Coefficient of kinetic friction on hockey puck

[ttk3]

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

 

[ttk3]

I know the masses cancel... What am I missing?

 

[Kurdt]

How did you work out the acceleration?

 

[ttk3]

0 = 22.3 ^2 + 2 * a * 249.3

and i solved for a

 

[Kurdt]

Ok. Now you know the masses cancel, which leaves you with:

$$\mu_k = \frac{a}{g}$$

What do you get for your answer, knowing that g is the acceleration due to gravity?

 

[plutoisacomet]

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?

 

[Kurdt]

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.