1. ## physics help...

"A hockey puck on a frozen pond with an initial speed of 16.3 m/s stops after sliding a distance of 220.5 m. Calculate the average value of the coefficient of kinetic friction between the puck and ice."

-i thought i knew what to do, it seemed simple enough. i found the deceleration, speed divided by time (time=distance divided by speed). And then it should be deceleration divided by gravity?(f=uN) i get the wrong answer so obviously not. am i doing something wrong??

2. Once it starts sliding, the only force acting on the puck is friction so:
$F = F_{fr} \ \iff \ m\vec{a} = \mu F_{N}$ where $F_{N}$ is the normal force

So first find the deceleration of the puck: $\vec{v}^{\ 2}_{f}= \vec{v}^{\ 2}_{i} + 2\vec{a}\vec{d}$ - You have the final velocity (it's stops!), the initial velocity, and its distance.

So after solving for $\vec{a}$, use the equation from the first line:
$\begin{array}{rcl} m\vec{a} & = & \mu F_{N} \\ m\vec{a} & = & \mu (mg) \end{array}$

3. Originally Posted by sarabolha
"A hockey puck on a frozen pond with an initial speed of 16.3 m/s stops after sliding a distance of 220.5 m. Calculate the average value of the coefficient of kinetic friction between the puck and ice."

-i thought i knew what to do, it seemed simple enough. i found the deceleration, speed divided by time (time=distance divided by speed). And then it should be deceleration divided by gravity?(f=uN) i get the wrong answer so obviously not. am i doing something wrong??
You need to find $\mu_k$

If you draw a free body diagram, you should get something like this:

To find acceleration, use the kinematic equation $v_f^2=v_i^2+2a(x_f-x_i)\implies 0=(16.3)^2+2a(220.5)\implies a=-.6~\tfrac{m}{s^2}$

Now let's do a FB analysis:

$\sum F_x=ma$:

Thus, $-F_{fr}=ma$

$\sum F_y=0$:

Thus, $N-mg=0\implies N=mg$

Therefore, $-F_{fr}=ma\implies -\mu_kN=ma\implies -\mu_kmg=ma\implies \mu_k=-\frac{a}{g}$ $\implies\mu_k=-\frac{-.6~\frac{m}{s^2}}{9.8~\frac{m}{s^2}}\implies \color{red}\boxed{\mu_k\approx 0.06}$

Does this make sense?

--Chris

4. ah that was too easy i'm embarrassed. but you explained it very very well, thanks!!