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: 60a

Answer

$\dfrac {\partial a}{\partial t}=-\dfrac {F}{m}\sin \left( \theta _{0}+\alpha t\right) \times \alpha =\dfrac {-20}{5}\sin \left( 0,4363rad\right) \times 3,49\times 10^{-4}\dfrac {rad}{s}\approx 1,475\times 10^{-4}\dfrac {m}{s^{3}}$

Work Step by Step

Le'ts say $\theta(t)$ is changing by time with $\theta \left( t\right) =\left( \theta _{0}+\alpha t\right) (1)$ where $\theta_{0}$ and $α$ is expressed in radians And we can say the object will not have any vertical movement because $mg=50N > F=20N$ So lets calculate horizontal acceleration of object at any given time $a=\dfrac {F\cos \theta \left( t\right) }{m}=\dfrac {F\cos \left( \theta _{0}+\alpha t\right) }{m}$ to find changing rate of acceleration : $\dfrac {\Delta a}{\Delta t}=\dfrac {\partial a}{\partial t}=\dfrac {\partial \left( \dfrac {F\cos \left( \theta _{0}+\alpha t\right) }{m}\right) }{\partial t}=-\dfrac {F}{m}\sin \left( \theta _{0}+\alpha t\right) \times \alpha $ $m=5kg;F=20N;\theta _{0}=25^{0}\approx 0,4363rad;\alpha =2\times 10^{-2}\dfrac {deg}{s}\approx 3,49\times 10^{-4}\dfrac {rad}{s}$ Using(1) when $\theta=25^{0}$ we get $t=0$ so we get $\dfrac {\partial a}{\partial t}=-\dfrac {F}{m}\sin \left( \theta _{0}+\alpha t\right) \times \alpha =\dfrac {-20}{5}\sin \left( 0,4363rad\right) \times 3,49\times 10^{-4}\dfrac {rad}{s}\approx -1,475\times 10^{-4}\dfrac {m}{s^{3}}$
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