Mechanics11/Seite11/Loesung4.2

m = 90 000 kg + 22 000 kg * 10 = 310 000 kg x = 4 000 m $$v_1 = 19,4 \frac{m}{s}$$ $$F_R = 30 000 N$$

$$F_r = F_R * x = 30 000 N * 4 000 m = 120 MJ$$

$$W_a = \frac{1}{2}mv^2 = \frac{1}{2} * 310 000 kg * (19,4\frac{m}{s})^2$$ = 60 000kJ = 60MJ

$$W_g = 60MJ + 120MJ = 180MJ$$

Alternative - "zweite" Lösung:

geg.: m1 = 90t = 90 000 Kg

m2 = 10*22t (Wagons) = 220 000 Kg

=> 310 000 Kg

x= 4,0 km = 4000 m

v= 70 km/h = 19,4 m/s

F(Reibung)= 30 kN = 30 000 N

Lös.:

Wa = 1/2*mv2 = 0,5 * 310 000 Kg * (19,4m/s)2

Wa=~58 000 J

Wr = Fr*x =120MJ

=> Wa + Wr = 178,3 MJ ---MoeM11 12:09, 16 February 2009 (UTC)