Answer:
Explanation:
a) the acceleration of the puck on the rough ice.
a = μg = 0.550(9.81) = 5.3955 = 5.40 m/s²
(comes from μ = F/N = ma/mg = a/g)
b) the distance from the end boards the puck is when it comes to a stop.
v² = u² + 2as
0² = 12.0² + 2(-5.40)s
s = 13.3 ft
so distance from the boards is
15.7 - 13.3 = 2.4 m
by the way...that's some VERY rough ice...more like sand.
