Calculating the coefficicent of friction

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A hockey puck starts at 12 m/s and slows to 6 m/s over 5 seconds, resulting in an acceleration of -1.2 m/s². The distance traveled during this time is calculated to be 75 meters, using the equation x = v₀t + (1/2)at². The coefficient of friction is determined to be 0.12, derived from the relationship between acceleration and gravitational force. The normal force is replaced with mg in the friction equation, confirming the calculations. This discussion emphasizes the importance of correctly applying physics equations to solve for friction and distance.
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Homework Statement



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?

Homework Equations



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


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:redface:
 
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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?
 
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.
 
haha thanks for the help! Forgot that the Normal force (N) can be replaced with "mg"
 
dang i could helped you with this one, we just finished are test on this. Cool
 
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