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Impulse and momentum for an athlete jumper

  1. Apr 26, 2013 #1
    1. The problem statement, all variables and given/known data
    Assume that the athlete approaches the take-off line from the left with a horizontal
    velocity of 10m/s, remains in contact with the ground for 0.1s, and take off at a 60°angle
    with a velocity for 10m/s, determine (1) the vertical component of the average impulsive
    force exerted by the ground on his foot. (2) the horizontal frictional force. The mass of
    the athlete is 100kg and the coefficient of kinetic friction (μk) is 0.1.
    image.jpg

    2. Relevant equations

    momentum 1 + ƩF Δt = momentum 2
    Friction force = Normal force X μk

    3. The attempt at a solution
    1)In the vertical Direction
    0+(Ry-Weight) Δt = mvsin60
    Ry = 9660 N

    2)
    Friction Force = Ry X μk = 966 N (I think this is wrong)
     
  2. jcsd
  3. Apr 26, 2013 #2

    Doc Al

    User Avatar

    Staff: Mentor

    Good.

    Use the same method as you used for part 1. (Is the coefficient of kinetic friction relevant?)
     
  4. Apr 26, 2013 #3
    So :
    m1v1 - Friction Force X Δt = m1v2 cos60
    Friction Force = 5000 N ??
     
  5. Apr 26, 2013 #4

    Doc Al

    User Avatar

    Staff: Mentor

    That's right.
     
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