Average Coefficient of Kinetic Friction between Ice and Puck

[K Wils]

1. The problem: A hockey puck is hit on a frozen lake and starts moving with a speed of 11.6 m/s. Five seconds later, its speed is 7.4 m/s. What is its average acceleration? The acceleration of gravity is 9.8 m/s. Answer in units of m/s. What is the average value of the coefficient of kinetic friction between puck and ice?

Homework Equations

: [/B]Sum of the forces' x-components equals mass times acceleration. Sum of the forces' y-components equals zero. Kinetic friction equals the coefficient of kinetic friction times the normal force.

The Attempt at a Solution

: [/B]I solved for the acceleration earlier and got -0.84 m/s squared. I did this by taking the difference in velocities and dividing by the time (7.4 - 11.6 = - 4.2. - 4.2 / 5 = - 0.84). I then applied what I knew about the normal force: FN=mg. I also knew that FK=μkFN. Therefore, FK=μk(mg). While ∑Fx=ma, μk(mg)=ma. The masses cancel out. Then I put in my numbers: μk(9.8 m/s squared)= - 0.84 m/s squared. From there I divided my acceleration by 9.8 and got μk= - 0.0857. I checked it and the system on the website I was using for homework (UT) deemed my answer incorrect. Can anyone help?

 

[TomHart]

You can't have a negative coefficient of friction.

P.S. Welcome to Physics Forums.

 

[zexxa]

Your work is correct but you have forgotten to take into account directions. Part (i) has the correct answer, as for part (ii), as @TomHart mentioned, you can't have a negative coefficient of friction. What do you think your mistake is then, and why?

 

[K Wils]

I had realized that shortly after posting the thread. Thank you for the help though! I really appreciate it!