1. Not finding help here? Sign up for a free 30min tutor trial with Chegg Tutors
    Dismiss Notice
Dismiss Notice
Join Physics Forums Today!
The friendliest, high quality science and math community on the planet! Everyone who loves science is here!

Kinetic energy

  1. Oct 19, 2015 #1
    http://imgur.com/luj32IE 1. The problem statement, all variables and given/known data
    A skier starts from rest at the top of a smooth incline of height 20 m as in Figure 2. At the bottom of the
    incline, the skier encounters a horizontal rough surface where the coecient of kinetic friction between the
    skis and the snow is 0.21.
    (a) How far does the skier travel on the horizontal surface before coming to rest?
    (b) Find the horizontal distance the skier travels before coming to rest if the incline is also rough with a coecient of kinetic friction equal to 0.21.

    2. Relevant equations
    Vf² = Vi² + 2a(Xf - Xi)
    K = MV²/2


    3. The attempt at a solution
    Yi =H= 20m
    θ = 20°
    uk = 0.21


    a) Vf² = Vi² - 2g(Yf-Yi)
    Vf² = 0 - 2g(0-20m)
    Vf = 19.8 m/s

    Ke2 - Ke1 = A(friction)

    Ke2 - Ke1 = 0 - MV²/2 = - MV²/2
    A(f) = F×S×cos(180°-20°)
    - MVf²/2 = F×S×cos(180°-20°)
    - MVf²/2 = (uk)(Mg)S(cos160°)
    S = 101m


    b) MVf²/2 = (uk)(Mg)S
    S= Vf²/2(ukg) = (19.8m/s) / 2(0.21)(9.8m/s²) = 95m
     
    Last edited: Oct 19, 2015
  2. jcsd
  3. Oct 19, 2015 #2

    BvU

    User Avatar
    Science Advisor
    Homework Helper
    Gold Member

    And your question is why are these answers not correct ?
    And you already have the correct answer ?
    And for the a) question it would be 95 m ?
    And the incline is 20 m high and the angle wrt horizontal is 20##^\circ## ?
     
  4. Oct 19, 2015 #3
    Could you please provide the figure or write what value θ represents?

    According to the text of the statement, the answer of the first question (a) should be the one you achieved in (b).
     
  5. Oct 19, 2015 #4
     

    Attached Files:

  6. Oct 19, 2015 #5
    I inserted a photo, thanks for reply!
     
  7. Oct 19, 2015 #6
    then what would b) be?
     
  8. Oct 19, 2015 #7

    BvU

    User Avatar
    Science Advisor
    Homework Helper
    Gold Member

    First we have to fix a) !
     
  9. Oct 19, 2015 #8
    if a) is 95m, then what would b) be? :cry::cry:
     
  10. Oct 19, 2015 #9

    BvU

    User Avatar
    Science Advisor
    Homework Helper
    Gold Member

    Your working shows 101 m for a). What has to be changed to get the book answer ?

    Oh, don't cry, please...
     
  11. Oct 19, 2015 #10
    I'm not sure... :( Can you enlighten me
     
  12. Oct 19, 2015 #11

    BvU

    User Avatar
    Science Advisor
    Homework Helper
    Gold Member

    All correct, except ...
    this happens in the right part of your picture. The trajectory is horizontal there
     
  13. Oct 19, 2015 #12
    I was thinking A(f) = F×S×cos(20°) seems right, but then I get S= - 21m....
     
  14. Oct 19, 2015 #13
    for a): The velocity of skier you calculated is correct (19.8 m/s).

    What is the position of the skier, when he reaches this velocity?
     
  15. Oct 19, 2015 #14
    tan20°/20m = (55m)i + (-20m)j @ velocity=19.8m/s
     
  16. Oct 19, 2015 #15
    Well, okay, I just wanted to know if he is still on the slope or already on the horizontal part, but I suppose you know, that he is leaving the slope and entering the horizontal part of his track at this point.

    On the horizontal part θ = 0°. What is the direction the friction force is pointing in?
     
  17. Oct 19, 2015 #16
  18. Oct 19, 2015 #17
    180°
     
  19. Oct 19, 2015 #18
    Okay then, work is the product of force and displacement. What is the direction of the displacement?
     
  20. Oct 19, 2015 #19
    towards positive x, 20° from x ?
     
  21. Oct 19, 2015 #20
    The skier is on a horizontal plane, so he only can move parallel to it (as well as the direction of the friction force). Towards positive x is correct, think again about θ. Maybe make a drawing of the situation, if you haven't already done that.
     
Know someone interested in this topic? Share this thread via Reddit, Google+, Twitter, or Facebook

Have something to add?
Draft saved Draft deleted



Similar Discussions: Kinetic energy
  1. Kinetic energy (Replies: 1)

  2. Kinetic Energy (Replies: 1)

  3. Kinetic Energy (Replies: 8)

Loading...