Physics: Principles with Applications (7th Edition)

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

Chapter 9 - Static Equilibrium; Elasticity and Fracture - Search and Learn - Page 259: 2

Answer

2.7 m.

Work Step by Step

The brick’s potential energy mgh is fully converted to kinetic energy. When the brick hits the floor, the floor does work of -mgh on the brick to bring it to rest. The amount mgh is the average stopping force (F) multiplied by the brick’s compression distance. Use equation 9–4 to write the compression distance in terms of the force. $$mgh=F\Delta \mathcal{l}=F(\frac{1}{E}\frac{F}{A}\mathcal{l_0})$$ $$mgh=\frac{F}{A}(\frac{A}{E}\frac{F}{A}\mathcal{l_0})$$ $$h=(\frac{F}{A})^2(\frac{A\mathcal{l_0}}{mgE})$$ Now replace the stress on the brick with the ultimate compressive strength of brick, and solve for the height needed to break the dropped brick. $$h=(35\times10^6\;N/m^2)^2(\frac{(0.040\;m)(0.150\;m)(0.060\;m)}{(1.2\;kg)(9.80\;m/s^2)(14\times10^9\;N/m^2)})=2.7m$$
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