Question
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?
Answers
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
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
Related Questions
A curling stone with a mass of 18 kg slides 38m across a sheet of ice in 8.0 s before it stops becau...
1. A stationary curling stone is struck in a glancing collision by a second curling stone of equal...
1. A stationary curling stone is struck in a glancing collision by a second curling stone of equal...
A curling stone with mass 19.00 kg is released with an initial speed v0 sliding on level ice. The co...