topΔΔKUKKmgL+=−+=000which, upon requiring Ktop= 0, gives K0= mgLand thus leads to 200222(9.80 m/s )(0.452 m)2.98 m/sKvgLm====.(b) We also found in the Problem 4 that the potential energy change is ΔU= –mgLin going from the initial point to the lowest point (the bottom). Thus, bottomKUmgL−−=000which, with K0= mgL, leads to Kbottom= 2mgL. Therefore, 2bottombottom244(9.80 m/s )(0.452 m)4.21 m/sKLm==.(c) Since there is no change in height (going from initial point to the rightmost point), thenΔU= 0, which implies ΔK= 0. Consequently, the speed is the same as what it was
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