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 224: 66



Work Step by Step

We can determine the required force as follows: According to the conservation of energy equation $\frac{1}{2}mv_1^2+mgh_1=\frac{1}{2}mv_2^2+mgh_2$ $\implies \frac{1}{2}(40)(0)^2+(40\times 9.81-2cos60)=\frac{1}{2}(40)v_2^2+(40\times 9.81(-2))$ This simplifies to: $v_2=4.429m/s$ We know that $T-mg=m\frac{v^2}{\rho}$ $\implies T-(40)(9.81)=40\times (\frac{(4.429)^2}{2})=784.8N$ We also know that $\Sigma F_y=0$ $\implies 2\times 2Fcos30-T=0$ $\implies 4Fcos30=784.8$ This simplifies to: $F=227N$
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.