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Spring Constant and Work (2 seperate problems)

  1. May 1, 2008 #1
    The spring of a spring gun has a constant of 6.0. It is compressed 0.0508 meters and a ball weighing 0.009 kg is placed in the barrel against the spring.

    a. What is the maximum speed of the ball as it leaves the gun?
    Vf = sqrt (k/m)x Xi = 1.31 m/s

    b. If it is shot off (horizontally) a 6 ft high table, how far from the base of the table will it land?

    Would I use x = vot + 1/2at2 ??? To find the time and then insert into the problem? I'm confused.

    And the second problem....

    I dead lift to a height of 30 inches 500lbs and proceed to walk the length of a football field (100 yds) and set the weight back down. Because I'm in great shape, I barely work up a sweat but how much work did I do?

    with 30 inches = .762m
    m= 500lbs, 2,224 N
    x= 100 yds, 91 m

    W = F change in X
    =2,224Nx(91m) = 202,384J

    This seems like an unreasonably high amount of work. I didn't take into consideration lifting the weight because it is perpendicular to the amount of work being done. Only the parallel force matters, right?
    Thanks for any and all help.
     
  2. jcsd
  3. May 1, 2008 #2

    Nabeshin

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    2b) Try solving for time in the y first and then using that to get distance.

    As for the second problem, consider that work is the dot product between force and displacement. This means work is F*D*Cos(). So there are three parts to this problem. Lifting the weight, walking, and then setting it down. Consider all three separately, keeping the angle between force and displacement in mind!
     
  4. May 2, 2008 #3
    Thank you for the reply. I will look over the problem this weekend and can hopefully figure it out. Wish me luck.

    So for the second problem cos*90 because of the right angle from dead lifting the weight and it being perpendicular to the ground?
     
  5. May 3, 2008 #4

    tiny-tim

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    Hi kudelko24! :smile:

    Work is force times distance moved parallel to the direction of the force. :smile:

    (not like torque, which uses perpendicular distance!)
     
  6. May 4, 2008 #5
    So the amount of work being done is 0, correct? Because Work = force * displacement * cos (). Cos = 1 because of the force being straight up and down, parallel. There's positive work being done and then that cancels out the negative work when he sets the weight back to the ground.
     
  7. May 6, 2008 #6

    tiny-tim

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    Hi kudelko24! :smile:

    Yes … the total work done on the weight is zero.

    Of course, the amount of energy you expend is very considerable … beats me where it all goes … :confused:
     
  8. May 6, 2008 #7
    Awesome... thank you! That was some what of a confusing question, but the more I looked into it, the more I understood it. Thanks again.
     
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