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Work/Kinetic Theory Problem

  1. Apr 17, 2006 #1
    I am a little confused by this problem. If someone could explain to me how to solve it, it would be greatly appreciated.

    A 100 kg crate is on a rough surface inclined at 30 degrees. A constant external force P= 800N is applied horizontally to the crate. The force pushes the crate a distance of 3.0m up the incline, in a time interval of 5.2s, and the velocity changes from V1= 1.4 m/s to V2= 2.8 m/s. The work done by the weight is closest to:
    A)+1400J
    B)zero
    C)-400J
    D)-1400J
    E)+400J
     
  2. jcsd
  3. Apr 17, 2006 #2

    Hootenanny

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    Welcome to PF Williams,

    Could you please any working or thoughts you have on the question.

    Regards,
    ~Hoot
     
  4. Apr 17, 2006 #3
    I just dont know where to start.
     
  5. Apr 17, 2006 #4

    Hootenanny

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    What is the equation for work done?

    HINT: Alot of the information given in the question is superfluous.

    Regards,
    ~Hoot
     
  6. Apr 17, 2006 #5
    Work done = Fs Cos(x)

    I kind of thought that there was some information in there that is irrelevant to solving the problem.
     
  7. Apr 17, 2006 #6

    Hootenanny

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    That's the one, so in this case the force is weight, which is mg. And you have the distance travelled and you can calculate your angle (take care), therefore you can obtain the work done by gravity.

    Regards,
    ~Hoot
     
  8. Apr 17, 2006 #7
    So that would be:
    100(9.8)(3.0)cos(30)?

    Cause that does not equal any of the choices.
     
  9. Apr 17, 2006 #8

    Hootenanny

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    You are almost right. Think about your angle, a sketch may be useful. I get an answer of -1471.5 J

    Regards,
    ~Hoot
     
  10. Apr 17, 2006 #9
    Well thanks for the help but I still do not understand this problem. I am pretty much screwed anyway. I have a take home test due tonight that I haven't been able to start until now. I have to somehow work out 10 problems when I can't even do this simple one. But thanks for helping.
     
  11. Apr 17, 2006 #10

    Hootenanny

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    If you use trigonometry you will find that it becomes cos60

    Regards,
    ~Hoot
     
  12. Apr 17, 2006 #11
    Thank you. But I still don't understand how that ends up being negative.
     
  13. Apr 17, 2006 #12

    Hootenanny

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    Work done is defined as force multiplied by the distance moved in the direction of that force. The weight acts down, but the crate is moving upwards. Therefore, the work done is negative.

    Regards,
    ~Hoot
     
  14. Apr 17, 2006 #13
    Wow, I really am an idiot. Can't believe I missed that. I have a question about another problem too.

    An object of mass 2 kg is repelled from the origin by a force in the +x-direction whose magnitude varies with x according to F=(7Nxm^2)x^-2. How much work is done by this force when the object moves from x=2 to x=3? (Be sure to say whether this work is positive or negative.)

    Do I just substitute for x for both and then take the difference?
     
  15. Apr 17, 2006 #14

    Hootenanny

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    Everyone's done it atleast once before :biggrin:

    No, you can't do that because the force is not constant. You have to integrate your function between the limits i.e.

    [tex]wd = \int^{3}_{2} \frac{7}{x^2} \;\;dx[/tex]

    Regards,
    ~Hoot
     
  16. Apr 17, 2006 #15
    Well thank you again but I think I am giving up on this test. No matter how hard I try I cannot grasp the concept of this stuff. My head is killing me right now and I've only completed 5 out of 12 problems.
     
  17. Apr 17, 2006 #16

    Hootenanny

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    Well, if you need anymore help don't hesitate to come back.

    Regards,
    ~Hoot
     
  18. Apr 17, 2006 #17
    Well I do need a lot of help but I feel like I would be a bother by asking for help on so many problems. Especially when it takes me forever to just comprehend one of them.
     
  19. Apr 17, 2006 #18

    Hootenanny

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    Not at all, all the homework helpers (and others) here would take pleasure in helping you. As long as you are willing to put the work in we will guide you through the questions.

    Regards,
    ~Hoot
     
  20. Apr 17, 2006 #19
    Im still not really sure what this question is asking...
     
  21. Apr 17, 2006 #20

    Hootenanny

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