Physics
posted by lola on .
A 20.0kg curling stone at the hog line and moves 28.35m [west] to sit on the button and score a point. If the coefficient of kinetic friction between the stone and the ice is 0.00200, what was the initial speed of the stone when it was released?

F =  20 * 9.8 * .002
F = m a
20 * 9.8 * .002 = 20 a
a = 19.6 *10^3 m/s^2
stops at time t at the button
v = Vi + a t
0 = Vi + a t
t = Vi/a
d = Vi t + (1/2) a t^2
28.35 = Vi (Vi/a) + (a/2) Vi^2/a^2
28.35 = Vi^2/a + (1/2) Vi^2/a
28.35 = (1/2) Vi^2/a
so
Vi^2 = 2 (28.35)(19.6*10^3)
Vi^2 = 1111
Vi = 33.3 m/s