sin the this angle Θ is the opposite divided by the hypotenuse so that's h over d and then we can solve because that d by multiply both sides by d over sin Θ. Us cancel the sin Θ's there and also cancel the d's end here and we have d amounts to h over sin Θ. Okay! This is important because we space gonna substitute in for d below in this formula because that the work-related done by the non-conservative force which is walk to be the an unfavorable and i put an unfavorable there just because, girlfriend know, job-related is pressure times displacement time cosine that the angle between them and cos of 180 levels is an adverse 1 or you can think of it together the displacement and the force are in opposite directions and also that's why there's a an adverse there. Alright. The other aspect here is this friction force and also it's going to be their coefficient of friction multiplied by the regular force applied by the slope and the normal force is walk to equal the component of gravity the is perpendicular to the slope. So heaviness is right down however there's this component right here which is the force of gravity times cos that Θ which will certainly be the perpendicular ingredient of gravity and also these two need to be equal due to the fact that the skier is not accelerating perpendicular come the slope. Okay! so we have μ kF N and also then substituting in the perpendicular component of heaviness in place of typical force and also we have μ k is times mgcos Θ is the friction force. For this reason we can substitute for both friction force— μ kmgcos Θ—and substitute because that the distance follow me the slope— which is h over sin Θ— and then we have this formula for the work-related done through friction. It's gonna be an unfavorable times coefficient that friction time mgh separated by tangent Θ— because cos end sin... There's an identification which says that cos Θ split by sin Θ is the reciprocal of tan Θ. Okay! So all of this it s okay plugged right into the job-related done through the non-conservative force here in ours conservation of power formula and then we can start plugging in for the other terms together well. So the initial kinetic energy is one-half mass times initial velocity squared plus the initial potential energy and we'll say that that is zero because we'll define this to it is in our recommendation level whereby h equates to 0 for this reason mgh would certainly be mg time 0 and then including to that... That need to be a minus or possibly a plus for the formula and also then a minus for the substitution but let's just resolve it into a solitary operation i beg your pardon is minus. Okay! and also that amounts to one-half mv f squared plus mgh— this is the potential power at the height of the slope and also the kinetic power at the top of the slope and this v f is what us are at some point trying come find. Therefore we'll make things look a little bit simpler by multiplying everything by 2 over m; it's sort of messy having actually fractions so we'll remove the fractions by multiply by 2 and also then we'll likewise get rid the the m i m sorry is a aspect in every one of the terms. Therefore we are left with, ~ you move the sides around, v f squared to add 2gh—the m is canceled and also this 2 is canceled however the 2 appears here due to the fact that it has to get distributed among both terms right into the brackets— and also that equates to v early squared minus 2μ kgh end tan Θ. Ns made a mistake right here with the to add sign yet nevertheless ns corrected it anyway and it's appropriately written as a minus here. Okay! so we are gonna subtract 2gh from both political parties to resolve for v f squared and also we have actually this line and I factored the end this common factor 2gh native both of these terms and also it becomes minus 2gh time 1 to add μ k end tan Θ and also then take the square root of both sides and also that's every the algebra the we have to do. Therefore v f is the square root of v early squared minus 2gh time 1 add to μ k over tan Θ. Therefore that's the square source of 12.0 meters per second squared minus 2 time 9.80 meters per second squared times 2.5 meter up the slope—that's the height of the slope— time 1 add to 0.0800—coefficient the friction— divided by tan the the slope edge of 35 degrees and also this provides 9.5 meters per second. Currently we space expecting part answer that is less than the initial speed so if we had had an answer an ext than 12, us would have actually said the doesn't happen the reality test however something less than 12 does. For this reason 9.5 is plausibly the correct answer.">


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This is college Physics Answers with Shaun Dychko. This skier has some early stage velocity of 12.0 meters per 2nd and climate they conference a 35 level slope. They walk up the slope—a complete height of 2.5 meters—and climate they space skiing horizontally in ~ the top and the inquiry is what speed will they have on this optimal level assuming the there"s a details amount that friction as result of a coefficient that friction that 0.0800 follow me this steep here. Therefore we are going to number out how much occupational is done by the non-conservative force, namely friction, follow me this slope and then that work is walk to be the energy that is taken away from kinetic and potential energy that the skier had actually initially and then as soon as they are at the optimal here, they will be left with just some potential and also kinetic energy however the full will not be the complete initial kinetic and potential energy since some of it will certainly be taken away by the occupational done by the non-conservative force. There"s a add to sign below which argues 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 contrary directions and so this work-related will be negative. Okay! for this reason the street that they take trip along the slope us can figure that the end from this triangle here: we recognize that sin of this edge Θ is the opposite divided by the hypotenuse so that"s h over d and also then we deserve to solve because that d by multiply both political parties by d over sin Θ. We cancel the sin Θ"s there and also cancel the d"s end here and we have d amounts to h end sin Θ. Okay! This is important because we room gonna substitute in because that d right here in this formula for the occupational done through the non-conservative pressure which is going to it is in the negative and ns put negative there simply because, you know, occupational is pressure times displacement times cosine that the angle in between them and cos the 180 levels is negative 1 or you might think of it together the displacement and the force 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 to be your coefficient the friction multiply by the typical force applied by the slope and the normal pressure is going to same the component of gravity the is perpendicular come the slope. So heaviness is right down yet there"s this component here which is the pressure of heaviness times cos the Θ which will certainly be the perpendicular ingredient of gravity and these two need to be equal because the skier is not speeding up perpendicular to the slope. Okay! therefore we have μ kF N and also then substituting in the perpendicular component of gravity in place of regular force and we have μ k is times mgcos Θ is the friction force. For this reason we have the right to substitute because that both friction force— μ kmgcos Θ—and substitute for the distance along the slope— i m sorry is h over sin Θ— and also then we have actually this formula because that the job-related done through friction. It"s gonna be an unfavorable times coefficient of friction times mgh separated by tangent Θ— because cos end sin... There"s an identification which says that cos Θ divided by sin Θ is the mutual of tan Θ. Okay! So all of this it s okay plugged into the occupational done through the non-conservative pressure here in our conservation of power formula and also then we can start plugging in because that the other terms together well. For this reason the early stage kinetic power is one-half mass time initial velocity squared plus the initial potential energy and also we"ll say the that is zero because we"ll specify this to be our reference level whereby h equals 0 so mgh would be mg times 0 and then adding to that... That have to be a minus or possibly a plus for the formula and then a minus because that the substitution but let"s simply resolve it right into a solitary operation i beg your pardon is minus. Okay! and also that amounts to one-half mv f squared add to mgh— this is the potential power at the height of the slope and also the kinetic power at the height of the slope and also this v f is what we are at some point trying come find. Therefore we"ll make points look a tiny bit easier by multiplying everything by 2 end m; it"s type of messy having actually fractions so we"ll remove the fractions by multiply by 2 and also then we"ll also get rid the the m which is a aspect in all of the terms. So we are left with, after ~ you move the sides around, v f squared add to 2gh—the m is canceled and also this 2 is canceled however the 2 shows up here since it has actually to acquire distributed among both terms right into the brackets— and that equals v early stage squared minus 2μ kgh over tan Θ. I made a mistake here with the to add sign however nevertheless ns corrected that anyway and it"s effectively written as a minus here. Okay! therefore we are gonna subtract 2gh indigenous both political parties to solve for v f squared and also we have actually this line and also I factored the end this common factor 2gh native both of this terms and it becomes minus 2gh times 1 plus μ k end tan Θ and then take it the square source of both sides and that"s all the algebra that we have to do. Therefore v f is the square source of v initial squared minus 2gh times 1 plus μ k over tan Θ. For this reason that"s the square source of 12.0 meter per 2nd squared minus 2 times 9.80 meter per 2nd squared time 2.5 meter up the slope—that"s the height of the slope— times 1 to add 0.0800—coefficient of friction— separated by tan the the slope edge of 35 degrees and also this provides 9.5 meters per second.

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Currently we space expecting part answer that is less than the initial rate so if we had had one answer more than 12, us would have said that doesn"t happen the fact test however something less than 12 does. Therefore 9.5 is plausibly the exactly answer.