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Force (3)

  1. Sep 22, 2007 #1
    1. The problem statement, all variables and given/known data
    Bob, who has a mass of 85 kg, can throw a 450g rock with a speed of 30m/s. The distance through which his hand moves as he accelerates the rock forward from rest until he releases it is 1.8m. a.) What constant force must Bob exert on the rock to throw it with this speed? b.) If Bob is standing on frictionless ice, what is his recoil speed after releasing the rock?


    2. Relevant equations

    F=ma

    3. The attempt at a solution

    I have no idea how to start this problem. There are way too many elements of the problem for me to know how to start.
     
  2. jcsd
  3. Sep 22, 2007 #2

    learningphysics

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    Think of it like a kinematics problem... find the acceleration, then you can find the force.
     
  4. Sep 23, 2007 #3
    Ok. I tried using Vxf^2 = Vxi^2 + 2ax(delta x) to find the acceleration, but that didn't work.
     
  5. Sep 23, 2007 #4

    learningphysics

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    What number did you get for acceleration?
     
  6. Sep 23, 2007 #5
    -250 m/s^2
     
  7. Sep 23, 2007 #6

    learningphysics

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    You shouldn't have the minus sign...
     
  8. Sep 23, 2007 #7
    Ok, I got that one, 112.5N. Now how do I start part b?
     
  9. Sep 23, 2007 #8

    learningphysics

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    Use conservation of momentum.
     
  10. Sep 23, 2007 #9
    We haven't gone over momentum yet.
     
  11. Sep 23, 2007 #10

    learningphysics

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    Oh... you can also do it this way. What is the force acting on Bob (use newton's third law). Then you can find Bob's acceleration... Also find the time over which Bob has contact with the rock (you can get this now easily, because you have acceleration of the rock)

    Using bob's acceleration and time, you can get bob's velocity at the end of the time period...
     
  12. Sep 23, 2007 #11
    Using Newton's third law would make the force acting on Bob -112.5N. So using that I found the acceleration to be -1.324m/s^2. Then I used Vxf = Vxi + ax (delta time). The answer I got was 30.079, which was wrong.
     
  13. Sep 23, 2007 #12

    learningphysics

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    How are you getting 30.079? Vxi = 0. ax = -1.324. delta t = 30/250=0.12s.
     
  14. Sep 23, 2007 #13
    I was using the wrong time. I didn't think to use acceleration = change in velocity/time. I got -0.159, which is still wrong.
     
  15. Sep 23, 2007 #14

    learningphysics

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    Did you try 0.159?
     
  16. Sep 23, 2007 #15
    The positive answer was right, which is good because that was my last guess at the problem.
     
  17. Sep 23, 2007 #16

    learningphysics

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    glad you got it.
     
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