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 - Page 257: 40


$v = 3.76~m/s$

Work Step by Step

We can find the work that gravity does on the block. $W_g = mg~d~cos(180^{\circ})$ $W_g = (1.50~kg)(9.80~m/s^2)(2.00~m)~cos(180^{\circ})$ $W_g = -29.4~J$ We can find the work done by tension. $W_T = T~d$ $W_T = (20~N)(2.00~m)$ $W_T = 40.0~J$ We can use the work energy theorem to find the kinetic energy of the block as it reaches 2.00 m. $KE_f = KE_0 + W_g+W_T$ $KE_f = 0 - 29.4~J+40.0~J$ $KE_f = 10.6~J$ We can find the speed of the block. $KE_f = 10.6~J$ $\frac{1}{2}mv^2 = 10.6~J$ $v^2 = \frac{(2)(10.6~J)}{m}$ $v = \sqrt{\frac{(2)(10.6~J)}{m}}$ $v = \sqrt{\frac{(2)(10.6~J)}{1.50~kg}}$ $v = 3.76~m/s$
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