22(167 J)12.9 m/s.2.00 kgfKvm===(b) If we proceed algebraically through the calculation in part (a), we find Kf= – ΔU= mghwhere h= yi– yfand is positive-valued. Thus, 22fKvghmas we might also have derived from the equations of Table 2-1 (particularly Eq. 2-16). The fact that the answer is independent of mass means that the answer to part (b) is identical to that of part (a), i.e.,12.9 m/sv=.(c) IfKi≠0, then we find Kf= mgh+ Ki(where Kiis necessarily positive-valued). This represents a larger value for Kfthan in the previous parts, and thus leads to a larger value
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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.