# Calculating the coefficicent of friction

#### sskk221

1. The problem statement, all variables and given/known data

A hockey Puck is hit and starts moving at 12m/s. Exactly 5 s later its speed is 6m/s. What is the coefficient of friction? How far has the pick traveled during the 5 s?

2. Relevant equations

$$\vec{F}_{net} = \Sigma \vec{F} = m \vec{a}$$
$$f_k = \mu_k N$$

3. The attempt at a solution

$$v = v_0 + a t$$
$$6m/s = 12m/s + a 5s$$
a= -1.2 m/s^2

$$x = x_0 + v_0 t + (1/2) a t^2$$
$$x = 0m + 12*5s + (1/2) (-1.2m/s^2) 5^2$$
x= 45m

Not sure how to precede. Answers are x =75m and coefficient of friction = 0.12. I apparently did the first part wrong

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#### cristo

Staff Emeritus
I think you're right. Your acceleration is definitely correct, and your substitution into the distance equation is correct. (To get x=75, one would put in 1.2 for the acceleration, but it is clearly decelerating!)

For the next part, use your two equations concerning the force. Put the mass in as m; it will cancel. Can you go on from here?

#### Astronuc

Staff Emeritus
coefficient of friction = 0.12
You have the correct magnitude of acceleration, which is consistent with 0.12.

The weight of the puck is mg, and the force of friction is mu*mg = ma, so

a = mu * g or mu = a/g, and g = 9.81 m/s2.

So one has to determine how far a puck travels while decelerating at a constant 1.2 m/s2, i.e. -1.2 m/s2 acceleration.

#### sskk221

haha thanks for the help! Forgot that the Normal force (N) can be replaced with "mg"

3

#### 3dsmax

dang i could helped you with this one, we just finished are test on this. Cool

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