K sysi u sysi k sysf u sysf 0 0 0 2 m g d 1 2 kd 2

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K sys,i + U sys,i = K sys,f + U sys,f 0 + 0 = 0 + (2 M ) g ( d ) + 1 2 kd 2 . Thus, with M = 2.0 kg, we obtain d = 0.39 m.
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15.P.8-110. We take the bottom of the incline to be the y = 0 reference level. The incline angle is 30 . The distance along the incline d (measured from the bottom) is related to height y by the relation y = d sin . (a) Using the conservation of energy, we have K U K U mv mgy 0 0 0 2 1 2 0 0 top top with v 0 50 . m s . This yields y = 1.3 m, from which we obtain d = 2.6 m. (b) An analysis of forces in the manner of Chapter 6 reveals that the magnitude of the friction force is f k = k mg cos . Now, we write Eq. 8-33 as K U K U f d mv mgy f d mv mgd mgd k k k 0 0 0 2 0 2 1 2 0 0 1 2 top top sin cos which upon canceling the mass and rearranging provides the result for d : . 5 . 1 ) sin cos ( 2 2 0 m g v d k (c) The thermal energy generated by friction is f k d = k mgd cos = 26 J. (d) The slide back down, from the height y = 1.5 sin 30º, is also described by Eq. 8-33. With E th again equal to 26 J, we have K U K U f d mgy mv k top top bot bot bot 0 1 2 0 26 2 from which we find v bot m s 21 . .
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