Fundamentals of Physics Extended (10th Edition)

Published by Wiley

Chapter 8 - Potential Energy and Conservation of Energy - Problems - Page 203: 10a

Answer

$E_{tot}=E_{k1}+E_{p01}=\dfrac {mv^{2}_{1}}{2}+mgh=mgH\Rightarrow v_{1}=\sqrt {2g\left( H-h\right) }\approx 12,91\dfrac {m}{s}$

Work Step by Step

The total Energy of the book at the top of the building is: $E_{tot}=E_{k0}+E_{p0}=0+mgH(1)$ since book doesn't have initial speed So when student catches the book the total energy of the book will be: $E_{tot}=E_{k1}+E_{p1}=\dfrac {mv^{2}_{1}}{2}+mgh\left( 2\right)$ where H is the height of the building and h is the height relative to the ground that student catches the book so from (1) and (2) we get: $E_{tot}=E_{k1}+E_{p01}=\dfrac {mv^{2}_{1}}{2}+mgh=mgH\Rightarrow v_{1}=\sqrt {2g\left( H-h\right) }\approx 12,91\dfrac {m}{s}$

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