1. Problem 6 V terminal = mg / b = (70kg)(9.8m/s^2) / 0.13 Ns^/m^2 = 5277 m/s Problem 1 If we assume that a salt shaker holds about 1/2 cup of salt than this would be approximately one million grains. Or 10^6 order.

2. 50 Kg F = 49 N Force gravity Force normal Force friction Problem 9 Force applied = 49 N Force gravity (Fg) = - mg = - 50*9.8 = - 490 N Force normal (Fn) = -Fg = -(-490N) = 490 N Force friction = Fn*coefficient of friction The force friction must equal force applied because the velocity is constant. 49N = 490N*coefficient of friction Coefficient of friction = 1/10 = 0.1

3. Problem 10 15 degrees 10 Kg 15 Fa Fn Fg Ff Force gravity (Fg) = - mg = - 10*9.8 = - 98 N Force normal (Fn) = -Fg*cos(15) = -98*cos(15) = -94.6 N Force applied = Fg*sin(15) = 98*sin(15) = 25.4 N Force friction = -Fn = -25.4 N Ff = Fn*coefficient of static friction 25.4N = 94.6N*coefficient of static friction Coefficient of friction = 25.4 / 94.6 = 0.268 Push of 50N 50 N – 25.4 N = 24.6N of force up incline F = ma a = F / m = 24.6N / 10kg = 2.46 m/s^2