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Particle problem

  1. Apr 5, 2008 #1
    1. The problem statement, all variables and given/known data
    A raindrop of mass 3.07 X 10^-5 kg falls vertically at constant speed under the influence of gravity and air resistance. Model the drop as a particle.

    a. As it falls 50m, what is the work done on the raindrop by the gravitational force?

    b. What is the work done on the raindrop by air resistance?


    2. Relevant equations



    3. The attempt at a solution

    a. Fg=-mgj=(3.07X10^-5)(-9.8m/s^2j)
    =-3.0086x10^-4

    delta r=???

    w=????

    Could someone please tell me if this is correct so far and help me find delta r?

    Thank you very much
     
  2. jcsd
  3. Apr 5, 2008 #2

    Shooting Star

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    What do you mean by delta r?

    Anyway, do you know the work done by gravity when a particle falls from a height h1 to a height h2?

    Also, note that the drop is falling at a constant speed. So, the KE is not increasing, but the gravitational PE is decreasing.

    Think for a while before rushing back here.
     
  4. Apr 6, 2008 #3
    Could you please give me a hint? Is what I did so far correct?

    Thank you
     
  5. Apr 6, 2008 #4

    nrqed

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    [tex] \Delta \vec{r} [/tex] is simply the final position minus the initial position so it's simply [tex] -50 m \hat{j} [/tex] (where m here is for meters, it's not the mass). To get the work done by gravity, take the dot product of the force with the vector delta r

    (another way of doing is is to take the difference of gravitational potential energy)
     
    Last edited: Apr 6, 2008
  6. Apr 6, 2008 #5

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    I specifically asked about the work done by gravity when a particle falls from a height h1 to a height h2, and you have blindly answered w=FΔrcosθ. To do this problem, you have to say what is the work done in terms of h1 and h2. Look at nrqed's post.
     
  7. Apr 7, 2008 #6
    Thank you very much

    Could you tell me if this is correct?

    a. As it falls 50m, what is the work done on the raindrop by the gravitational force?

    Wouldn't this be 0J? Because when it is perpendicular the work is 0, right?

    b. What is the work done on the raindrop by air resistance?

    Fg=-mgj=(3.07X10^-5)(-9.8m/s^2)j
    =-3.0086X10^-4=-.0003JN

    Δr=initial-final

    =0mi-50mj
    =-50mj

    w=(-50mj)(.003jN)
    =.015W

    Thank you
     
  8. Apr 8, 2008 #7

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    It is not zero. I was not able to understand which is perpendicular to what. Please clarify.

    If you are taking the initial position to be (0,0), then,
    Δr=initial-final = 0j - (-50)j. (What is the m doing there?)

    This is, in fact, the work done by gravity on the body, although in this particular problem, the force due to resistance is equal to the force of gravity, because the drop is falling at a constant speed. The work done by gravity is dissipated into other forms of energy, mostly heat, due to the frictional force.

    Work done by gravity = -(work done by air resistance).
     
  9. Apr 30, 2008 #8
    Thank you very much

    Regards
     
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