Fundamentals of Physics Extended (10th Edition)

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

Chapter 7 - Kinetic Energy and Work - Problems - Page 173: 29b

Answer

The work done by the spring on the block is $~~2.1~J$

Work Step by Step

The work that we do on the block must be equal in magnitude to the work done on the block by the spring. When $d = 3.0~cm,$ the work done on the block is $0.9~J$ We can find the spring constant $k$: $\frac{1}{2}kd^2 = 0.9~J$ $k = \frac{(2)(0.9~J)}{d^2}$ $k = \frac{(2)(0.9~J)}{(0.030~m)^2}$ $k = 2000~N/m$ We can find the work done by the spring on the block as the block moves from $x_i = 5.0~cm$ to $x_f = -2.0~cm$: $W = \frac{1}{2}kx_i^2-\frac{1}{2}kx_f^2$ $W = \frac{1}{2}k~(x_i^2-x_f^2)$ $W = \frac{1}{2}(2000~N/m)~[(0.050~m)^2-(-0.020~m)^2)]$ $W = 2.1~J$ The work done by the spring on the block is $~~2.1~J$.
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