Calculating Coefficient of Kinetic Friction on a Slope Using Newton's Laws

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SUMMARY

The discussion focuses on calculating the coefficient of kinetic friction (uk) between snow and a snowboard using Newton's Laws. The user describes an experiment where the snowboard takes twice as long to slide down the slope compared to sliding up. Key equations derived include the net force equations in both the i-hat and j-hat directions, leading to expressions for acceleration (au and ad). The user expresses confusion regarding the signs of friction and acceleration, particularly when uk exceeds 1, which raises questions about the physical feasibility of the results.

PREREQUISITES
  • Understanding of Newton's Laws of Motion
  • Familiarity with Free Body Diagrams (FBD)
  • Knowledge of trigonometric functions related to angles
  • Basic algebra for solving equations
NEXT STEPS
  • Study the derivation of acceleration equations in inclined plane problems
  • Learn about the implications of coefficients of friction exceeding 1
  • Explore the concept of Free Body Diagrams in more complex scenarios
  • Investigate the relationship between angle of incline and frictional forces
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Students studying physics, particularly those focusing on mechanics and dynamics, as well as educators looking for practical examples of applying Newton's Laws in real-world scenarios.

jmonkey
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Homework Statement



Note - I have used (uk) as the symbol for co-efficient of kinetic friction (mew k?).

You want to measure (uk) between snow and your snowboard. You measure the angle x the slope makes with the horizontal. You kick your snowboard so it slides up the slope, and then back down. You notice that it takes twice as long to complete the downward portion of it's slide than the upward portion. Determine (uk). The steps to do so are as follows:

1) FBD of snowboard, using variables f and a together with i-hat in the direction up the slop, to represent the friction and acceleration vectors. j-hat is chosen perpendicular to the slope then. What are the signs of f and a on the upward/downward portion of the motion?

2) Use Newton's 2nd law in the i-hat and j-hat directions to solve for a (on the way up and on the way down, denoted (au) and (ad) respectively) in terms of (uk), x and g. Be careful about the sign of f.

...

The question after this point I understand. My problem is coming up with equations for (au) and (ad) that make sense to me. Please help :)

Homework Equations



NetForce = mass*acceleration
NetForce(i-hat) = mass*accel(i-hat)
NetForce(j-hat) = mass*accel(j-hat)

The Attempt at a Solution



When I worked this out, I am saying that a is negative all the time (remember (i-hat) is in the direction up the slope), and f is negative on the way up, and positive on the way down. I obtained...

Fnet = Normal*(j-hat) + f(i-hat) - mgsinx(i-hat) - mgcosx(j-hat)
= m*accel = m*a(i-hat)

(i-hat) direction: f-(mgsin x) = ma
(j-hat) direction: N - (mgcos x) = 0

f = (uk) * N = (uk) * mgcos x

I make f negative to solve for (au), and get au = g(-(uk)cos x - sin x). I keep f positive to solve for ad = g( (uk)cos x - sin x). My problem is this. If (uk) happens to be greater than 1, and the angle x is less than 45 degrees, won't (uk)*cos x be greater than sin x. And then (ad) would have to be positive, which means the snowboard is accelerating "up" the slope...which cannot happen obviously. So I assume I have done something wrong. If anything is unclear, please let me know. I realize using (uk) and (ad) creates a lot of clutter. Thanks for your time,
 
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Bump...any suggestions?
 
Another bump. Sorry to be persistent, but 39 views now, and no one has thoughts (I don't need the problem solved completely, I'm pretty sure my issue is with grasping the variables f and a)? If I'm being too vague/unclear, please let me know.
 

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