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 10 - Energy - Exercises and Problems - Page 272: 14

Answer

a) $392\;\rm N/m$ b) $17.5\;\rm cm$

Work Step by Step

a) Let's assume that the stretching of the spring due to its own weight when suspended from the ceiling is negligible. After attaching the 2-kg mass, the spring stretched 5 cm and stopped. This means that the net force exerted on this 2-kg mass is zero since it is at rest. $$\sum F_y=F_{sp}-mg=ma_y=m(0)=0$$ Hence, $$F_{sp}=mg$$ We know that the spring force is given by $-kx$ where $x$ is the stretched or compressed distance. Thus, $$-kx=mg\tag 1$$ Thus, the spring constant is given by $$k=\dfrac{mg}{-x}$$ Plugging the known and recall that the spring with the block stretched down its equilibrium point by 5 cm, so $x=-0.05$ m. $$k=\dfrac{2\times 9.8}{-(-0.05)}=\color{red}{\bf 392}\;\rm N/m$$ --- b) Now we know that the mass attached is now a 3-kg mass. Using (1) to find the stretched distance below the equilibrium point. Then we can find the final length of the spring $$x=\dfrac{mg}{-k}$$ Plugging the known; $$x=\dfrac{3\times 9.8}{-392}=-0.075\;\rm m=\bf -7.5\;\rm cm$$ The negative sign is due to the direction. Then the final length of the spring is $$L_f=L_i+|x|=10+7.5=\color{red}{\bf17.5}\;\rm cm$$
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