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 196: 6



Work Step by Step

We know that $\frac{1}{2}mv_1^2+\Sigma U_{1\rightarrow 2}=\frac{1}{2}mv_2^2$ We plug in the known values to obtain: $\frac{1}{2}m(\frac{40\times 1000}{3600})2-\mu_k (9.81\times 3)=\frac{1}{2}m(0)^2$ This simplifies to: $\mu_k=2.097$ Now if the speed of the truck is $80Km/h$, then $\frac{1}{2}(\frac{80\times 1000}{3600})^2-(2.097\times 9.81\times d)=\frac{1}{2}(0)^2$ This simplifies to: $d=12m$
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