Essential University Physics: Volume 1 (4th Edition) Clone

Published by Pearson
ISBN 10: 0-134-98855-8
ISBN 13: 978-0-13498-855-9

Chapter 7 - Exercises and Problems - Page 133: 67

Answer

$=\frac{mgh}{2d}\sqrt{2g(h-d)}$

Work Step by Step

We know that the power is equal to the work over the change in time. We know that work done will equal the potential energy after the jump, which is given by $mgh$. We now need to find an expression for change in time. We know the following equations: $ \frac{v}{2}=\frac{d}{t}$ To find v, we need to use how high the animals get. At the top of their path, their velocity is 0, so it follows: $ \frac{1}{2}mv^2 = mg(h-d)$ $v = \sqrt{2g(h-d)}$ Plugging this into our equation for t, it follows: $t = \frac{2d}{\sqrt{2g(h-d)}}$ We plug this into the equation for power to find: $P=\frac{mgh}{2d}\sqrt{2g(h-d)}$
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