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Physics Forces

  1. Feb 28, 2015 #1
    • Member warned about misplacing homework questions in technical forum sections
    Kangaroos have large, powerful legs for jumping. A male kangaroo that has a mass of 66.5 kg can accelerate to a vertical velocity of 6.08 before his feet leave the ground (at a height of 1m). What force is the kangaroo's legs able to exert on the ground in order to do this?
     
  2. jcsd
  3. Feb 28, 2015 #2
    Assuming that the height is 0 initially,

    6.08^2 = 2a
    F = 66,5 (a + 9,81)
     
  4. Feb 28, 2015 #3

    Suraj M

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    You have the final velocity and also the distance covered before attaining that velocity, so use 3rd equation of motion and find acceleration, i think you can do the rest!
     
  5. Feb 28, 2015 #4
    thank you! my only question is, how did you know this part: 6.08^2=2a?
     
  6. Feb 28, 2015 #5

    Suraj M

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    v²-u²=2as and your u=0 and s=1
     
  7. Feb 28, 2015 #6
    Timeless velocity equation: v(final)^2 = 2ax if there is uniform acceleration. It is easy to derive from a simple velocity vs. time graph.
     
  8. Feb 28, 2015 #7
    and now to figure out how high he can jump. im not sure what to use, because f=ma does not deal with distance
     
  9. Feb 28, 2015 #8
    Use the same formula.
     
  10. Feb 28, 2015 #9
    the formula for the previous is the same as vf^2=vi^2+2ax, correct?
     
  11. Feb 28, 2015 #10

    Suraj M

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    You know both the quantities on the RHS.
     
  12. Feb 28, 2015 #11

    Suraj M

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    yes
     
  13. Feb 28, 2015 #12
    okay.. 6.08^2=2(18.48)(x)
    im getting x=1..where am i going wrong?
     
  14. Feb 28, 2015 #13

    Suraj M

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    No, you found ## a## from the equation v²=2ax... ;you already know## x=1 m## you should find ##a## , which you have done very nicely,now substitute in the formula ##F=ma##
     
  15. Feb 28, 2015 #14
    i found the force to be 1881. Now i am trying to find how far he can jump, or x
     
  16. Feb 28, 2015 #15

    Suraj M

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    For that you need to use a=g and vintial= 6.08 units, because the kangaroo is in the air and gravity is acting on it.
    substitute in the equation##v²_{final}-v²_{initial}=2ax## be careful while you choose the sign for ##a## when you substitute.
     
  17. Feb 28, 2015 #16
    so my equation setup would be:
    (0)^2 - (6.08)^2=2(18.48)(x)
     
  18. Feb 28, 2015 #17

    Suraj M

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    how far, or how high??
     
  19. Feb 28, 2015 #18
  20. Feb 28, 2015 #19

    Suraj M

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    choose your## a## well, its in the air, gravity acts on it so it should be ##a=g=-9.8 m/s²## can you tell me why the -ve sign?
     
  21. Feb 28, 2015 #20
    thats right, gravity acts on him once he jumps. okay i solved (0)^2 - (6.08)^2=2(9.8)(x) and got x=1.886
     
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