C13R2 - C13R.2 This is a conceptually tricky problem, and...

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Unformatted text preview: C13R.2 This is a conceptually tricky problem, and the book’s answer is incorrect. I will instruct the grader to grade your problem on effort only. To set the stage, let us first consider the situation where the wheels of the jetliner are locked, so there is no rotation. Then, if we treat the jetliner plus wheels as our system, there is a constant frictional force vector f acting over a distance d . If the jet moves in the + x direction, this gives rise to a constant acceleration in the − x direction: a x = − f M . (1) where M is the mass of the jet (including wheels). This causes the velocity of the jetliner to decrease: v 2 f,x = v 2 i,x + 2 a x d = v 2 i,x − 2 fd M . (2) The translational kinetic energy of the jet therefore decreases: Δ K trans = 1 2 Mv 2 f − 1 2 Mv 2 i = − fd. (3) (Here we are using a constant acceleration kinematic equation that we have not yet covered in this course.) Since microscopically energy is conserved, this energy must be converted into heat. However, this heat may be in the tires or in the runway. We do not know how much goes into the wheels and how much goes into the tires. (The book’s answer supposes that all of it goes into the tires, but that is not the most important mistake in that solution!) We will avoid this issue by computing the totalimportant mistake in that solution!...
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This note was uploaded on 03/01/2012 for the course PHY 9HA taught by Professor Markusluty during the Fall '10 term at UC Davis.

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C13R2 - C13R.2 This is a conceptually tricky problem, and...

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