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Conservative force help

  1. Jun 15, 2008 #1
    a bullet with a mass of .02kg and a velocity of 800m/s is fired horizontally into a tree. the tree exerts a resistive force and brings the bullet to rest after it penetrates .56m into the tree. given this information fins the constant resistive force exerted by the tree on the bullet and the power generated in bringing the bullet to a halt....
     
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  3. Jun 15, 2008 #2

    rock.freak667

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    Where is your attempt?

    and think energy.
     
  4. Jun 15, 2008 #3
    i been on this problem for the past 3 hours man/// i need help plz plz plz
     
  5. Jun 15, 2008 #4

    rock.freak667

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    In your 3 hour attempt, what methods did you try?

    Did you try considering energy as I posted before?
     
  6. Jun 15, 2008 #5
    i dont know how to set up the problem i know that

    (1/2)mv^2 + mgy + (1/2)kx^2= (1/2)mv^2 + mgy + (1/2)kx^2

    so what do i do
     
  7. Jun 15, 2008 #6

    rock.freak667

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    Initially the bullet has kinetic energy,[itex]\frac{1}{2}mv^2[/itex]. When it hits the tree, all that k.e. is used to cause the bullet to move a distance of 0.56m into the tree.

    So Work done by tree= change in Kinetic energy of bullet.

    Do you know a formula for Work done by a force,F, which moves it point of application a distance,s?
     
  8. Jun 15, 2008 #7
    is it (1/2)MV^2 - (1/)MV^2
     
  9. Jun 15, 2008 #8

    rock.freak667

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    [itex]\frac{1}{2}mv^2-\frac{1}{2}mu^2[/itex] is the change in kinetic energy.


    Do you know the definition of Mechanical Work done by a force?
     
  10. Jun 15, 2008 #9
    know i do after i read that thing from wikipedia... then wat do i do
     
  11. Jun 15, 2008 #10

    rock.freak667

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    Fs=change in ke


    you know change in ke. s=0.56. Find F
     
  12. Jun 15, 2008 #11
    i did (1/2)(.02)(800)^2 / (.56) =11428
     
  13. Jun 15, 2008 #12

    rock.freak667

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    Yep.
    Now how do you find power with a force and the velocity of 800m/s?
     
  14. Jun 15, 2008 #13
    aw man i dont know....
     
  15. Jun 15, 2008 #14

    rock.freak667

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    P=Fv.


    You do know these formulas right?
     
  16. Jun 15, 2008 #15
    yeah but i cant use those beacuse the teacher said to use the principles of energy of conservation
     
  17. Jun 15, 2008 #16

    rock.freak667

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    I think you'll need to use P=Fv though.
     
  18. Jun 15, 2008 #17
    i did and i get 9142400 a huge number....but i wont get any credit if dont use the conservation of energy principles.....
     
  19. Jun 18, 2008 #18

    dynamicsolo

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    This won't help much here -- since the velocity is not constant over the path, you would have to perform an integration to find either an average power or a function for power. (The problem probably wants the average power, but in the OP's course, they probably want the students to use kinematics to find the time it takes to stop the bullet.)
     
  20. Jun 18, 2008 #19

    dynamicsolo

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    All right, the conservation of energy equation you are using is

    KE_init + W_non-conservative = KE_final ,

    with the non-conservative work being given by

    W_non-con = -F · (delta_x) .

    Since the final KE is zero, you have

    W_non-con = -KE_init and

    -F · (delta_x) = -KE_init gives F = KE_init / delta_x .

    You did this and found F = 11,430 Newtons.

    That is the part you can do by conservation of energy principles. To find the power involved in stopping the bullet, you need to use the definition of power

    P = W / delta_t , where delta_t is the time over which the force you found has acted.

    You know the initial speed (800 m/sec) and final speed (0 m/sec) of the bullet. If you can find the acceleration that acted on the bullet while the trunk of the tree was stopping it, you can find the time it took to bring the bullet to rest. You also now know the force that acted on the bullet. Can you find the bullet's acceleration?

    (Actually, you're getting close: the number you gave in post #17 is off by a factor of about 2...)
     
    Last edited: Jun 18, 2008
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