Physics for Scientists and Engineers: A Strategic Approach with Modern Physics (3rd Edition)

Published by Pearson
ISBN 10: 0321740904
ISBN 13: 978-0-32174-090-8

Chapter 6 - Dynamics I: Motion Along a Line - Exercises and Problems - Page 163: 45

Answer

$2.73\;\rm m/s$

Work Step by Step

First of all, we need to draw a force diagram of the box, as you see below. We analyzed the weight of the box into its $x$ and $y$ components, as you see in the second figure below. We need to make the nail box moves for 5 meters toward your friend at the edge of the roof. This means that its final speed at the edge of the roof is zero. $$\overbrace{v_{fx}^2}^{0}=v_{ix}^2+2a_x\Delta x$$ Solving for $v_i$; and plugging the known. $$v_{ix}^2=-2a_x\Delta x=-2a_x\cdot 5=-10a_x$$ $$v_{ix} =\sqrt{-10a_x }\tag 1$$ Now we need to find the box's acceleration. We know that the net force exerted on the object in the $y$-direction is zero since it has no motion in this direction. $$\sum F_y=F_n-mg\cos25^\circ=ma_y=m(0)=0$$ Thus, $$F_n=mg\cos25^\circ\tag 2$$ The net force exerted on the nail box in the $x$-direction is not zero, since it is slowing down; and is given by $$\sum F_x=mg\sin25^\circ -f_k=ma_x$$ Recalling that $f_k=\mu_kF_n$, and from (2), we got $f_k=\mu_kmg\cos25^\circ$ Thus, $$mg\sin25^\circ -\mu_kmg\cos25^\circ=ma_x$$ Thus, $$ g\sin25^\circ -\mu_k g\cos25^\circ= a_x$$ $$a_x= g\left[\sin25^\circ -\mu_k \cos25^\circ\right] $$ Plugging the known; $$a_x= 9.8\left[\sin25^\circ -0.55 \cos25^\circ\right] =\bf -0.743\;\rm m/s^2$$ Plugging into (1); $$v_{ix} =\sqrt{-10 \times-0.743 } =\color{red}{\bf 2.73}\;\rm m/s$$
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