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Force on object with position functions

  1. Feb 25, 2006 #1
    Hello. I'm studying for my AP Physics C exam upcoming, so I'v e been doing practice problems out of this moderately difficult book. I ran into this question and wasn't too sure how to deal with it. Any hints or advice would be very much appreciated!

    "The displacement of an object of 1 kg mass as a function of time t in the x-y plane is given by [tex] x = 20t + 30t^2 [/tex] and [tex] y = 25 - 40t [/tex]. in the units of meters. The units of x and y are in meters and t is in seconds.
    The force acting on the object is?"

    The answer is 60 N in the constant x-direction.

    I initially thought [tex] F = dp/dt [/tex]. I figured that since I knew the mass, I could say [tex] F = m dv/dt [/tex]. To get the velocity, I figured I could find [tex] dx/dt [/tex] and [tex] dy/dt [/tex]. I wasn't too sure what to do then. I thought perhaps about finding the magnitude of the velocity, but wasn't too sure what to do.

    Any hints?

    Thank you.

    Edit:

    I think I may have figured it out. I realized that [tex] d^2x/dt^2 = 60 [/tex] and [tex] d^2y/dt^2 = 0 [/tex]. Thus, [tex] F = ma [/tex] and therefore, [tex] F = 60 N[/tex] in the x direction (as that is the only way it seems to be accelerating. Is my logic correct?
     
    Last edited: Feb 25, 2006
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  3. Feb 25, 2006 #2

    Hootenanny

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    Yes, it appears your logic is correct.
     
    Last edited: Feb 25, 2006
  4. Feb 25, 2006 #3
    The first derivative of the position with respect to time is velocity, the second derivative is acceleration. You can then use F=ma.
     
  5. Feb 25, 2006 #4
    Awesome! I figured it out almost immediately after posting.

    However, another question that was posed to me.

    (With the same functions above), the magnitude and direction of the velocity of the object from the positive x-axis at the end of 1 second would be?

    The answer is 89 m/s and 333.5 degrees.

    I was able to get 89 m/s by using [tex] |V| = \sqrt(v_x^2 + v_y^2 ) [/tex]

    I thought the angle could be found via [tex] tan^-1 (v_y /v_x) [/tex]. However, this yielded 26.5 degrees,which is not correct.
     
  6. Feb 25, 2006 #5

    Hootenanny

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    It depends were the angle of reference is taken from. Yours is taken from the x-axis in an clockwise direction. I assume the quoted answer (333.5) is taken from the x-axis but in an anti-clockwise directon. (333.5 + 26.5 = 360)
     
    Last edited: Feb 25, 2006
  7. Feb 25, 2006 #6
    That's because the angle you need is outside the regularly used range of the inverse tangent. Mathematically speaking it has many branch cuts. See here for more discussion.
     
  8. Feb 25, 2006 #7
    Ahh, I see. Thank you very much. I can't believe I forgot how to do that.
     
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