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 9 - Work and Kinetic Energy - Exercises and Problems - Page 229: 43


(a) $W_g = -9.8\times 10^4~J$ (b) $W_T = 1.08\times 10^5~J$ (c) $KE_2 = 1.0\times 10^4~J$ (d) $v = 4.47~m/s$

Work Step by Step

(a) We can find the work that gravity does on the elevator. $W_g = mg~d~cos(180^{\circ})$ $W_g = (1000~kg)(9.80~m/s^2)(10~m)~cos(180^{\circ})$ $W_g = -9.8\times 10^4~J$ (b) We can find the magnitude of the tension. $\sum F = ma$ $T - mg = ma$ $T = m(g+a)$ $T = (1000~kg)(9.80~m/s^2+1.0~m/s^2)$ $T = 10,800~N$ We can find the work done by tension. $W_T = T~d$ $W_T = (10,800~N)(10~m)$ $W_T = 1.08\times 10^5~J$ (c) We can use the work energy theorem to find the kinetic energy of the elevator as it reaches 10 m. $KE_2 = KE_1 + W_g+W_T$ $KE_2 = 0 + -9.8\times 10^4~J+1.08\times 10^5~J$ $KE_2 = 1.0\times 10^4~J$ (d) We can find the speed of the elevator. $KE = 1.0\times 10^4~J$ $\frac{1}{2}mv^2 = 1.0\times 10^4~J$ $v^2 = \frac{(2)(1.0\times 10^4~J)}{m}$ $v = \sqrt{\frac{(2)(1.0\times 10^4~J)}{m}}$ $v = \sqrt{\frac{(2)(1.0\times 10^4~J)}{1000~kg}}$ $v = 4.47~m/s$
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