Fundamentals of Physics Extended (10th Edition)

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

Chapter 8 - Potential Energy and Conservation of Energy - Problems - Page 212: 119b

Answer

The speed when it is 3.0 meters below the window is $~~11~m/s$

Work Step by Step

We can use conservation of energy to find the speed when it is 3.0 meters below the window: $K_f+U_f = K_0+U_0$ $K_f = K_0+U_0-U_f$ $\frac{1}{2}mv_f^2 = \frac{1}{2}mv_0^2+mg(h_0-h_f)$ $v_f^2 = v_0^2+(2g)(3.0)$ $v_f = \sqrt{v_0^2+(2g)(3.0)}$ $v_f = \sqrt{(8.0~m/s)^2+(2)(9.8~m/s^2)(3.0~m)}$ $v_f = 11~m/s$ The speed when it is 3.0 meters below the window is $~~11~m/s$
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