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Questions about work.

  1. Feb 13, 2012 #1
    1. The problem statement, all variables and given/known data
    We will start with the work formula ##A=Fd##, can that formula be defined as:
    ##A=(ma)Δd=(m\frac{v}{t})vt=(m\frac{v}{t})v_0 t + \frac{1}{2} at^2## ??

    2. Relevant equations
    ##A=Fd##

    3. The attempt at a solution
    Yes, it can be defined.

    I think the answer will be yes, but I will be sure.


    ALSO: Can we define the work by velocity , mass or any other..?
    If so, what's the formula
     
    Last edited: Feb 13, 2012
  2. jcsd
  3. Feb 13, 2012 #2
    The equation is incorrect. ALWAYS check your units. When things are added, units must be homogeneous. 1/2at^2 has the units of length, not work.
     
  4. Feb 13, 2012 #3
    Thanks, anyways:

    , can you answer?
     
  5. Feb 13, 2012 #4
    In certain cases you can equate work to kinetic energy. For instance if you pushed a block (initially stationary) on a frictionless horizontal surface with a constant force of F for a distance d, the following equation can be written:

    F * d = 0.5 * m * v^2

    where

    F is force
    d is distance
    m is mass
    v is velocity

    Note that the units to the left and right of equals sign are the same. This always is the case.
     
  6. Feb 13, 2012 #5
    thanks, but can I know why ##0.5## is put there and why is the velocity squared ##v^2##? What's the effect of that?
     
  7. Feb 13, 2012 #6
    ##v^2=v_0+2aχd## would give us the equation ##E_k=\frac{1}{2}mv^2=\frac{1}{2}aFt^2## E_k is Kinetic Energy, Now I understand the 0.5, But What I don't understand is why did you multiply the force ##F## with distance ##d##?
     
  8. Feb 14, 2012 #7
    See derivation of kinetic energy formula : http://www.indiastudychannel.com/projects/1644-Derivation-Kinetic-energy.aspx

    http://answers.yahoo.com/question/index?qid=20110530195816AAVPotW

    Work is simply defined as mathematical product of force times displacement in direction of force or resolved in component of that force. What to say when you do work ? Push wall and it does not move. You do no work. That's the concept which implies meaning of the word "work" in physics.

    By the way , your equation is also incorrect...
    Correct will be to write :
    v2=v02+2ad
     
  9. Feb 14, 2012 #8
    thank you. got it now :)
     
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