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 2 - Kinematics in One Dimension - Exercises and Problems - Page 66: 55

Answer

$9.9\;\rm m/s$

Work Step by Step

As you see in the figure below, in the second sketch, we analyzed the forces exerted on Santa to its $x$ and $y$ components. We choose down the inclined to be our positive $x$-direction. As you can see there are only two forces acting on him which are the normal force and his own weight due to Earth's gravitational pull. The net force exerted on him in the $y$- direction is zero since he has no motion in this direction. $$\sum F_y=F_n-mg\cos\theta=0$$ thus, $$F_n=mg\cos\theta $$ The net force exerted on him in the $ x$ direction is given by $$\sum F_x=mg\sin\theta =ma_x$$Thus, $$a_x=g\sin\theta\tag 1$$ Now we know that Santa slides from rest to 10 m away, so we can find the final velocity by applying the kinematic formula of velocity squared. $$v_{fx}^2=\overbrace{ v_{ix}^2 }^{0} +2a_x\Delta x$$ $$v_{fx}= \sqrt{2a_x\Delta x}=\sqrt{2g\sin30^\circ\Delta x}\tag { from (1) }$$ Plugging the known; $$v_{fx}= \sqrt{2\cdot 9.8 \sin30^\circ\cdot 10}=\pm 9.9\;\rm m/s$$ We can neglect the negative root since we chose down the inclined to be the positive direction. Thus, $$v_{fx}= \color{red}{\bf 9.9}\;\rm m/s$$
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