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a hockey puck is hit on a frozen lake and starts moving with a speed of 12m/s exactly 5.0 sec later its speed is 6.0 m/s
now i have the acceleration figured out-correct me if i'm wrong:
vi=vf+a*t 6.0=12+5.0a -12+6.0=5.0a - 6.0=5.0a a= -1.2 m/s2
now what is the coefficient of kinetic friction between the puck and the ice ?

2007-03-05 13:47:56 · 1 answers · asked by skybluu 2 in Education & Reference Homework Help

1 answers

For friction, the equation is:
F = μN
where μ is the coefficient of Friction, and N is the normal force exerted on the surface.

In this case, N is the weight of the puck (since it's on a flat horizontal surface. We don't have the mass of the puck, but we don't need it, because the mass appears on both sides of the equation.

Start with F = μN.
F = ma and N = mg so:
ma = μmg
a = μg (m cancels)
-1.2 = 9.8μ (substitute)
μ = 0.1224 (solution!)

2007-03-06 01:30:28 · answer #1 · answered by ³√carthagebrujah 6 · 0 0

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