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Power input of a Force acting on a particle moving with Velocity

  1. Oct 27, 2012 #1
    1. The problem statement, all variables and given/known data

    Find the power input of a force [itex]\vec{F}[/itex] acting on a particle that moves with a velocity [itex]\vec{V}[/itex] for each of the following situations.

    • [itex]\vec{F}[/itex] = 4[itex]\hat{i}[/itex] N  + 3[itex]\hat{j}[/itex] N , [itex]\vec{V}[/itex] = 7[itex]\hat{i}[/itex] m/s 
    • [itex]\vec{F}[/itex] = 7[itex]\hat{i}[/itex] N  - 5[itex]\hat{j}[/itex] N , [itex]\vec{V}[/itex] = -5[itex]\hat{i}[/itex] m/s + 4[itex]\hat{j}[/itex] m/s 
    • [itex]\vec{F}[/itex] = 2[itex]\hat{i}[/itex] N + 10[itex]\hat{j}[/itex] N , [itex]\vec{V}[/itex] = 2[itex]\hat{i}[/itex] m/s  + 3[itex]\hat{j}[/itex] m/s 

    2. Relevant equations

    P = [itex]\frac{dW}{dt}[/itex] = [itex]\vec{F}[/itex] • [itex]\vec{v}[/itex]

    3. The attempt at a solution

    • [itex]\sqrt{4^2 + 3^2}[/itex] [itex]\times[/itex] cos(arctan(3/4)) [itex]\times[/itex] 7
      For this problem, I got the correct answer: 28 W.
    • [itex]\sqrt{7^2 + (-5)^2}[/itex] [itex]\times[/itex] [itex]\sqrt{(-5)^2 + 4^2}[/itex]
      I also got the correct answer for this problem: 55 W, except that the answer is negative, I figure because the velocity vector's [itex]\hat{i}[/itex] and [itex]\hat{j}[/itex] components are in the opposite direction of the force's components (individually). Is this supposition correct?
    • [itex]\sqrt{2^2 + 10^2}[/itex] [itex]\times[/itex] sin(arctan(10/2)) [itex]\times[/itex] [itex]\sqrt{2^2 + 3^2}[/itex] [itex]\times[/itex] sin(arctan(3/2))
      I cannot figure out what I am doing wrong on this one. (The correct answer is 34.)

    I am having a hard time wrapping by head around dot products. I'm guess that is my problem here. Any and all help is greatly appreciated.
     
  2. jcsd
  3. Oct 27, 2012 #2
    When you do dot products, you take the sum of the products of each component.
    For the first one:
    P = 4N*7m/s (the i components) + 3N*0m/s (the j components) = 28 W.
    For the second one:
    P=7N*-5m/s (i components) + -5N*4m/s (j components) = -55W.
    Try the third one out this way.
    P= (product of i components) + (product of j components) = ?
     
  4. Oct 27, 2012 #3
    What you are doing is finding it using magnitudes and angles, which you must first calculate. When calculating angles, you also have to think about the fact that it may be more than 90 degrees, something that arctan will not provide.
     
  5. Oct 27, 2012 #4
    OK, thanks. I completely get how to do this in the future.

    Regarding your second comment though -- I understand about arctan. I drew a diagram so I knew where my vectors were. Was this comment just FYI or was my attempted method something that would work, I was just simply doing it incorrectly?
     
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