Engineering Mechanics: Statics & Dynamics (14th Edition)

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

Chapter 15 - Kinetics of a Particle: Impulse and Momentum - Section 15.2 - Principle of Linear Impulse and Momentum for a System of Particles - Fundamental Problems - Page 246: 5

Answer

$45m/s;3375N$

Work Step by Step

The required speed and tension can be calculated as follows: $a=\frac{F_D}{m_1+m_2}$ $\implies a=\frac{9000}{2500+1500}$ $\implies a=2.25m/s^2$ Now $v=at$ Substituting the known values, we obtain: $v=2.25(20)$ $\implies v=45m/s$ The tension can be determined as $m_1a=F_D-T$ $\implies T=F_D-m_1a$ $\implies T=9000-2500(2.25)$ $\implies T=3375N$
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