Physics Technology Update (4th Edition)

Published by Pearson
ISBN 10: 0-32190-308-0
ISBN 13: 978-0-32190-308-2

Chapter 8 - Potential Energy and Conservation of Energy - Problems and Conceptual Exercises - Page 249: 75

Answer

(a) $53.0MJ$ (b) $218\frac{m}{s}$

Work Step by Step

(a) According to law of conservation of energy $\Delta E=\Delta U+\Delta K.E$ $\Delta E=mg(y_f-0)+\frac{1}{2}(v_f^2-(0)^2)$ We plug in the known values to obtain: $\Delta E=(1865)(9.81)(2420)+\frac{1}{2}(1865)(96.5)^2$ $\Delta E=4.43\times 10^7+8.68\times 10^6$ $\Delta E=5.30\times 10^7J=53.0MJ$ (b) $\frac{1}{2}mv_f^2=mgy_f$ This simplifies to: $v_f=\sqrt{2gy_f}$ We plug in the known values to obtain: $v_f=\sqrt{2(9.8)(2420)}$ $v_f=218\frac{m}{s}$
Update this answer!

You can help us out by revising, improving and updating this answer.

Update this answer

After you claim an answer you’ll have 24 hours to send in a draft. An editor will review the submission and either publish your submission or provide feedback.