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Homework Help: Parachute motion question

  1. Mar 22, 2010 #1
    1. The problem statement, all variables and given/known data
    An 80.0kg skydiver jumps out of balloon at an altitude of 1000m and opens the parachute at an altitude of 200m. a) Assume that the total retarding force on the diver is constant at 50.0N with the parachute closed and constant at 3600N with the parachute open, what is the speed of the diver when he lands on the ground?


    2. Relevant equations
    Work Done equation
    Kinetic Energy equation
    Potential energy equation


    3. The attempt at a solution

    i have tried to solve this question but my answers does not seem correct and logically not possible. So i need some help on how to start this question.

    P.S
     
  2. jcsd
  3. Mar 23, 2010 #2
    Since force is constant and motion is one dimensional, our best friend in this case is the F*d = W, mgh = PE, and 0.5mv^2 = KE equations. There is "positive" energy from the used potential energy in the fall and there is "negative" energy (force is acting opposite to direction of motion) from 800m of constant 50N force and then 200 m of constant 3600N force thereafter. Sum up the "positive" and "negative" energies and you will be able to solve for velocity with 0.5mv^2 = KE (all of the energy is converted to KE).
     
  4. Mar 24, 2010 #3
    so is the 3600N and 50N Potential Energy or is it the total force exerted within that distance?
     
  5. Mar 24, 2010 #4
    Force. The potential energy incolves the 1000 m of height.
     
  6. Mar 24, 2010 #5
    this is what i have done:

    PE = mgh
    PE = 80 * 9.8 * 1000
    PE = 1784J

    Now to find the W done for 800m and 200m

    W = 50 * 800
    W = 4000J negative because it is opposite to the downward direction

    W = 3600 * 200
    W = -720000J negative because it is opposite to the downward direction

    So if i make Total Force=KE then

    -4000 + -720000 + 1784 = 0.5* 80 * v^2

    722216 = 40 * v^2

    v=134.37m/s
     
  7. Mar 24, 2010 #6
    Thats quite a big error you have in calculating for potential energy there. The concept is right. The calculations are wrong.
     
  8. Mar 24, 2010 #7
    fixed i get the correct answer :) thanks.
     
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