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 - Problems - Page 166: 49

Answer

(a) v = 15.3 m/s (b) The average force of air resistance is 1.03 N upwards.

Work Step by Step

(a) Without air resistance, the original potential energy will equal the kinetic energy at the bottom. $KE = PE$ $\frac{1}{2}mv^2 = mgh$ $v = \sqrt{2gh}$ $v = \sqrt{(2)(9.80 ~m/s^2)(12.0 ~m)}$ $v = 15.3 ~m/s$ (b) The original potential energy plus the work done by air resistance will equal the kinetic energy at the bottom. $PE + Work = KE$ $Work = KE - PE$ $Fd = \frac{1}{2}mv^2 - mgh$ $F = \frac{\frac{1}{2}mv^2 - mgh}{d}$ $F = \frac{\frac{1}{2}(0.145 ~kg)(8.00 ~m/s)^2 - (0.145 ~kg)(9.80 ~m/s^2)(12.0 ~m)}{12.0 ~m}$ $F = -1.03 ~N$ Note that the force is negative because air resistance did negative work on the ball as it fell. The magnitude of the average force of air resistance is 1.03 N upwards.
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