A curling stone with mass 19.00 kg is released with an initial speed v0 sliding on level ice. The coefficient of kinetic friction between the curling stone and the ice is 0.01203. The curling stone travels a distance of 37.29 m before it stops. What is the initial speed of the curling stone?
2 answers
if i knew why would i be here :P (This is a joke lol)
friction force = -0.01203 * 19.00 * 9.81 = -2.24 N
That is the only force
so
a = -F/m = 0.118 m/s^2
v = Vo - 0.118 t
0 = Vo - 0.118 t so Vo = 0.118 t
average speed during stop = Vo/2 = 0.059 t
distance= average speed * t
37.29 = 0.059 t^2
so t = 25.1 seconds to stop
Vo = 0.059 t
= 1.48 m/s
That is the only force
so
a = -F/m = 0.118 m/s^2
v = Vo - 0.118 t
0 = Vo - 0.118 t so Vo = 0.118 t
average speed during stop = Vo/2 = 0.059 t
distance= average speed * t
37.29 = 0.059 t^2
so t = 25.1 seconds to stop
Vo = 0.059 t
= 1.48 m/s
1 answer