Calculating Coefficient of Kinetic Friction for a Hockey Puck

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SUMMARY

The discussion focuses on calculating the coefficient of kinetic friction (μ) for a hockey puck sliding on ice. The puck, initially traveling at 12.3 m/s, comes to a stop after sliding 198.9 m. The user initially miscalculated acceleration by dividing speed by distance, resulting in an incorrect time. The correct approach involves using the equation v² = u² + 2as to find acceleration, leading to a more accurate calculation of μ using the formula μ = a/g, where g is 9.81 m/s².

PREREQUISITES
  • Understanding of kinematic equations, specifically v² = u² + 2as
  • Basic knowledge of friction and its relationship with normal force (N = mg)
  • Familiarity with acceleration and its calculation in physics
  • Concept of uniform acceleration in motion
NEXT STEPS
  • Study kinematic equations in detail, focusing on uniform acceleration scenarios
  • Learn about the derivation and application of the coefficient of friction in various materials
  • Explore real-world examples of friction calculations in sports physics
  • Investigate the impact of surface conditions on kinetic friction coefficients
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Physics students, sports scientists, and engineers interested in the dynamics of motion and friction in sports applications.

Psyguy22
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A hockey puck on a frozen pond with an initial speed of 12.3 m/s stops after sliding a distance of 198.9 m. Calculate the average value of the coefficient of kinetic friction between the puck and the ice.

So I started with finding the acceleration. I divided 12.3m/s by 198.9m then took the inverse (so units would go back seconds) and came up with 16.17s. So now I have Δv and Δt so I divided 12.3m/s by 16.17s to get an acceleration of .76m/s^2

I also know that friction equals μN and that F=ma. Since N=mg and friction is the only force acting on the object, i divided the mass out. So I had μg=a, or μ=a/g. (using 9.81m/s^2 as gravity) So μ=.0774 which seemed reasonable for ice. But I don't know how to find the actual force without knowing the mass. Am I even heading in the right direction?
 
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Psyguy22 said:
So I started with finding the acceleration. I divided 12.3m/s by 198.9m then took the inverse (so units would go back seconds) and came up with 16.17s. So now I have Δv and Δt so I divided 12.3m/s by 16.17s to get an acceleration of .76m/s^2
That does not work.
To travel 198.9m in 16.17s, it would have to travel at 12.3m/s the whole time, which is clearly not the case. You need a different approach here.

I also know that friction equals μN and that F=ma. Since N=mg and friction is the only force acting on the object, i divided the mass out. So I had μg=a, or μ=a/g. (using 9.81m/s^2 as gravity) So μ=.0774 which seemed reasonable for ice. But I don't know how to find the actual force without knowing the mass. Am I even heading in the right direction?
That part is good, once you fix your acceleration value. You cannot, and do not have to, calculate the force.
 
Then how do I find acceleration? I don't have time..
 
But you do know lots of other things. For example, you know the initial and the final speeds. I think it's okay to assume uniform acceleration here. You probably have a list of equations for dealing with this sort of motion that look something like, for example, v = u + at.

What you need to do is find an equation that includes the quantities that you know along with the acceleration. Then there's some algebra to find a.
 

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