ch08-p012 - 12. We use Eq. 8-18, representing the...

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top ΔΔ KU KK m g L += −+ = 0 0 0 which, upon requiring K top = 0, gives K 0 = mgL and thus leads to 2 0 0 2 2 2(9.80 m/s )(0.452 m) 2.98 m/s K vg L m == = = . (b) We also found in the Problem 4 that the potential energy change is Δ U = – mgL in going from the initial point to the lowest point (the bottom). Thus, bottom K U m g L −− = 0 0 0 which, with K 0 = mgL , leads to K bottom = 2 mgL . Therefore, 2 bottom bottom 2 4 4(9.80 m/s )(0.452 m) 4.21 m/s K L m = = . (c) Since there is no change in height (going from initial point to the rightmost point), then Δ U = 0, which implies Δ K = 0. Consequently, the speed is the same as what it was
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