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CcujccdJ 0the man releases the case of tuna and thereafter it m 80 the slide on its own: Air resistance (in pounds) is nun equal to one-third the velocity (in feetl...

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CcujccdJ 0the man releases the case of tuna and thereafter it m 80 the slide on its own: Air resistance (in pounds) is nun equal to one-third the velocity (in feetlsec) of the cas down the slide_ 1 of The slide has a slope of 8/15,and th inters of sliding friction between the case and the slide is vis (but DON'T SOLVE) a BVP whose solution gives th the case in terms of the time t that has elapsed sinc was released Answers: 1. (x-5)? tan(3v)=K(2x-15)} . 2. In|x/2|-arcsinly/x) =-w/6. 3. xy+x

CcujccdJ 0 the man releases the case of tuna and thereafter it m 80 the slide on its own: Air resistance (in pounds) is nun equal to one-third the velocity (in feetlsec) of the cas down the slide_ 1 of The slide has a slope of 8/15,and th inters of sliding friction between the case and the slide is vis (but DON'T SOLVE) a BVP whose solution gives th the case in terms of the time t that has elapsed sinc was released Answers: 1. (x-5)? tan(3v)=K(2x-15)} . 2. In|x/2|-arcsinly/x) =-w/6. 3. xy+x sinlv)=K. Xy =yezy ~ezy +Key 15.17.3 feetlsec SS 5 on d is )ve 6. 21.9 feet: al nt 2218 , 7. (a) v=V1ox-32+32e (b) 2.9 feet. 8. (a) v=26-26e-16tp5_ (b) 91 feet .. ^ . ` 1, ~ . {"' e is m; Jer ^- * COeu' < ' S, % 3Y076 Tiino; < 0 ` ds. Ia



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A child of mass $m$ starts from rest and slides without friction from a height $h$ along a curved waterslide (Fig. P5,46). She is launched from a height $h / 5$ into the pool. (a) Is mechanical energy conserved? Why? (b) Give the gravitational potential encrgy associated with the child and her kinetic energy in terms of $m g h$ at the following positions: the top of the waterslide, the launching point, and the point where she lands in the pool. (c) Determine her initial speed $v_{0}$ at the launch point in terms of $g$ and h. (d) Determine her maximum airborne height $y_{\max }$ in terms of $h, g$, and the horizontal speed at that height, $v_{0 x^{-}}$ (c) Use the $x$ -component of the answer to part (c) to eliminate $v_{0}$ from the answer to part (d), giving the height $y_{\text {eaax }}$ in terms of $g, h$, and the launch angle $\theta$. (f) Would your answers be the same if the waterslide were not frictionless? Explain.

So apart A. Can the conservation of mechanical energy be applied? Yes. So we can say conservation of mechanical energy can be applied okay? In this case the child slides down a frictionless surface. Uh There are two forces that are acting on the child. Of course one is the normal force, one is the gravitational force. And uh the only non conservative force is the normal force and that's always acting perpendicular to the plane of contact with the child. And so the work done by the normal forces zero. It's the normal forces, of course, acting perpendicular to the direction of motion. In this case, uh In general, the normal force acts perpendicular to the plane of contact, and in this case it's acting perpendicular to the plane of contact and perpendicular to the displacement vector. And so it is there's no work being done by the normal force. So there's no So in this case the conservation of mechanical energy can be applied. And we can say that then for part B, we have then the initial kinetic energy plus the initial potential energy equaling the final kinetic energy plus the final potential energy. And we can say that then we have 1/2 times the mass times the initial velocity squared plus M. G. H. Or we can say mg. Why initial equaling 1/2 um The final square plus she? Why final? The masses all cancel out. And we can see that then the mass of child uh is not a factor. So the masses trial is not a factor in the frictionless case, again, in the frictionless case. So we can say that then for part C, solving for the final velocity from the equation above this would be equal to the square root of the initial velocity squared Plus two G. multiplied by the initial height minus the final height. And in this case we can see that um Given that the initial speed of the child is zero In both the cases that the initial velocity is zero and so in both cases the child has the same change in latitude. So with this equaling zero and in both cases the change in height being the same, we can say that then the final or the final velocity is same in both cases. So here for part D. We have friction. So here is not frictionless friction. Now does some work, friction is non conservative. So instead of using the conservation of mechanical energy, will be using the work energy theorem. And it's very similar except that now we're going to be with the work energy serum, you're adding a term. And that term that you're adding would essentially be the work done by friction. Ah and everything else is pretty much the same thing. And then for part E we have the final velocity Equalling then The Square Root of two G. Y. Initial rather minus Y. Final. And in this case and in this case we are already eliminating the initial velocity, we know that the initial velocity is zero. And so we can say that then the final velocity for part E would be equaling to the square root of two, Multiplied by 9.8 m/s squared, multiplied by 12 0.0. Oh more and more 12.0 m. There we go. And this is giving us then 15 .3m/s approx. That is the end of the solution. Thank you for watching.

So the last part off the motion is projectiles. We can use that to figure out the speed with which the person comes off the ramp and in projectile motion. We know that the horizontal velocity remains constant, so we can actually say he's equal to five meter divided by 0.5 2nd So that comes out to be equal to me. 10 meters per second and the initial speed here we not was equal to zero. So over this distance, what we can do now is we confined first distance this height, age and once we have this height age, then we can find this distance, each one and then we can add them together. So let's first figure out each. So we know that the initial kinetic energy, plus the initial potential energy equals the final kinetic energy plus the initial potential energy. Okay, So initial kinetic energy plus initial potential energy equals final kinetic energy plus final potential energy. So we can take the potential energy at this Louis point to be equal to zero, and the person starts from rest of the initial kinetic energies. Also zero. So we can say M g H one is equal to 1/2 and times 10 squared across of the two m's and then you can solve for each one. Each one comes out to be equal to 5.10 meter. Now for the second part, if we analyzed the vertical motion in projectile. So in vertical motion, the initial velocity is so The initial velocity in the vertical direction is zero, and the acceleration in the vertical direction is 9.8 meter per second. And we already know that the time is 0.5 seconds. So we can use why is equal to we not t plus 1/2 a T squared, which gives me zero plus 1/2 times 9.8 times 0.5 squared. So this comes out to be equal to 0.5 times 9.8 times 0.5 squared is equal to 1.23 meters. So I'm just gonna go ahead and write a disease. So this is 1.23 meters. So we have each is equal to 5.10 meter and we have each one is equal to 1.23 meters, so each would be equal to 5.10 plus 1.23 and that is my 0.10 plus 1.23 equals 6.33 leaders.

Question Number 27. That child. It's slipping on this life, the slightest friction less so the system will be or isolated system. And since any non conservative force is not acting on the system, so we can say there, Don said is no non conservative forces are not a thing. As for action is absent in the situation. What barred? See the initial energy at the top of this light is and G that because the cannon tick energy at the top of the slide zero. So only gravitational potential energy exists there in the pardon me, uh, total energy is the some What can it take? Energy up And the Esquire less the potential energy at this point, which is, um, G by five. Because here I did X y by in by e, the potential energy will be MGI. Why? And the kinetic energy is, um and only component off We lost in the horizontal direction. The X It's what this is the triple energy off the child. Because in the projectile motion at the top, most point on leader horizontal component of velocity exists in the part f we can write m g at is equal to, uh, and the spread less MGI Bye bye. This is by conservation off energy. From this a question we will solve for we the Quito on the road It g it's bye bye. This is the required philosophy in the five G from the other parts we can write m v squared less MGI by five is a widow and G. Why megs less? Uh And we spread the excess web, for example, proponent of velocity squared in tow Because it's why he, uh you know, that horizontal component off the the Lord speeds Vico straight up so we can write that can take energies, help and the square What's a Squire? Trita? No such shooting the value of B and they, uh um into a GH by five less MGH This is a quote and G Why I m less Oh, and a G yet by five forces fired Pekka Mark and selling MGI from both sides. We get why am is equal. That's minus for by five at course, inspired He uh this is that require expression no end up by edge. If that slide is not friction less then the frictional forces also do work on the J and hands. We cannot apply the conservation off in a deep principle here, in the presents of the frictional forces, the conservation off energy doesn't apply.

So for Part of the key word is actually frictionless and frictionless means that there is no energy loss to friction. This means that conservation of mechanical energy can the applied. And this is again because there is no friction. Therefore, she doesn't lose. The child doesn't lose any energy to friction. Also, all of her, um, all of her potential energy initial is going to be converted into, in this case, uh, kinetic energy. So we can then say that what all of the potential energy initial would be converted into kinetic and potential energy. And so we consent. Say that here. Four part beef. We know that the potential energy, the gravitational potential energy is equaling MGH and we can say that at the top of the slide, the child is at rest. So we can say that this would be equaling the potential energy initial where the kinetic energy initial is equal in zero jewels. We can then say that if the total initial mechanical energy is conserved the total, this would be the total energy initial. Now we know that Ah, the launching point is going to be at some. Why initial equaling that the maximum height divided by five. And so we can say that then at launching point, the potential energy of the child is unequal ng m g h over five. At this point, we can say that the kinetic energy of the child act the launching point kinetic energy at launching rather at launch is gonna be equaling the potential energy initial M G H minus M g h over five. And so this would be essentially four m g h over five and once for party once at level surface. So there is no height here. The potential energy is going to be zero Jules, And this means that the kinetic energy at this point is gonna equal the total initial potential energy. Given that now, all of the initial potential energy has been converted into a kinetic energy. And so, for part C, we can say that the kinetic energy of the child is 1/2 and the initial squared and at the launching point, this is equaling four m g. H over five, and so we can cancel out the masses and we find that then the initial velocity is equaling eight g h over five. This would be the initial velocity at the launching point. So at launch point, and we can then say this would be your answer for part C for part D. Then, as the child Lee leaves the launching point, then it will simply be projectile emotion. The child is in free fall and so we can say 1/2 and the squared plus and G y equals M g h. Of course, the masses all cancelled out and we have 1/2 most supplied by these squared plus G. Why equaling g h Now we're going to split this velocity into X and Y components, so this would be 1/2 times the square root. But this term is being squared, so we're gonna lose the square root. And we're simply going to say velocity initial in the ex direction squared, plus velocity initial in the UAE direction squared, uh, plus G y equaling gh. And so we know that at the maximum height, the vertical component of the velocity is gonna be zero. And so we can say that that's gonna be zero, and we can say that 1/2 times the initial velocity in the ex directions squared, plus G. Why Max is gonna be equaling G h. And so we have that. Why Max is equaling H minus the initial in the ex direction squared, divided by two G. And so the maximum height obtained by the child is there going to be a check minus the initials in the X. The initial velocity in the extraction squared divided by two times the acceleration due to gravity on Earth's surface. Now we can then ah, four part east substitute. We can say that why? Max is equaling H minus v initial co sign of fada multi squared, divided by two G. And we're going to then substitute in the initial equaling the square root of eight G h over five from what we found earlier. And we can then say that why Max is gonna be equaling H minus the square root eight g h over five co sign of fate, a quantity squared divided by two G. And this is gonna be equally H. We're gonna factor at an H, and it'll be one minus four co sign squared of data over five. This would be the maximum height obtained by the child during freefall and four part if It's asking us if the waterside were not frictionless. And as you can see, these these are only possible with this, uh, this side, this slide being frictionless. So for part F, um, essentially all results would change if the slide was not frictionless and then we could say because, uh, the conservation of mechanical energy would not apply two this system. So if this if this slide was not frictionless than yes, we would have to account for this work done by friction in all of our subsequent at the calculation. So, yes, all results would change if the slide was not frictionless. That is the end of the solution. Thank you for watching.


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