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 14.3 - Principle of Work and Energy for a System of Particles - Problems - Page 199: 19



Work Step by Step

The required velocity can be determined as follows: We know that $d=\sqrt{(0.7)^2+(0.4)^2}-\sqrt{(0.2)^2+(0.2)^2}=0.523m$ Now, according to the principle of work and energy $\frac{1}{2}mv_A^2+\Sigma U_{1\rightarrow 2}=\frac{1}{2}mv_B^2$ $\frac{1}{2}mv_A^2+(-mgd+Fd)=\frac{1}{2}mv_B^2$ We plug in the known values to obtain: $\frac{1}{2}(15)(0)^2+(-15\times 9.8\times 0.5)+(300\times 0.523)=\frac{1}{2}(15)v_B^2$ This simplifies to: $v_B=3.34m/s$
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.