Physics for Scientists and Engineers: A Strategic Approach with Modern Physics (4th Edition)

Published by Pearson
ISBN 10: 0133942651
ISBN 13: 978-0-13394-265-1

Chapter 10 - Interactions and Potential Energy - Exercises and Problems: 52

Answer

(a) $v = \sqrt{\frac{2gh~(m-M~\mu_k)}{M+m}}$ (b) $v = \sqrt{\frac{2mgh}{M+m}}$

Work Step by Step

(a) The kinetic energy of the system just before the block of mass $m$ hits the floor will be equal to the sum of the initial potential energy of block $m$ plus the work done by friction. We can find an expression for the speed of the block just before it hits the floor. $KE = PE+W_f$ $\frac{1}{2}(M+m)v^2 = mgh-Mg~\mu_k~h$ $v^2 = \frac{2mgh-2Mg~\mu_k~h}{M+m}$ $v = \sqrt{\frac{2gh~(m-M~\mu_k)}{M+m}}$ (b) The kinetic energy of the system just before the block of mass $m$ hits the floor will be equal to the initial potential energy of block $m$. We can find an expression for the speed of the block just before it hits the floor. $KE = PE$ $\frac{1}{2}(M+m)v^2 = mgh$ $v^2 = \frac{2mgh}{M+m}$ $v = \sqrt{\frac{2mgh}{M+m}}$
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