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With what speed does the stone leave the slingshot?

  1. Apr 18, 2005 #1
    A slingshot consists of a light leather cup containing a stone that is pulled back against two parallel rubber bands. It takes a force of 15 N to stretch either one of the bands 1.0 cm. A)What is the potential energy in the two bands together when a 47g stone is placed in the cup and pulled back.12 m from the equilibrium position? B) With what speed does the stone leave the slingshot? I'm not sure where to start!!!!
     
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  3. Apr 18, 2005 #2

    quasar987

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    you have to consider the rubber band has happy Hookian springs. I.e. such that F = kx.

    By "It takes a force of 15 N to stretch either one of the bands 1.0 cm", what they mean is that if you exert a force of 15N on one of the rubber band, you're not gonna be able to stretch it farther than 1cm. At this point, equilibrium will have been attained. this means that at x = 1cm, the force of the rubber band is of 15N too. From this, you get k = 15 [N/cm].

    You go from there.
     
    Last edited: Apr 18, 2005
  4. Apr 18, 2005 #3
    so... then would 7.5 be the correctanswer to part A?
     
  5. Apr 18, 2005 #4

    quasar987

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    Nope. How did you get that answer?

    (Btw - I fixed some dimension errors in my first post. I replaced all meters with centimeters)
     
    Last edited: Apr 18, 2005
  6. Apr 18, 2005 #5
    I used the equation PE=.5kx^2and put 15 infor k and 1 in for x.
     
  7. Apr 18, 2005 #6

    quasar987

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    That's the right idea, but pay more attention.

    The actual problem talks about a rock that is pulled back.12 m (12 cm, not 1cm). Moreover, 15 [N/m] is the spring constant of ONE rubber band. You got two.
     
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