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This is university Physics Answers v Shaun Dychko. This skier has actually some initial velocity of 12.0 meters per second and then they conference a 35 level slope. They go up the slope—a full height the 2.5 meters—and then they room skiing horizontally in ~ the top and the question is what rate will they have on this peak level assuming that there"s a certain amount that friction due to a coefficient the friction that 0.0800 along this slope here. So we are going to number out exactly how much occupational is excellent by the non-conservative force, namely friction, follow me this slope and then that work-related is walking to be the energy that is taken far from kinetic and potential energy that the skier had actually initially and also then once they are at the peak here, they will be left with just some potential and kinetic energy but the total will no be the total initial kinetic and potential energy because some of it will be taken away by the job-related done by the non-conservative force. There"s a add to sign right here which says that perhaps this thing is including energy however in fact, it"s going to be an adverse because the friction and the displacement are in the opposite directions and also so this work-related will be negative. Okay! therefore the street that they take trip along the slope us can figure that out from this triangle here: we understand that sin that this edge Θ is the opposite divided by the hypotenuse therefore that"s h over d and also then we deserve to solve for d by multiply both sides by d end sin Θ. Us cancel the sin Θ"s there and also cancel the d"s over here and also we have actually d amounts to h over sin Θ. Okay! This is important because we are gonna instead of in for d right here in this formula because that the work-related done through the non-conservative force which is walking to it is in the an unfavorable and i put an unfavorable there just because, you know, work-related is pressure times displacement time cosine of the angle in between them and cos of 180 levels is negative 1 or you might think that it as the displacement and also the pressure are in opposite directions and that"s why there"s a an unfavorable there. Alright. The other aspect here is this friction force and also it"s going come be their coefficient that friction multiply by the typical force applied by the slope and also the normal force is walk to same the ingredient of gravity the is perpendicular to the slope. So gravity is right down but there"s this component below which is the pressure of gravity times cos that Θ which will be the perpendicular component of gravity and these two need to be equal since the skier is not accelerating perpendicular come the slope. Okay! for this reason we have actually μ kF N and then substituting in the perpendicular component of gravity in location of typical force and we have actually μ k is times mgcos Θ is the friction force. Therefore we deserve to substitute for both friction force— μ kmgcos Θ—and substitute because that the distance follow me the slope— i beg your pardon is h over sin Θ— and then we have actually this formula for the work done through friction. It"s gonna be an adverse times coefficient of friction time mgh divided by tangent Θ— since cos over sin... There"s an identification which says that cos Θ split by sin Θ is the mutual of tan Θ. Okay! So every one of this gets plugged right into the occupational done by the non-conservative pressure here in our conservation of energy formula and also then we have the right to start plugging in because that the various other terms as well. So the early stage kinetic energy is one-half mass time initial velocity squared plus the early potential energy and we"ll say that that is zero since we"ll specify this to be our recommendation level where h equates to 0 for this reason mgh would be mg time 0 and then including to that... That should be a minus or perhaps a plus for the formula and then a minus for the substitution but let"s just resolve it into a solitary operation i m sorry is minus. Okay! and that equals one-half mv f squared add to mgh— this is the potential power at the peak of the slope and the kinetic energy at the top of the slope and also this v f is what us are at some point trying come find. For this reason we"ll make things look a tiny bit much easier by multiplying everything by 2 end m; it"s kind of messy having fractions therefore we"ll eliminate the fountain by multiplying by 2 and also then we"ll also get rid the the m which is a aspect in every one of the terms. So we space left with, ~ you move the sides around, v f squared plus 2gh—the m is canceled and also this 2 is canceled yet the 2 appears here due to the fact that it has to get distributed amongst both terms into the brackets— and also that equals v early squared minus 2μ kgh over tan Θ. I made a mistake here with the to add sign but nevertheless i corrected the anyway and also it"s appropriately written together a minus here. Okay! for this reason we space gonna subtract 2gh indigenous both sides to settle for v f squared and we have actually this line and also I factored the end this typical factor 2gh indigenous both of these terms and it i do not care minus 2gh times 1 add to μ k end tan Θ and also then take the square source of both sides and that"s every the algebra that we need to do. For this reason v f is the square root of v early stage squared minus 2gh times 1 to add μ k over tan Θ. So that"s the square root of 12.0 meter per second squared minus 2 time 9.80 meter per second squared time 2.5 meters up the slope—that"s the elevation of the slope— times 1 add to 0.0800—coefficient of friction— split by tan the the slope edge of 35 degrees and this provides 9.5 meters per second.
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Now we room expecting some answer that is much less than the initial speed so if we had actually had one answer an ext than 12, we would have said the doesn"t happen the fact test but something less than 12 does. Therefore 9.5 is plausibly the exactly answer.