Answer to Question #168937 in Physics for Froilan

Question #168937

The massless spring of a spring gun has a force constant k=12N/cm. When the gun is aimed vertically, a 15-g projectile is shot to a height of 5.0 m above the end of the expanded spring. How much was the spring compressed initially?


1
Expert's answer
2021-03-14T19:15:44-0400

We can find the initial compression of the spring from the law of conservation of energy:


12dx2=mgh,\dfrac{1}{2}dx^2=mgh,d=2mghk,d=\sqrt{\dfrac{2mgh}{k}},d=20.015 kg9.8 ms25 m12 Ncm100 cm1 m=0.035 m=3.5 cm.d=\sqrt{\dfrac{2\cdot0.015\ kg\cdot9.8\ \dfrac{m}{s^2}\cdot5\ m}{12\ \dfrac{N}{cm}\cdot\dfrac{100\ cm}{1\ m}}}=0.035\ m=3.5\ cm.

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