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Coursework in tomorrow, Completely Stuck

  1. Jan 9, 2007 #1
    1. The problem statement, all variables and given/known data

    a body mass of 5kg starts off from rest and is acted upon a force of 5N over a distance of 2m. if there is no frictional forces make use of the principle of the conservation of energy to determinei ts final speed


    2. Relevant equations

    above is all i have.

    3. The attempt at a solution

    didnt know where to start? any help?
     
  2. jcsd
  3. Jan 9, 2007 #2
    Use the work-energy theorem
     
  4. Jan 9, 2007 #3
    sorry, new to the whole engineering thing, my first week today, could someone give me a lot more help with this sorry.
     
  5. Jan 9, 2007 #4
    Remember that [tex]F=ma[/tex]. Where force is F, m = mass, a = acceleration.Thus you can find out the acceleration.

    Then you need to find out which kinematics (The study of motion in one direction) equation you need to use. Because your assignment is due tommorow, I'll pick it for you.

    Try using [tex] v_{f}^{2}=v_{i}^{2} + 2ad [/tex]

    Where [tex]v_{f}[/tex]=final velocity , [tex]v_{i}[/tex]=initial velocity, a = acceleration, and d = distance.

    Here are the other three kinematics equations that you will need to memorize, just in case you haven't done so already.

    [tex]d=v_{i}t+\frac{1}{2}at^{2}[/tex]
    [tex]v_{f}=v_{i}+at[/tex]
    [tex]d=\frac{v_{i}+v_{f}}{2}*t[/tex]

    memorize them!

    Most of these formulas can be understood quite easily by looking at a velocity-time graph. Since acceleration is constant, velocity will be linear. The slope of the velocity time graph is the acceleration, so rise/run = a = [tex]\frac{v_{f}-v_{i}}{t}[/tex]. Etc. In fact that may be the best way to memorize them.
     
    Last edited: Jan 10, 2007
  6. Jan 10, 2007 #5
    That's one way to do it, but his question calls for it to be done with conservation of energy equations. Remember that work done by a force is given by multiplying the magnitude of the force by the distance the object moves in the direction of the force:

    [tex]
    W = Fs
    [/tex]

    And like turdferguson said, use the work-kinetic energy theorem, which states that the amount of work done is the change in kinetic energy of the system

    [tex]
    W = \Delta K
    [/tex]

    and kinetic energy of an object is given by

    [tex]
    K = \frac{1}{2}m v^2
    [/tex]

    So, work is done by the 5 N force to increase the kinetic energy of the object. You can find how much work was done, and therefore how much the kinetic energy increased--and therefore how much the velocity increased.
     
  7. Jan 10, 2007 #6
    Ahhhh true enough, thanks gabee. I should probably stick to learning the stuff before I start to help with homework problems. :-)
    [​IMG]
     
    Last edited: Jan 10, 2007
  8. Jan 10, 2007 #7
    i know how to get kinetic energy but dont know how use that to find its final speed. i have 3 hours to go! ahhhh!
     
  9. Jan 10, 2007 #8

    Hootenanny

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    How do you define the work done by a [constant] force?
     
  10. Jan 10, 2007 #9
    am i trying to work out work done? WD = FD, WD = 5N x 2m WD = 10J?
     
  11. Jan 10, 2007 #10

    Hootenanny

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    Correct! So where does all this work done (energy) go to?
     
  12. Jan 10, 2007 #11
    into moving the 5kg mass (accelerating from 0?)
     
  13. Jan 10, 2007 #12

    Hootenanny

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    Yes, good! So the work done is converted into ________ energy.
     
  14. Jan 10, 2007 #13
    kinetic energy? what formula do i use? 1/2 mv^2 is what i have, but i dont have v^2,
     
  15. Jan 10, 2007 #14
    Thats the point. You need to work out v^2.
     
  16. Jan 10, 2007 #15
    is that v^2 = 1/2 mass divided by Kintetic energy
     
  17. Jan 10, 2007 #16

    Hootenanny

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    Okay, you know that the work done by the force is 10J and that this is converted in the kinetic energy of the body. Therefore;

    [tex]10 = \frac{1}{2}mv^2\hspace{1cm}\text{Find }v[/tex]

    Do you follow?
     
  18. Jan 10, 2007 #17
    i not sure how to rearrange this but is it somehting along the lines of sqrt of 1/2ke / m
     
  19. Jan 10, 2007 #18

    Hootenanny

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    Close;

    [tex]v = \sqrt{\frac{2E_k}{m}}[/tex]
     
  20. Jan 10, 2007 #19
    so its final velocity is 2m/s?
     
  21. Jan 10, 2007 #20

    Hootenanny

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    Correct! text too short
     
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