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WILL GIVE BRAINLIEST!!! NO LINKS PLZ!!!!

A 225 g hockey puck is sliding on ice in an arena towards the end boards that are 15.7 m away. The puck is travelling 12.0 m/s when it slides into some rough ice (coefficient of kinetic friction= 0.550).
Determine:
a) the acceleration of the puck on the rough ice.
b) the distance from the end boards the puck is when it comes to a stop.

Please show work.​

Sagot :

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.

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