Fundamentals of Physics Extended (10th Edition)

Published by Wiley
ISBN 10: 1-11823-072-8
ISBN 13: 978-1-11823-072-5

Chapter 5 - Force and Motion-I - Problems - Page 122: 68

Answer

$$ F=334.8 \mathrm{\ N} $$

Work Step by Step

We first use Eq. $4-26$ to solve for the launch speed of the shot: $$ y-y_{0}=(\tan \theta) x-\frac{g x^{2}}{2\left(v^{\prime} \cos \theta\right)^{2}} $$ With $\theta=34.10^{\circ}, \quad y_{0}=2.11 \mathrm{m},$ and $(x, y)=(15.90 \mathrm{m}, 0),$ we find the launch speed to be $v^{\prime}=11.85 \mathrm{m} / \mathrm{s}$ . During this phase, the acceleration is $$ a=\frac{v^{\prime 2}-v_{0}^{2}}{2 L}=\frac{(11.85 \mathrm{m} / \mathrm{s})^{2}-(2.50 \mathrm{m} / \mathrm{s})^{2}}{2(1.65 \mathrm{m})}=40.63 \mathrm{m} / \mathrm{s}^{2} $$ since the acceleration along the slanted path depends on only the force components along the path, not the components perpendicular to the path, the average force on the shot during the acceleration phase is $$ F=m(a+g \sin \theta)=$$ $$(7.260 \mathrm{kg})\left[40.63 \mathrm{m} / \mathrm{s}^{2}+\left(9.80 \mathrm{m} / \mathrm{s}^{2}\right) \sin 34.10^{\circ}\right]$$ $$=334.8 \mathrm{\ N} $$
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