College Physics (4th Edition)

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

Chapter 6 - Problems - Page 228: 30

Answer

(a) The brick is moving with a speed of $4.43~m/s$ (b) The brick is moving with a speed of $4.03~m/s$

Work Step by Step

(a) We can use the conservation of energy to find the final speed of the brick: $KE_2+U_2 = KE_1+U_1$ $KE_2 = 0+U_1-U_2$ $\frac{1}{2}mv^2 = -mg~\Delta h$ $v^2 = -2g \Delta h$ $v = \sqrt{-2g~\Delta h}$ $v = \sqrt{-(2)(9.80~m/s^2)(-2.00~m)~sin~30.0^{\circ}}$ $v = 4.43~m/s$ The brick is moving with a speed of $4.43~m/s$ (b) We can use work and energy to find the speed of the brick: $KE_2+U_2 = KE_1+U_1+Work$ $KE_2 = 0+U_1-U_2+Work$ $\frac{1}{2}mv^2 = -mg~\Delta h-mg~cos~\theta~\mu_k~d$ $v^2 = 2gd~sin~\theta -2g~cos~\theta~\mu_k~d$ $v^2 = 2gd~(sin~\theta -cos~\theta~\mu_k)$ $v = \sqrt{2gd~(sin~\theta -cos~\theta~\mu_k)}$ $v = \sqrt{(2)(9.80~m/s^2)(2.00~m)~(sin~30.0^{\circ} -(0.10)~cos~30.0^{\circ})}$ $v = 4.03~m/s$ The brick is moving with a speed of $4.03~m/s$
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