You are watching: Find her final speed right at the top
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.