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Work/Energy - Incline Plane w/ Friction

  1. Nov 24, 2015 #1
    1. The problem statement, all variables and given/known data

    A 74.5 kg snowboarder heads down a 16.0° hill that has a height of 78.4 m. If the hill is assumed to be frictionless and there is horizontal wind with a force of 93 N acting against the snowboarder, find the speed of the snowboarder as they reach the bottom of the hill using work and energy.


    2. Relevant equations

    W = FΔd
    Ep = mgΔy
    Ek = 1/2mv2


    3. The attempt at a solution

    So far, I tried calculating potential energy at top:

    Ek = mgΔy
    = (74.5)(9.8)(78.4)
    = 57,239.84 J

    Work done by wind:
    W = FΔd
    = (93N)(78.4/sin16.0°)
    = 26,452.14753 J

    Then, I subtracted work done by wind from potential energy, then used the kinetic energy to solve for final velocity.
    What am I doing wrong?
     
  2. jcsd
  3. Nov 24, 2015 #2

    TSny

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    Hello, and welcome to PF!
    Can you explain why you used the sine of the angle when finding Δd?
     
  4. Nov 24, 2015 #3
    MG10C15_003.png

    Opposite = 78.4 m
    Angle = 16.0°


    Solving for hypothenuse:
    sin16° = 78.4/h
    h = 78.4/sin16°

    Is this not correct?
     
  5. Nov 24, 2015 #4

    TSny

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    It's correct if Δd represents the distance along the slope. But, then, is the formula W = FΔd correct? What is the direction of the wind force?
     
  6. Nov 24, 2015 #5
    Horizontal wind force opposing the skateboarder's motion.
     
  7. Nov 24, 2015 #6

    TSny

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    So, the force of the wind is not parallel to the displacement Δd. How do you find the work when the force is not parallel to the displacment?
     
  8. Nov 24, 2015 #7
    I must consider the incline plane, should I not?

    Would I need to find the adjacent side?

    Edit: Would the distance be 78.4/tan16?
     
  9. Nov 24, 2015 #8

    TSny

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    In general, how do you find the work done by a force that is not parallel to the displacement?
     
  10. Nov 24, 2015 #9
    W = FΔdcosθ.
     
  11. Nov 24, 2015 #10

    TSny

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    Right. Can you apply that to this problem? What would you use for θ? Use a carefully drawn diagram.
     
  12. Nov 24, 2015 #11
  13. Nov 24, 2015 #12

    TSny

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    No. What does θ stand for in the formula W = FΔdcosθ?
     
  14. Nov 24, 2015 #13
    The angle between the Force and Distance. Would it be 16° because of the Z Pattern?
     
  15. Nov 24, 2015 #14

    TSny

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    Yes

    I'm not sure what you mean by the Z pattern. Did you draw vectors representing the wind force and the displacement? Does the angle between the force and the displacement look like it's less than 90o or greater than 90o?
     
  16. Nov 24, 2015 #15
    I might be picturing the problem incorrectly. The wind would be directed west while displacement would be down the hill, correct?
     
  17. Nov 24, 2015 #16

    TSny

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    Yes, if west means to the left in your figure back in post #3.
     
  18. Nov 24, 2015 #17
    The angle would be less than 90 then.
     
  19. Nov 24, 2015 #18

    TSny

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    When getting the angle between two vectors, you should draw the two vectors from the same point. That is, the "tails" of the vectors should be at the same point.
     
  20. Nov 24, 2015 #19
    Even if you do that, can't you move the vector around and displace the tail of the second vector to the head of the first vector? Resulting in the same angle of 16°*?
     
  21. Nov 24, 2015 #20

    TSny

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