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Tuesday, August 7, 2012

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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