Engineering Mechanics: Statics & Dynamics (14th Edition)

Published by Pearson
ISBN 10: 0133915425
ISBN 13: 978-0-13391-542-6

Chapter 14 - Kinetics of a Particle: Work and Energy - Section 14.6 - Conservation of Energy - Problems - Page 226: 77

Answer

$h=23.75m$, $v_C=21.6m/s$

Work Step by Step

We can determine the required height and speed as follows: We apply the equation of conservation of energy between A and B $\frac{1}{2}m_A^2+mgh_A=\frac{1}{2}mv_B^2+mgh_B$ We plug in the known values to obtain: $0+9.81h=\frac{1}{2}(8.578)^2+(9.81)(20)$ This simplifies to: $h=23.75m$ Now, we apply the equation of conservation of energy between B and C to determine the speed $\frac{1}{2}mv_B^2+mgh_B=\frac{1}{2}mv_C^2+mgh_C$ We plug in the known value to obtain: $\frac{1}{2}(8.578)^2+(9.81)(20)=\frac{1}{2}v_C^2+0$ This simplifies to: $v_C=21.6m/s$
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.