Final 17 - Springs and Tables

Given:

• $\mbox{A 1 kg mass is connected to a spring with a constant of 350 newtons/meter.} $
• $\mbox{The spring is compressed 25 centimeters and let go from 2 meters.}$

Question:

• $\mbox{How far away will the mass impact the floor?} $

Rationale:

• $PE_{spring} = KE_{mass} $
• $.5 K x^2 = .5 m V_x^2   $ 
• $V_x^2 = \dfrac{(.5 * K * x^2 )}{(.5 * m)}$
• $t_y^2 = \dfrac{2 * \Delta{y}}{g}   $
• $t_y  = t_x $
• $\Delta{x} = V_x * t_x $
• $\Delta{x} = \sqrt{\dfrac{(.5 * K * x^2 )}{(.5 * m)}} *  \sqrt{\dfrac{2 * \Delta{y}}{g}}    $

Calculate:

• $\Delta{x} = \sqrt{\dfrac{(.5 * 350 * .25^2 )}{(.5 * 1)}} *  \sqrt{\dfrac{2 * 2}{10}}    $ 
• $\Delta_{x} = \sqrt{10.9375 * .5} *  \sqrt{.4} = 4.6770717 * .632455 = 2.9580374$ 
• $\Delta_{x} \approx 2.96 \; \mbox{meters}$

References:

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