Mechanics11/Seite10/Loesung3.19

$m = 52 k g$ $a = 22 g r a d$ $m u e = 0,06$ $x = 25 m$

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

$= 191 N$

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

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

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

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

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

b)

$F N = 52 k g * c o s (22)$

$F N = 473 N$

$F R = 0,06 * 473 N$

$F N = F H - F R$

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

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

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

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

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

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

c)

$F T = F H + F R$

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

d)

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

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

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

$F = 323 N$

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