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Problem 40

(II) A roller-coaster car shown in Fig. 6-41 is p…

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Problem 39

(II) A vertical spring (ignore its mass), whose spring constant is 875 N/m, is attached to a table and is compressed down by 0.160 m. ($a$) What upward speed can it give to a 0.380-kg ball when released? ($b$) How high above its original position (spring compressed) will the ball fly?

Answer

a) $7.47 \mathrm{m} / \mathrm{s}$
b) $3.01 \mathrm{m}$


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Video Transcript

in this question. There are three situations at the beginning. There was an un compress the spring. Then someone compressed that spring and put a bowl on top of the platform, which formed Situation number two. And then someone releases the spring so that it can un compress and that the process, it launches the ball upwards. So then we have situation. Treat the first item off this question asks us. What is the velocity off the ball after it's launched by the spring and sold it? We have to remember about energy conservation. Well, you can't destroy on. You can't create energy. You can only transform energy. So to solve this question, we have to use energy conservation in between situations true and three so that the total energy is conserved in these situations. In situation number two, there are two kinds off energy. We have the elastic energy from the spring, and we have the gravitational potential energy from the bolt. So we have the energy that comes from the ball, which is potential gravitational plus this spring's energy, which is an elastic booking show energy. So I write it like this for situation three. We only have two kinds off energy this spring now is on compressed, so there is no energy. Store it in the spring. The reform that is no elastic putting show energy. On the other hand, the ball is moving, so it has a kinetic energy and at the same time it still have a potential reputational energy. And this is the potential reputational energy off situation. Three. Let you write it like this now, in order to calculate these values, we have to choose a reference frame for our situation. I would choose that this height represents why equals zero and that my eye axis points upwards. Now we are able to write expressions for those energies. Remember that the gravitational potential energy is given by the mass times acceleration of gravity, times the height that the elastic potential energy is given by 1/2 off the spring. Constant times that information squared and finally, that the kinetic energy is given by 1/2 the mass times velocity squared. Then using these expressions in these reference frame for situation liberty, we have the following. The potential gravitational energy is just zero because the height in the situation, according to my reference frame zero to it vanishes. And then we have the elastic potential energy, which is given by 1/2 elastic, constant deformation squared on the right hand side, we have something else. For the right hand side, we have the kinetic energy, which is 1/2 m the squared plus the gravitational potential energy, which is AM times G times the height off the ball, inspiration on retreat for the height. Remember that we choose a reference frame to be like this. So here we have Why equals 20 our Y axis points of ports on then noticed that in the situation this spring is un compressed and their afford the ball is at the position Delta l with respect to the origin. Surely height is delta L Then we have to stop this equation for V in order to discover what is the velocity off the boat. For that, we have to do the following. We begin by standing this term toe the left hand side so that we have 1/2 off k times Delta l squared minus m times g times delta l is equals to 1/2 times m times v squared. Now we multiply both sides of this equation by True. So now we have Kate Delta L squared minus two times M times g times Delta L Being equals two m times b squared Then divide both sides of the equation by am so OK divided by AM Delta l squared minus two times g times Delta, Ill. Is He goes to V squared to V easy. Of course, to the square it off k divided by em times. They'll tail squared minus two times g times, don't l? No. We have to plug in the values that were given by the problem. And by doing that we get the following we is given by the square it Off K, which is 875 divided by m, which is 0.380 times the deformation squared, which is zero 0.116 now square it minus two times G times 0.160. And these wolfing is under a square root sign and these results in a velocity off approximately seven point 47 meters per second. Sir, this is the speed off the ball. As soon as it leaves the table for the second item, we have to organize my board. Let me do it before proceeding. Okay, so now the situation is the following. The ball leaves the surface off the table so it will go up. But gravity is trying to push it down so it will go up and then it becomes wars lowered and then it starts to go down again. We have to get her mind. What is the maximum height that this bowl reaches? In order to do that, we have to use energy conservation again. We have to use it because of the following. These will be the situation Number four. The boat had just left the surface off the table. And then we have situation number five, where the bolt reaches the highest point off its trajectory. Now that has choose deems to be our reference frame. So these will be why equals zero. And then we have our Y axis pointing upwards. Let us call the height off the bowl. The maximum height, age. What is the energy off the ball in both situations? Situation number for the energy off the ball is purely kinetic because it's at why equals zero and the situation number five. It's purely potential because the ball has a velocity equals 20 at its maximum point before. By using energy conservation, we got the following energy situation for must be the same US energy in situation five. The situation for we only have kinetic energy. So 1/2 off the mask off the ball times its initial velocity squared and this must be equal to the energy of the ball situation. Five. Which is purely potential and given by the mass off the ball times, acceleration of gravity times its height and that's it. We have to solve this equation for age. For that, we can simplify the masses and then divide by G sending these to the other side of this equation and these results in age being equals two v squared divided by true times. G. Then we can use the velocity from Item A, which is 7.47 divided by a trial times. Acceleration of gravity, which is 9.8 meters per second, squared these results in the height off approximately two point 85 meters. But this isn't the final answer yet because this height age is with respect to the surface off the table In the situation on the problem, ask this. What is the maximum height with respect to the situation number two. So we have to discover what is the maximum height with respect to this origin. For that, we just have to increase these value by Delta L, which amounts to some 0.16. So the true height is equal to 3.85 plus 0.16 and these results in 3.1 meters.

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