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 - Problems - Page 250: 18

Answer

$-5.68m/s$ and $21.14m/s$

Work Step by Step

The required speed of the crate can be determined as $F=150+(\frac{450-150}{6-0})t$ $\implies F=150+50t$ According to impulse momentum principle $mv_1+\Sigma \int _{t_1}^{t_2} F_y dt=mv_2$ $\implies 40(-10)+2\int _{t_1}^{t_2}(150+50t)dt-\int_{t_1}^{t_2}W dt=40v$ $\implies -400+2(150t+25t^2)|^t_0-(40\times 9.81)\Delta t=40v$ $\implies v=-10-2.31t+1.25t^2$ Now when $t=3 s$, then $v=-10-2.31(3)+1.25(3)^2=-5.68m/s$ and when $t=6s$, then $v=-10-2.31(6)+1.25(6)^2=21.14m/s$
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