9. We neglect any work done by friction. We work with SI units, so the speed is converted:v= 130(1000/3600) = 36.1 m/s. (a) We use Eq. 8-17: Kf+ Uf= Ki+ Uiwith Ui= 0, Uf= mghand Kf= 0. Since Kmvi=122, where vis the initial speed of the truck, we obtain 22221(36.1m/s)66.5 m222(9.8 m/s )vmvmghhg=¡===.IfLis the length of the ramp, then Lsin 15° = 66.5 m so that L= (66.5 m)/sin 15° = 257
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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.