kind of technology off this element? You may right, hop um el que into the square and square deal toe. Find that total. We have to integrate it. After integration, you will get M B square basics. So this is the answer off a part. Now be part energy can be defined. That&#x27;s toe on being toe the V a quantity potential energy K X. Do you accept quantity? That must be true. So we can write AM a plus k x. It&#x27;s called zero. So from here Omega, you can find a skull toe minus K upon them into X on age. Um, you guys square X so miserable Be route off the upon up by energy consideration because energy is constant. So it&#x27;s the derivative is you Don&#x27;t I have written Did this equation directly? If you would like, you can do it here like I am doing, total energy will be half and be sweat plus half X squared since total energy is always constant So it&#x27;s derivative is Toby zero. If you differentiated, you will get and be duly appointed plus k x de except quantity. That must be zero. So that equation we have written there directly, so use it. But it from here. No sea parte if spring having the mosque, bank it, um, to be replaced by m three. So omega is card too. Three. Give I am Or you may right que upon MDs. Um, this will be, And by three, that&#x27;s all for it. Thanks for watching

Derek Walkama · Answer

Okay, so we have a spring with mass m spring constant K equilibrium like l not The end of the spring is moving with the velocity V said we&#x27;ll call Well, define another velocity you This is the velocity that changes along the length of the spring. So you have l equals V when little l Which is to find from the from the end here this is l equals zero l equals capital l wherever the spring extended Thio. So you&#x27;ve l equals V when l equals big l knew of l equal zero when little well, equal zero So at the end of the spring, it&#x27;s moving with speed V At the beginning. In the spring, it&#x27;s moving with speed zero. The next thing we needed to find is the linear mass density. So this is how mass changes would length row l. And this is just em over l the total mass divided by the total extended length. And we&#x27;re gonna turn this into a differential so d m equals m over l d l. So as we move incrementally with little l r mass changes by big M Big M over big L, which is the total extended length of the spring. Our potential energy kay is going Eagle 1/2 M U of l squared. Let&#x27;s take a derivative here, So d k legal 1/2 d m times u of l squared. Um, let&#x27;s plug in what we know for u of l c U of l equals l over l a times V. Um, since we have our boundary conditions that u of l when little l equals big l has TV and has to be zero when little while equal zero. And a linear equation of Celeste fits all of our conditions. So we go and plug this in here, all right? And then we want to replace D. M with, um, this guy. Okay, so we&#x27;ll combine all this and get 1/2 m the so sorry. This is little b squared over l cute times. Little l squared d l Now we can integrate the kinetic energy. So, uh, the integral of decay, which is gonna equal kay equals 1/2 Mm. The squared over l cubed times the integral of l squared rum. 02 big l. So we&#x27;re integrating. Were instigating over our length of our spring from zero to the total length of this spring. So this is 1/2 m the squared over l cubed l cubed over three from zero to l. This equals 1/2 m the squared over l cubed times l cubed over three. It&#x27;s just equals 1/6 Um, the square. So, you see, we get about 1/6 of the kinetic energy. I&#x27;m gonna go to a new page for the second part. All right, so we&#x27;re gonna get the equation for the total energy e equals 1/2. Mm. You squared plus 1/2. Okay. X squared where X is defined as l minus l not. So it&#x27;s a distance from equilibrium. We could drop the one halfs for simplicity because we&#x27;re just going to solve because we&#x27;re saying, is that the secret of a constant and differentiate it So we have m u squared plus K X squared eagles a constant. Now, if we differentiate both sides, we get to m d to emu Dutt once to k x dx DT equals the differential of a constant, which is zero. If you differentiate a constant, you get zero so we can replace Dutt with a and e x city with the Are you sorry in this case? So we have to m u a plus two k x, you equals zero. Take a you out. We have you to m a waas two. Okay, X equals zero. We, of course, could take the two out and divide by the two. So you can just get rid of it here on. And we see that furthest equal zero m a plus k x must equal zero or a equals negative k x over at which we know is true for simple harmonic motion as well. And finally we can do part c No. So omega equals k o gram and we&#x27;re at the kinetic energy again is 1/2 m the squared. Um, but we know our kinetic energy is actually 1/2 of me. Sorry. 1/6 Bigham B squared. So? So if you compare the two equations, we see that little M would equal big M over three. So if we replace that into omega to see how this changes things, we get Omega equals route kay over Big M over three, which equals route three. Okay, Over M. So the, uh, the angular frequency increases by a factor of Route three. If we consider the mass of the spring

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