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Average force and center of mass

  1. Oct 23, 2005 #1
    A 44 kg woman leaps vertically from a crouching position in which her center of mass is 48 cm above the ground. As her feet leave the floor her center of mass is 81 cm above the ground; it raises to 124 cm at the top of her leap.

    What average force was exerted on her by the ground during the jump?

    the initial force is the normal force (44 * 9.81) = 431.64 N
    but i don't know what the final force is.

    after i find the final force the average is (Fi+Ff)/2.
     
  2. jcsd
  3. Oct 23, 2005 #2

    Päällikkö

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    Think about velocity. What initial velocity must she have to reach 124 cm?
     
    Last edited: Oct 23, 2005
  4. Oct 23, 2005 #3
    the velocity is 2.905 m/s
     
  5. Oct 23, 2005 #4

    Päällikkö

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    Now it should be down to kinematics.
     
  6. Oct 23, 2005 #5
    i got the acceleration of the woman to be 3.57 m/s/s so i add that to the 9.81 gravitational force and then multiply that by 44 kg. I got 588.9 N. so (588.9+431.64)/2 =510.3 N ... this is wrong.. so i don't know what to do now
     
  7. Oct 23, 2005 #6

    Doc Al

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    Why are you dividing by 2? 588 N is already the average net force during the jump.
     
  8. Oct 23, 2005 #7
    so its 588.9 + 431.64 = 1021. this is "close but not within 1% of correct answer" did i round wrong somewhere?
     
  9. Oct 23, 2005 #8

    Doc Al

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    Recalculate the net force (588.9 N); I get a different value.
     
  10. Oct 23, 2005 #9
    is it 562.58 N? just checking because it is my last guess.. thanks by the way
     
  11. Oct 24, 2005 #10

    Doc Al

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    Looks OK to me.
     
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