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 227: 10


(a) $W_g = -25,700~J$ (b) $W_T = 25,900~J$

Work Step by Step

(a) We can find the work done by gravity. $W_g = F\cdot d$ $W_g = F ~d~cos(\theta)$ $W_g = (mg) ~d~cos(180^{\circ})$ $W_g = (750~kg)(9.80~m/s^2)(3.5~m)(-1)$ $W_g = -25,700~J$ (b) We can use the work energy theorem to find the work done by tension. $KE_1+W_{ext} = KE_2$ $KE_1+W_T+W_g = KE_2$ $W_T = KE_2 -KE_1 - W_g$ $W_T = \frac{1}{2}mv_2^2-\frac{1}{2}mv_1^2- W_g$ $W_T = \frac{1}{2}m(v_2^2-v_1^2)- W_g$ $W_T = \frac{1}{2}(750~kg)[(0.75~m/s)^2-(0.25~m/s)^2]- (-25,700)$ $W_T = 25,900~J$
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