# A hockey puck is given an initial speed of 16m/s .?

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A hockey puck is given an initial speed of 16m/s .If it comes to rest in 60m , what is the coefficient of kinetic friction

• There is only only force, friction. By Newton’s 2nd Law we have ug=a so will need to find the acceleration (deceleration) which from kinematics we find is a=v^2/2s=2.13m/s^2. The coefficient of kinetic friction is u=a/g=0.217

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• Coeff. of Kinetic Friction= Kinetic Friction/Normal Force

We know

u= 16 m/s, s= 60 m, m= M, v= 0, Normal Force= 9.81M

We want

a=?, Kinetic Friction

…………

–Solution–

Using

2as = v^2 – u^2

a= 0-256/120

a= 2.14 ms^2 (the negative is for retardation, we can ignore it)

Now.

F=ma

F(Friction)=2.14M

Now using the formula stated first we have,

Coeff of Kinetic Friction: 2.14M/9.81M (M cancels off 😉 )

Therefore the answer is: 0.05, quite slippery surface…

• I believe the answer is off a bit. it should be 0.5 instead…

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