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Coefficient of kinetic friction on hockey puck

  • Thread starter ttk3
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  • #1
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The way I approached the problem I found acceleration using V^2=vo^2 + 2ax. Then I set uk*mg = ma. The masses cancel... so its uk*g = a. I keep getting the wrong answer. What am I doing wrong?




Homework Statement



A hockey puck on a frozen pond with an initial speed of 22.3 m/s stops after sliding a distance of 249.3 m. Calculate the average value of the coefficient of kinetic friction between the puck and the ice.

Homework Equations



uk*m*g = ma

The Attempt at a Solution



0.0093
 
Last edited:

Answers and Replies

  • #2
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I know the masses cancel... What am I missing?
 
  • #3
Kurdt
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How did you work out the acceleration?
 
  • #4
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0 = 22.3 ^2 + 2 * a * 249.3

and i solved for a
 
  • #5
Kurdt
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Ok. Now you know the masses cancel, which leaves you with:

[tex] \mu_k = \frac{a}{g} [/tex]

What do you get for your answer, knowing that g is the acceleration due to gravity?
 
  • #6
Try this:

1 ∑F_{x}=-f_{k}=ma_{x}
2 ∑F_{y}=n-mg=0
note: n=mg becomes(n/g)=m
3 if puck is moving right then:
4 -u_{k}n=-u_{k}mg=ma_{x}
5 a_{x}=-u_{k}g
6 V_{xf}²=V_{xi}²+2a_{x(X_{f}}-x_{i)}
7 so now 0=V_{xi}²+2a_{x}x_{f} becomes
8 V_{xi}²-2u_{k}gx_{f}
9 u_{k}=((V_{xi}²)/(2gx_{f}))
10 u_{k}=(((22.3(m/s))²)/(2(9.80(m/(s²)))(249.3m))) =0.1017727
11 how's that?
 
  • #7
Kurdt
Staff Emeritus
Science Advisor
Gold Member
4,812
6
Try this:

1 ∑F_{x}=-f_{k}=ma_{x}
2 ∑F_{y}=n-mg=0
note: n=mg becomes(n/g)=m
3 if puck is moving right then:
4 -u_{k}n=-u_{k}mg=ma_{x}
5 a_{x}=-u_{k}g
6 V_{xf}²=V_{xi}²+2a_{x(X_{f}}-x_{i)}
7 so now 0=V_{xi}²+2a_{x}x_{f} becomes
8 V_{xi}²-2u_{k}gx_{f}
9 u_{k}=((V_{xi}²)/(2gx_{f}))
10 u_{k}=(((22.3(m/s))²)/(2(9.80(m/(s²)))(249.3m))) =0.1017727
11 how's that?
Please don't give away full solutions. It defeats the object of helping students to learn for themselves.
 

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