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Find power given:velocity,CoFriction,Mass,slope angle

  1. Nov 2, 2008 #1
    1. The problem statement, all variables and given/known data
    What power must a woman expend on a 50 kg wagon that she is pulling up a hill at a speed of 0.40 m/s. The hill makes an angle of 20° with the horizontal and the coefficient of friction is 0.35. <<<this was copied and pasted straight from web ct

    2. Relevant equations

    Fnet=Fp-Ffr
    P=Fvcostheta
    p=w/t
    3. The attempt at a solution
    I havent really gotten anywhere I just dont see how I could find power without knowing any time or distance and since the woman is moving at a constant velocity the net force must be zero. right? however I know if an object is moved work is done. and considering there is no distance given I dont see how I could find a time. I have a feeling im missing something very obvious or it is unsolvable. I am stuck and I am in a distance course and my instructor doesnt seem to want to check his email so im getting a bit stressed out whihc as you know tends to cloud the mind this is the last question of my assignment whihc is due tonight please help me.
     
  2. jcsd
  3. Nov 2, 2008 #2
    There are two types of "work", the one where the velocity changes, and the one where the position of the particle is lifted. Even if the person is pulling at a constant speed, since she's going uphill, that means she's doing work. And since power is just the change of work, the problem is solvable.

    One of your three formulas actually provides you with what you need, as you have everything avaliable.

    Remember, just because the net force is 0 does not mean the work done on the box is zero.

    If you're still stuck, draw a free body diagram of the system. It should help you in figuring out what's going on.
     
    Last edited: Nov 2, 2008
  4. Nov 2, 2008 #3
    Ok thanks Im thinkning it through right now
     
  5. Nov 2, 2008 #4
    if work =change in kinetic energy than wnet=25kg(0.40m/s)^2
    Wnet=4J if force is balanced I should ignore the friction correct?
     
  6. Nov 2, 2008 #5
    Hmmm, I was looking more at the "P= F*V" equation

    No, friction should not be ignored here. The friction is what is keeping the total acceleration here 0, thus making a net force of 0 for the system.

    You must keep in mind that friction is doing negative work on the system itself. If you are going to play with energy instead of kinematics, you will need to do the entire equation for energy:

    PE i (Potential energy initial) + KE i (Kinetic energy initial) = PE f (final) + KE f

    The problem with your assumption, is that you assume there is no final potential energy, but there is, because she is going uphill.

    Try approaching it kinematically instead of with energy. Once you find the force the woman exerts, you just plug it in to the equation and you got your power.

    Edit: Your free body diagram should aid in helping you figure out what the force is.
     
    Last edited: Nov 2, 2008
  7. Nov 2, 2008 #6
    YES! I think it just clicked all I need to do is find the Force of friction which will be a negative value and then I know that since there is no acceleration the forces are balanced so I simply take the positive value of the force of friction and that is the force the woman is exerting on the box therefore. Fp=Ffr so -Ffr times velocity equals power. Ughh I hate it when I goof up on easy problems and over think them to the point of mind numbing frustration. I hope i am on the right path here I think I am. Thank you very much
     
  8. Nov 2, 2008 #7
    Ah, not quite. The woman's going uphill. Friction's not the only force acting against her, there's also gravity to consider. (Assuming that you have your x axis along the slope of the hill.)

    Other then that, you're on the right track now. :)
     
  9. Nov 2, 2008 #8
    haha yep thank you. like I said mind numbing frustration I tend to lose my common sense
     
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