Physics: Principles with Applications (7th Edition)

Published by Pearson
ISBN 10: 0-32162-592-7
ISBN 13: 978-0-32162-592-2

Chapter 6 - Work and Energy - General Problems - Page 168: 85


(a) v = 42 m/s (b) $P = 3.2 \times 10^5 ~W$

Work Step by Step

(a) We can use conservation of energy to solve this question. The kinetic energy at the bottom is equal to the potential energy at the top: $KE = PE$ $\frac{1}{2}mv^2 = mgy$ $v^2 = 2gy$ $v = \sqrt{2gy} = \sqrt{(2)(9.80 ~m/s^2)(88 ~m)}$ $v = 42 ~m/s$ (b) 55% of the initial potential energy will be transferred to the turbine blades. Note that 680 kg of water fall each second. As a result, $P = \frac{0.55PE}{t}$ $P = \frac{0.55mgy}{t}$ $P = \frac{(0.55)(680 ~kg)(9.80 ~m/s^2)(88 ~m)}{1 ~s}$ $P = 3.2 \times 10^5 ~W$
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