ch08-p071

# ch08-p071 - 71. This can be worked entirely by the methods...

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KU E m g d fk =− = '' th (sin cos ) θ μ which leads to the speed at the bottom of the ramp v K m gd f k == = 2 25 5 ms . sin cos . θμ b g (b) This speed begins its horizontal motion, where f k = k mg and Δ U = 0. It slides a distance d' before it stops. According to Eq. 8-31 (with W = 0), 0 0 1 2 0 1 2 2 2 =++ ++ + ΔΔ KUE mv mgd gd gd k kk Δ th θ μ sin cos b g ch where we have divided by mass and substituted from part (a) in the last step. Therefore, ′ = = d d k k cos .. bg 54m (c) We see from the algebraic form of the results, above, that the answers do not depend on mass. A 90 kg crate should have the same speed at the bottom and sliding distance across the floor, to the extent that the friction relations in Ch. 6 are accurate. Interestingly, since g does not appear in the relation for d' , the sliding distance would seem to be the
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## This note was uploaded on 05/17/2011 for the course PHY 2049 taught by Professor Any during the Spring '08 term at University of Florida.

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