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