Answer to Question #168937 in Physics for Froilan

Question

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?

Answer 1

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

$$\dfrac{1}{2}dx^2=mgh,$$ $$d=\sqrt{\dfrac{2mgh}{k}},$$ $$d=\sqrt{\dfrac{2\cdot0.015\ kg\cdot9.8\ \dfrac{m}{s^2}\cdot5\ m}{12\ \dfrac{N}{cm}\cdot\dfrac{100\ cm}{1\ m}}}=0.035\ m=3.5\ cm.$$