Mechanics11/Seite10/Loesung3.19

$$m = 52kg$$ $$a = 22 grad $$ $$ mue = 0,06$$ $$x= 25m$$

a) $$F_H = sin (22) *52kg*9,81\frac{N}{m}$$

$$= 191N$$

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

$$\frac{191N}{52kg}= 3,7\frac{m}{s*s}$$

$$v =(2*a*x)^\frac{1}{2} $$

$$v= (2*3,7\frac{m}{s*s}*25$$

$$v=13,6\frac{m}{s}$$

b)

$$F_N =52kg*cos(22)$$

$$F_N = 473N$$

$$F_R = 0,06 *473N$$

$$F_N=F_H-F_R$$

$$F_N=191N-28,3N =162,7N$$

$$\frac{f}{m}=a$$

$$\frac{162,7N}{52kg} = 3,13\frac{m}{s*s}$$

$$v=(2*a*x)^\frac{1}{2}$$

$$v = (2*3,13\frac{m}{s*s}*25m)^\frac{1}{2}$$

$$v= 12,5\frac{m}{s}$$

c)

$$F_T=F_H+F_R$$

$$191N+28,3N= 219,3N$$

d)

$$a=\frac{F-219,3N}{m}$$

$$F= a*m+219,3N$$

$$F= 2,0\frac{m}{s*s}*52kg+219,3N$$

$$F= 323N$$

--Max 12:18, 1 February 2008 (CET)