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Mechanics11/Seite9/Loesung3.9

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Aufgabenlösung 3.9

geg.: m = 330000kg ; Fmax = 9,0 *10^5N  ; F-start = 8,2*10^5N

       Fr= 2,7*10^5N ;  v-start = 300km/h

a) ges.: t

Lsg.: F-start - Fr = 8,2*10^5N - 2,7*10^5N = 5,5*10^5N 

     F=m*a  a=\frac{F}{m}  

     a=\frac{5,5*10^5N}{3,3*10^5kg}=1,66\frac{m}{s^2}

     a=\frac{v}{t}  t=\frac{v}{a} = \frac{83,33\frac{m}{s}}{1,66\frac{m}{s^2}} = 50,1s


b) ges.: x

Lsg.: x = 0.5 * a * t^2 = 0.5 * 1.66\frac{m}{s^2} * (50s)^2 = 2075m


c) zusätzlich geg.: x = 3000m

ges.: F

Lsg.: v^2 = 2ax 

     a = \frac{v^2}{2x}

       = \frac{(83,33\frac{m}{s})^2}{2 * 3000m}

       = 1,15\frac{m}{s^2}

     F = m * a 
       = 3,3*10^5kg * 1,15\frac{m}{s^2}
       = 3,795 * 10^5N = 3,8*10^5N

--Nikih 13:17, 7 December 2007 (CET)

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