What is the average value of the coefficient of kinetic friction between puck and ice?

Question

A hockey puck is hit on a frozen lake and starts moving with a speed of 12.0 m/s. Five seconds later it's speed is 6.00 m/s. What is the average value of the coefficient of kinetic friction between puck and ice?

Answer 1

Hi Amelia!

 

Once the puck is moving, we can assume the only force acting on it horizontally is kinetic friction, as it is sliding over the ice.  Thus, in the horizontal direction, we have, for Newton's 2nd law:

 

F_net-h = F_k = ma

 

which we can immediately rewrite as:

 

μ_kN = ma   (μ_k = coefficient of kinetic friction)

 

 

In the vertical direction, only the normal force, N (up), and gravity, mg (down), act, so we have:

 

F_net-v = N - mg = 0  (since the object is stationary vertically)

 

Thus, N = mg, and we can rewrite the horizontal direction as:

 

μ_kmg = ma -->  a = μ_kg --> μ_k = a/g

 

Since we know g, we could calculate a value for μ_k if we had a value for a, the acceleration.  What the problem gives us is a way to calculate an average value for acceleration, using the definition of acceleration.

 

a = (v-v_o)/Δt

 

v = final velocity, after our interval of interest = 6 m/s

v_o = initial velocity = 12 m/s

Δt = 5 s

 

[NOTE:  I have implicitly taken the direction of motion for the puck to be positive, which means I will calculate a negative acceleration, because the puck is slowing down.  The friction force must also then point in the negative direction in Newton's 2nd law for the horizontal, above.  So, technically, we have:

 

-μ_kN = -ma

 

to start above.  But the negative signs just cancel out in this case, and we deal with magnitudes.  So I will only worry about the magnitude of a in what we calculate below, and forget about the sign, which we just took care of]

 

Then, just calculate a, and put it (without worrying about sign) into the expression for μ_k above, along with g, and solve.

 

[NOTE:  This gives the AVERAGE coefficient of kinetic friction because, when solving for acceleration with the expression above, we actually solved for the AVERAGE acceleration.  The acceleration may have been changing from moment to moment, but the expression we used assumed it was constant over that 5 seconds, thus giving us the average value.... which is all we usually worry about at this level]

 

I hope this helps!  Just let me know if you have any more questions about this.