Fundamentals of Physics Extended (10th Edition)

Published by Wiley
ISBN 10: 1-11823-072-8
ISBN 13: 978-1-11823-072-5

Chapter 8 - Potential Energy and Conservation of Energy - Problems - Page 205: 32

Answer

$$W=0.0010 \mathrm{\ J}$$

Work Step by Step

$$ d U=(m / L) g|y| d y=-(m / L) g y d y $$ since $y$ is negative valued (we have $+y$ upward and the source is at the table-top). The total potential energy change is $$ \Delta U=-\frac{m g}{L} L_{L / 4}^{0} y d y=\frac{1}{2} \frac{m g}{L}(L / 4)^{2}=m g L / 32 $$ The work required to pull the chain onto the table is therefore $$W=\Delta U=m g L / 32=(0.012 \mathrm{kg})\left(9.8 \mathrm{m} / \mathrm{s}^{2}\right)(0.28 \mathrm{m}) / 32$$$$=0.0010 \mathrm{\ J}$$
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