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: 48

Answer

a) $ 3.79 \;\rm m $ b) $7\;\rm m/s$

Work Step by Step

a) To find the vertical height, we need to find the distance traveled up the ramp. See the figures below. We can use the kinematic formula of $$\overbrace{v_{fx}^2}^{0}-v_{ix}^2=2a_x\Delta x$$ We know that the box will stop after a while and then moves down the ramp. This means that the final velocity when it stops is zero. $$ -v_{ix}^2=2a_x\Delta x$$ Thus, $$\Delta x=\dfrac{ -v_{ix}^2}{2a_x }$$ From the figure below, we can see that $$\sin30^\circ =\dfrac{h}{\Delta x}$$ Thus, $$\Delta x=\dfrac{h}{\sin30^\circ}=2h$$ Hence, $$2h=\dfrac{ -v_{ix}^2}{2a_x }$$ $$ h=\dfrac{ -v_{ix}^2}{4a_x }\tag 1$$ Now we need to find $a_x$; The net exerted on the box in the $y$-direction is given by $$\sum F_y=F_n-mg\cos30^\circ=ma_y=m(0)=0$$ Thus, $$F_n=mg\cos30^\circ\tag 2$$ The net exerted on the box in the $x$-direction is given by $$\sum F_x=-f_k-mg\sin30^\circ =ma_x$$ and $f_k=\mu_kF_n$ $$ -\mu_kF_n-mg\sin30^\circ =ma_x$$ Plugging from (2); $$ -\mu_k g\cos30^\circ- g\sin30^\circ= a_x$$ $$a_x= -g\left[ \mu_k \cos30^\circ+\sin30^\circ \right]=-9.8\left[ 0.2\cos30^\circ+\sin30^\circ \right]$$ $$a_x=\bf -6.6\;\rm m/s^2$$ Plugging into (1) and plug the known as well; $$ h=\dfrac{ -10^2}{4\times -6.6 } =\color{red}{\bf3.79}\;\rm m$$ b) Now we know that its initial velocity down the ramp is zero and that the friction force now is upward the ramp, see the third figure below. We need to find $a_x$ here as well by the same approach above. $$\sum F_x=f_k-mg\sin30^\circ=ma_x$$ $$f_k-mg\sin30^\circ=ma_x$$ $$\mu_kF_n-mg\sin30^\circ=ma_x$$ $$\mu_k mg\cos30^\circ-mg\sin30^\circ=ma_x\tag{From (2)}$$ Thus, $$a_x=g\left[\mu_k \cos30^\circ- \sin30^\circ\right]=9.8\left[0.2 \cos30^\circ- \sin30^\circ\right]$$ $$a_x=\bf -3.2\;\rm m/s^2$$ So, the final velocity of the box at the end of the ramp is given by $$v_{fx}^2=\overbrace{v_{ix}^2}^{0}+2a_x\Delta x=2a_x\cdot 2h$$ $$v_{fx} =\pm \sqrt{4a_xh}=\pm\sqrt{ 4\times -3.2\times -3.79 }$$ and since we need the speed, $$v_{fx}=\color{red}{\bf 7.0}\;\rm m/s$$
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