Answer
$v=4.08m/s$
Work Step by Step
We know that
$T+\Sigma U_{1\rightarrow 2}=T_2$
$\implies 0+50=\frac{1}{2}mv^2$
$\implies 50=\frac{1}{2}(6)v^2$
This simplifies to:
$v=4.08m/s$
Engineering Mechanics: Statics & Dynamics (14th Edition)
by Hibbeler, Russell C.
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 197: 11
Update this answer!
You can help us out by revising, improving and updating this answer.
Update this answerAfter 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.
