ch08-p010

# ch08-p010 - 10 We use Eq 8-17 representing the conservation...

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2 2(167 J) 12.9 m/s. 2.00 kg f K v m == = (b) If we proceed algebraically through the calculation in part (a), we find K f = – Δ U = mgh where h = y i y f and is positive-valued. Thus, 2 2 f K vg h m as 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/s v = . (c) If K i 0 , then we find K f = mgh + K i (where K i is necessarily positive-valued). This represents a larger value for K f than 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.

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