College Physics (4th Edition)

Published by McGraw-Hill Education
ISBN 10: 0073512141
ISBN 13: 978-0-07351-214-3

Chapter 6 - Problems - Page 234: 110


The maximum speed of the block will be $1.57~m/s$

Work Step by Step

We can find the spring constant $k$: $kx = mg$ $k = \frac{mg}{x}$ $k = \frac{(0.20~kg)(9.80~m/s^2)}{0.010~m}$ $k = 196~N/m$ The maximum kinetic energy of the block will be equal to the elastic potential energy stored in the spring when it is stretched a distance of 5.0 cm: $\frac{1}{2}mv_{max}^2 = \frac{1}{2}kx^2$ $v_{max}^2 = \frac{kx^2}{m}$ $v_{max} = \sqrt{\frac{k}{m}}~x$ $v_{max} = (\sqrt{\frac{196~N/m}{0.20~kg}})~(0.050~m)$ $v_{max} = 1.57~m/s$ The maximum speed of the block will be $1.57~m/s$.
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